6. Introduction to Ket notation

Before we dive deeper into the different layers of a quantum computer, we first want to introduce you to ket notation. Ket notation is a specific quantum method used to perform operations and measurements and will introduce the basics of ket notation.

Note that this is a densely packed section and don’t expect to understand everything after the first viewing. Much more time is needed to grasp the basics of Ket notation, and we advice you to review the following passages. Please do not let this subject deter you from following further. It is crucially important to understand the basics, because that will be used in the continuation. This is a difficult topic!

Typo:

$\mid \psi\rangle=\cos(\theta/2)|0\rangle+e^{i\phi}\sin(\theta/2)|1\rangle$
The highlited line should read: $$\frac{1}{\sqrt{2}}(|0\rangle\langle0\mid0\rangle\otimes I|0\rangle+|0\rangle\langle0\mid1\rangle\otimes I|0\rangle+|1\rangle\langle1\mid0\rangle\otimes X|0\rangle+|1\rangle\langle1\mid1\rangle\otimes X|0\rangle)\end{array}$$

We would like to control quantum degrees of freedom in order to implement quantum computing, among other goals that we have. To talk about these degrees of freedom, we’re going to use some notation that you might not be familiar with, but it’s all very close to linear algebra. You are likely already familiar with linear algebra, which has column vectors denoted with an arrow, or with boldface type.

Column & row vectors: $$\vec{x}=\boldsymbol{x}=\left[\begin{array}{c} 10\\ 11\\ 12 \end{array}\right],\,\,\,\vec{y}^{T}=\boldsymbol{y}^{T}=\left[\begin{array}{ccc} 1 & 2 & 3\end{array}\right]$$

It also has a transpose operation that maps column vectors to row vectors, and vice versa. We can calculate the inner product of two such vectors, either by transposing one of them and multiplying, or by writing the dot product explicitly. Either way, we get a scalar by taking the inner product of these two vectors.

\[\begin{split}\overrightarrow{y}^{T}\overrightarrow{x}=\boldsymbol{y}\cdot\boldsymbol{x}=\left[\begin{array}{ccc} 1 & 2 & 3\end{array}\right]\left[\begin{array}{c} 10\\ 11\\ 12 \end{array}\right]=68\end{split}\]

In addition, we can transform one vector into another using a matrix, which can be written using boldface type or a hat, your preference.

\[\begin{split}\boldsymbol{A}\boldsymbol{x}=\left[\begin{array}{ccc} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{array}\right]\left[\begin{array}{c} 10\\ 11\\ 12 \end{array}\right]=\left[\begin{array}{c} 66\\ 167\\ 266 \end{array}\right]\end{split}\]

There is a slightly different notation that we use for vectors and matrices in quantum computing, which we often call ket notation.

In ket notation, a quantum state is expressed using a column vector with complex coefficients. For example, we can express the state of a qubit, a two-level system, as a linear combination of two basis vectors, which we call 0 and 1.

Column vectors $$\rightarrow kets:\mid\psi\rangle=\alpha\mid0\rangle+\beta\mid1\rangle=\left[\begin{array}{c} \alpha\\ \beta \end{array}\right]$$

In a departure from regular linear algebra, there is a well-defined dual vector for each quantum state, called a bra, which we can obtain by taking the complex conjugate and the transpose of the ket vector.

Dual vectors $$\text{Bra:}\langle\psi\mid=\left[\alpha^{*}\beta^{*}\right]$$

This is important for calculating inner products, which we always do by multiplying the bra for one state by the ket of the other, forming a bra-ket.

Inner products:

\[\begin{split}\equiv brackets \langle\psi\mid\phi\rangle=\left[\begin{array}{cc}\alpha^{*} & \beta^{*}\end{array}\right]\left[\begin{array}{c}\gamma\\\delta\end{array}\right]=\alpha^{*}\gamma+\beta^{*}\delta\end{split}\]

A central feature of quantum mechanics is that, when we perform a measurement to determine whether a state is $0$ or $1$, for example, we get a random answer, and the probability of measuring a state to be $0$ is given by the squared magnitude of its $0$ coefficient.

\[\mid0\rangle/\mid1\rangle$ meaurement yields $\mid0\rangle$ with probability $\mid\langle\psi\mid0\rangle\mid^{2}=\mid\alpha\mid^{2}\]

One consequence of this is that, since such a measurement on a qubit state must result in $0$ or $1$, these squared magnitudes must sum to $1$, since they are probabilities. This is called Born’s rule and the constraint that the probabilities must sum to $1$, is called normalization.

Normalization:

\[\begin{split}\langle\psi\mid\psi\rangle=\left[\begin{array}{cc} \alpha^{*} & \beta^{*}\end{array}\right]\left[\begin{array}{c} \alpha\\ \beta \end{array}\right]=\mid\alpha\mid^{2}+\mid\beta\mid^{2}=1 \end{split}\]

We can also express qubit states in different bases. Consider the often-used plus-minus basis, which consists of the normalized sum and difference of the 0 and 1 ket vectors. Given a state expressed in the 0/1 basis, we can calculate the coefficients required to express the same state in the plus-minus basis, as I have done here.

States can be expressed in different bases:

\[\mid\pm\rangle=\frac{\mid0\rangle\pm\mid1\rangle}{\sqrt{2}}\,\,\,\alpha\mid0\rangle+\beta\mid1\rangle=\left(\frac{\alpha+\beta}{\sqrt{2}}\right)\mid+\rangle+\left(\frac{\alpha-\beta}{\sqrt{2}}\right)\mid-\rangle\]

This leaves us with a small question: if any basis is just as good as any other, is there a basis that we should use as the default? Fortunately, the devices that we use to store and manipulate qubit states provide us with just such a default basis.

alt text

One such device, which we see quite frequently in quantum mechanics, is the harmonic oscillator, shown on the left. These devices are described by a Hamiltonian, which is a matrix that assigns an energy E_k to each of its eigenstates, or preferred basis states, psi_k.

Every device has a Hamiltonian:

\[H=\sum_{k}E_{k}\mid\psi_{k}\rangle\langle\psi_{k}\mid\]

These states $\mid\psi_{k}\rangle$ are orthogonal:

\[\langle\psi_{j}\mid\psi_{k}\rangle=\delta_{j,k}\]

and normalized (so we say that they’re orthonormal), so the inner product of $psi_j$ with any $psi_k$ other than itself is $0$.

This allows us to use these states as a computational basis, replacing any detailed knowledge of the wavefunction psi_k with a simple label k that indicates which state we’re talking about.

They form a comutational basis

$\mid\psi_{k}\rangle\mapsto\mid k\rangle$

Some devices have finite-dimensional state spaces, unlike the harmonic oscillator.

alt text

6.1. Operations & Observables

Logic gates \equiv unitary matrices $\equiv$ changes of basis

\[U=\sum_{k}\mid\psi_{k}\rangle\langle k\mid\,\,\,\,\,\left(\langle\psi_{j}\mid\psi_{k}\rangle=\delta_{j,k}\right)\]

The logic gate is the smallest classical computing system, and all classical computations can be expressed as a large sequence of these gates. The quantum counterpart to a logic gate is a unitary matrix, which replaces the computational basis with a new basis that depends on the operation that we want to perform. Probabilistic measurements are also described by matrices, but these matrices are hermitian, so in ket notation, they are just weighted sums of these ket-bra terms, where the psi_k states form an orthonormal basis for the space.

Readout $\equiv$ measurement operators $\equiv$ hermitian matrices

\[A=\sum_{k}r_{k}\mid\psi_{k}\rangle\langle\psi_{k}\mid\,\,\,\,\,\left(r_{k}\,\,real\right)\]

If a measurement results in the state $\psi_{k}$, the value $r_{k}$ becomes known to the experimentalist.

The expected value for this measurement is given by sandwiching the measurement operator with the state that we’re measuring, and as you can see, the expectation value is just a weighted sum of the $r_{k}$ terms, with the probabilities given by Born’s rule.

Average experimental outcomes $\equiv$ ‘sandwich’ prodducts

\[\langle\phi\mid A\mid\phi\rangle=\sum_{k}r_{k}\langle\phi\mid\psi_{k}\rangle\langle\psi_{k}\mid\phi\rangle=\sum_{k}r_{k}\mid\langle\phi\mid\psi_{k}\rangle\mid^{2}\]

6.1.1. Unitary operations

That’s probably a lot to take in all at once, so let’s take a look at a few examples. Here, we have a unitary operation that exchanges states in the computational basis, which we call X, or the Pauli X if you’re already familiar with Pauli matrices. $$ X=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]=\mid1\rangle\langle0\mid+\mid0\rangle\langle1\mid\,\,\,\,\,X\mid0\rangle=\mid1\rangle\,\,\,\,\,X\mid1\rangle=\mid0\rangle $$

Here it is decomposed into ket-bra terms, and here’s what happens when we use it to transform one of the computational basis states. We just get the other state. 0 goes to 1, and 1 goes back to 0. Not so exciting. Now let’s take a look at a more interesting operation, the Hadamard gate, H.

\[\begin{split} H=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right]=\mid+\rangle\langle0\mid+\mid-\rangle\langle1\mid\,\,\,\,\,H\mid0\rangle=\mid+\rangle\,\,\,\,\,H\mid1\rangle=\mid-\rangle \end{split}\]

As you can see, this changes the basis from the 0/1 basis to the +/- basis that we discussed earlier. Put in a 0, get out a +, put in a 1, get out a -. There is another gate, called the phase gate, which only multiplies the one state by a factor of i, changing the basis in a very subtle way. $$ P=\left[\begin{array}{cc} 1 & 0\\ 0 & i \end{array}\right]=\mid0\rangle\langle0\mid+i\mid1\rangle\langle1\mid\,\,\,\,\,P\mid0\rangle=\mid0\rangle\,\,\,\,\,P\mid1\rangle=\mid1\rangle $$

Now let’s take a look at a few measurement operators. In an interesting coincidence, the Pauli X shows up again, since it’s both unitary and hermitian, we can use it both as an operation and a measurement.

Hermitian observables: $$ X=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]=\mid+\rangle\langle+\mid-\mid-\rangle\langle-\mid\,\,\,\,\,\langle+\mid X\mid+\rangle=1\,\,\,\,\,\langle-\mid X\mid-\rangle=-1 $$

Here, we have decomposed it into its eigenbasis, so the ket-bra terms are different than before, and we can see that its output values are +/- 1. Here’s another matrix which is both unitary and hermitian, the pauli Z. $$ Z=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]=\mid0\rangle\langle0\mid-\mid1\rangle\langle1\mid\,\,\,\,\,\langle0\mid Z\mid0\rangle=1\,\,\,\,\,\langle1\mid Z\mid1\rangle=-1 $$

It returns the exact same values as X, just for states in the 0/1 basis instead of the +/- basis.

And finally we have the identity, which is also unitary and hermitian.

\[\begin{split} I=\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]=\begin{array}{ccc} \mid0\rangle\langle0\mid & - & \mid1\rangle\langle1\mid\\ & or\\ \mid+\rangle\langle+\mid & + & \mid-\rangle\langle-\mid \end{array}\,\,\,\,\,\langle\psi\mid I\mid\psi\rangle=1\,\,\,\text{for all}\mid\psi\rangle \end{split}\]

We can note that the identity doesn’t change the basis at all, so that indeed it can be expressed in any basis. Also when used as a measurement operator, it always returns 1, no matter what state is input.

6.1.2. The Bloch Sphere

Coefficient in polar co-ordinates:

normalization constrains qubits’ states further

Now that we have a few examples done, I would like to focus a little on a useful geometric representation of qubit states, called the Bloch Sphere. To show how qubit states can be mapped to the surface of a sphere, let’s start by expressing the coefficients alpha and beta in polar co-ordinates.

\[ |\psi\rangle=r_{0}e^{i\phi_{0}}|0\rangle+r_{1}e^{i\left(\phi_{0}+\phi\right)}|1\rangle \]

Next, we can note that we can get rid of the phase on the alpha coefficient, since a ket multiplied by a phase produces the same measurement results as the ket itself, for any potential experiment, so these are not different in any physical sense.

(global phases don’t matter??

\[ \mid\psi\rangle\mapsto e^{i\phi_{0}}\mid\psi\rangle \]

\[ \langle\psi\mid A\mid\psi\rangle\mapsto\left\langle \psi\left|e^{-i\alpha}Ae^{i\alpha}\right|\psi\right\rangle =\langle\psi\mid A\mid\psi\rangle \]

\[ \mid\psi\rangle=r_{0}\mid0\rangle+r_{1}e^{i\phi}\mid1\rangle \]

This results in a simpler expression for the state, but it’s not yet as simple as it can be.

To get it even simpeler, we need to recall that these states have to be normalized, so the sum of the squares of the two radii involved has to be 1.

\[ \langle\psi\mid\psi\rangle=r_{0}^{2}+r_{1}^{2}\equiv1 \]

\[ \therefore\mid\psi\rangle=\cos(\theta)|0\rangle+e^{i\varphi}\mathrm{sin}(\theta)|1\rangle \]

This implies that we can express them as the cosine and sine of an angle theta, since cos(theta)^2 + sin(theta)^2 is always 1.

This family of angles theta and phi also describes the set of points on the surface of a sphere of unit radius in 3d space.

alt text

\[ \begin{align}\begin{aligned}\mid\psi\rangle=\cos(\theta)|0\rangle+e^{i\phi}\sin(\theta)|1\rangle$$ anti-parallel on the Bloch sphere\\We can see that, setting theta to 0, and phi to whatever we want (here it's 0), we get $cos(theta) = 1$, $sin(theta) = 0$, so the corresponding state is simply the $0$ basis state.\\$$\mid\theta=0,\phi=0\rangle=\mid0\rangle,|\theta=\pi,\phi=0\rangle=\mid1\rangle\end{aligned}\end{align} \]

If we set theta to $\pi$ however, we get the 1 state regardless of the setting of phi. Likewise if we set theta to $pi/2$, and $\phi$ to either $0$ or $\pi$, we get one of the states from the $+/-$ basis on the equator of the sphere.

\[\mid\theta=\pi/2,\phi=0\rangle=|+\rangle,|\theta=\pi/2,\phi=\pi\rangle = \mid -\rangle\]

Unitary operations, which change the basis we are working in, effect1ively rotate the sphere. For example, the Hadamard operation from earlier rotates the +/- states on the equator by 90 degrees until they’re at the poles. Also note that the states on opposite sides of the sphere are actually orthogonal, so this mapping does not preserve the angle between states, but it’s still useful for describing single-qubit states and operations.

Unitary operations rotate the sphere $$\left\langle \psi\mid\psi^{\prime}\right\rangle =0\leftrightarrow\text{anti-parallel on the Bloch sphere}$$

But how do we describe multi-qubit states and operations? Specifically, how do we build them up from operations on smaller subsystems of a many-qubit state? To accomplish this, we use a different kind of matrix product, called the tensor product, or Kronecker product. To take the tensor product of two matrices A and B, we write out a block matrix where each block is equal to B times the appropriate element of A.

Multi-qubit states and operations: tensor (or Kronecker) products:

\[\begin{split}A\otimes B=\left[\begin{array}{ccc} A_{0,0}B & \cdots & A_{0,n-1}B\\ \vdots & \ddots & \vdots\\ A_{m-1,0}B & \cdots & A_{m-1,n-1}B \end{array}\right] Compatible with matrix-matrix product (B)(C\otimes D)=(AC)\end{split}\]

The upper left block is B times the upper left element of $A$, and so on.

The interesting thing about the tensor product is that it’s compatible with the regular matrix product.

\[(A\otimes B)(C\otimes D)=(AC)\otimes(BD)\]

That is to say if I take the matrix product of two tensor products, I get the same matrix as if I took the matrix products first, then evaluated the tensor product. This is also easier to understand if we take a look at a few examples.

7. Multi-Qubit States & Operations

Here we see the tensor product of two states, + and 0, resulting in a state which we often call +0. Each 2-by-1 block of the state vector we’re calculating is proportional to the vector for the zero state, and it’s multiplied by one of the elements of the plus state.

\[\begin{split}\mid+\rangle\otimes|0\rangle=\mid+0\rangle=\frac{1}{\sqrt{2}}\left[\begin{array}{lc} 1 & \left(\begin{array}{l} 1\\ 0 \end{array}\right)\\ 1 & \left(\begin{array}{l} 1\\ 0 \end{array}\right) \end{array}\right]=\frac{1}{\sqrt{2}}\left[\begin{array}{l} 1\\ 0\\ 1\\ 0 \end{array}\right]\end{split}\]

We can also take tensor products of operations and observables, here we take the tensor product of X and the identity, making a block matrix whose blocks are all equal to the identity, multiplied by one of the elements of X.

\[\begin{split}X\otimes I=\left[\begin{array}{ll} 0\left(\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right) & 1\left(\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right)\\ 1\left(\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right) & 0\left(\begin{array}{ll} 1 & 0\\ 0 & 1 \end{array}\right) \end{array}\right]=\left[\begin{array}{llll} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{array}\right]\end{split}\]

7.1. Example: Bell State Preparation

Now that we have seen the basic formalism and notation, we’re ready to calculate the results of a small sequence of quantum operations, which prepares a Bell state. First, we introduce a two-qubit gate which cannot be written as a tensor product of one-qubit gates, the controlled not, or cnot for short.

\[\begin{split} \begin{array}{l} \text{ controlled-NOT: }\text{CNOT}=\left[\begin{array}{llll} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]=|0\rangle\langle0|\otimes I+|1\rangle\langle1|\otimes X\\ \\ \end{array} Bell states are entangled: cannot be written as tensor product \end{split}\]

As we can see, it can be decomposed into a sum of two tensor products that says if the first qubit is in the zero state, do the identity, and if the first qubit is in the one state, perform an X on the second qubit. Here we see the Bell state, which we’re going to prepare using the cnot.

\[\begin{split} \text{ Bell state: }\mid\Omega\rangle=\frac{\mid00\rangle+\mid11\rangle}{\sqrt{2}}=\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right] \end{split}\]

We can write it out using either ket notation, or as a column vector.

7.2. Example: Bell State Preparation

Note that the Bell state, just like the cnot, cannot be written as a tensor product. The first step in preparing the Bell state is to perform a Hadamard operation on the first qubit, which has been initiliazed in the zero state. So we calculate the tensor product of the Hadamard with the identity, since we’re not going to do anything to the second qubit immediately.

\[\begin{split} H\otimes I=\frac{1}{\sqrt{2}}\left[\begin{array}{cccc} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & -1 & 0\\ 0 & 1 & 0 & -1 \end{array}\right] \end{split}\]

Now we’re ready to set up our sequence of operations. First, we prepare the state 00, then we apply the Hadamard, then the cnot. This results in a Bell state, which is written out here as a column vector.

\[\begin{split} \begin{array}{l} \text{ CNOT }\times(H\otimes I)(|0\rangle\otimes|0\rangle)\\ =\frac{1}{\sqrt{2}}\left[\begin{array}{llll} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]\left[\begin{array}{cccc} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & -1 & 0\\ 0 & 1 & 0 & -1 \end{array}\right]\left[\begin{array}{l} 1\\ 0\\ 0\\ 0 \end{array}\right]=\left[\begin{array}{c} \frac{1}{\sqrt{2}}\\ 0\\ 0\\ \frac{1}{\sqrt{2}} \end{array}\right] \end{array} \end{split}\]

Now this is the correct answer, but it was a little tedious to come to, and the matrices involved are a little large. 3x3 matrices are typically big enough for any phisicist, so 4x4 is overdoing it a little.

Let’s try and do it the easy way, using some more ket notation. First, we use the compatibility of the Kronecker and regular products to show that the state we get from executing the Hadamard on the first qubit is simply +0, without having to use any matrices.

\[ \begin{array}{l} H\otimes I(\mid0\rangle\otimes\mid0\rangle)=\mid+\rangle|0\rangle=\mid+0\rangle=\frac{1}{\sqrt{2}}\mid00\rangle+\mid10\rangle\end{array} \]

Now, all we have to show is that the cnot will take our state, which is 00 + 10, to the 00 + 11 state that we’re after.

\[ \Omega\rangle=CNOT\times\left(\frac{1}{\sqrt{2}}(\mid00\rangle+\mid10\rangle)\right) \]

If we insert the decomposition of the cnot into tensor products that we saw earlier, we can see that the resulting state has coefficients proportional to these inner products of the 0 & 1 states.

\[ =\frac{1}{\sqrt{2}}(\mid0\rangle\langle0\mid0\rangle\otimes I\mid0\rangle+\mid0\rangle\langle0\mid1\rangle\otimes I\mid1\rangle\quad+\mid1\rangle\langle1\mid0\rangle\otimes X\mid0\rangle+\mid1\rangle\langle1\mid1\rangle\otimes X\mid0\rangle) \]

Of course, the 0/1 basis is orthonormal, so the products 00 and 11 evaluate to 1, and the products 01 and 10 evaluate to 0, leaving us with two remaining terms.

\[ =\frac{1}{\sqrt{2}}\left(\mid00\rangle+\mid11\rangle\right) \]

These terms are exactly the 00 and 11 terms that we were after. And that’s how ket notation van help us to describe the effects of operations and measurements on quantum states.

7.2.1. Main takeaways

In ket notation, a quantum state is expressed using a column vector with complex coefficients. The squared magnitudes of these coefficients are the probabilities of measuring that particular outcome.
Devices used for storing and manipulating qubit states are described by a Hamiltonian, which is a matrix that assigns energies to each of its eigenstates or preferred basis states.
Quantum logic gates are described by unitary matrices. Probabilistic measurements are described by Hermitian matrices.
An entangled state is a multi-qubit state that cannot be written as a tensor product of one-qubit states.
Qubit states are geometrically represented using the Bloch sphere.
Multi-qubit states and operations can be described using the tensor product. Note, an actual two-qubit gate cannot be described with a tensor product.

8. Advantages and dissadvantages of Ket notation

In the previous ket notation we discussed three ways to represent quantum states:

column vectors with complex coefficients,
kets themselves and
the geometric representation,

which uses the Bloch sphere. Each one of these has their own little advantages and disadvantages, which we are going to see in the course of these videos. First, let’s take a look at some of the advantages of ket notation. So here’s ket…notation. Is very nice for…sparse…states. That is to say: states… …which don’t have that many non-zero entries in some bases. Take for example this state here, which I will call psi, along with most other states.

\[\mid\psi\rangle=\frac{\mid00000\rangle+\mid11111\rangle}{\sqrt{2}}\]

It has got two nonzero terms. There is a one over root two term for the component in $00000$ and a one over root two term for the all-ones as well. That would be very cumbersome to write out in terms of a full vector, because that would require $2^5$ or $32$ entries, but here you can do it very compactly. Now, if you don’t have a compact discription for the state like this, if you just have some arbitrary coefficients, it can be very painstaking and time-consuming to write out something like this.

\[\alpha\mid00\rangle+\beta\mid01\rangle+\gamma\mid10\rangle+\delta\mid11\rangle\]

$\beta\otimes 01 + \gamma \otimes 10 + \delta \otimes 11$. Here you would be better of to use something more like a column vector. $\alpha, \beta, \gamma, \delta$.

\[\begin{split} \left[\begin{array}{c} \alpha\\ \beta\\ \gamma\\ \delta \end{array}\right] \end{split}\]

And in some other times, especially if you’re considering operations - transformations on the set of states – it can be beneficial to consider the Bloch sphere, where you have the reference state zero, some other states one, plus, minus and like so…

8.1. Synthesising rotations

In some setups, experimentalists can only rotate along two axes of the Bloch sphere. In this video Ben will show you an example of how you can use these two rotation axes to achieve rotations around a third axis. He will do it both in vector form and in the Bloch sphere representation. Which do you think is the most convenient?

Let’s show an example of something that’s a little bit difficult to do when looking at the matrix based picture of quantum mechanics, but if we put everything on the Bloch sphere it makes perfect sense.

In many experiments, experimentalists can only rotate along two axes of the Bloch sphere. That is the x-axis, which goes in this direction and the y-axis, which is going in this way.

Many algorithms rely on rotations around the z-axis, which is independent from these other two. And we would like to see how to synthesize the z-rotations out of the tools that we have available: x and y rotations.

Now, I happen to know the answer to this question and I can begin by writing down the matrices that describe the operations that we would like to do. So, here is z and I’ll say this is z(theta). For those of you who are really in the know, this is

\[\begin{split} e^{i*z*theta}. But if this doesn’t make sense to you, then this is just a diagonal matrix with Z(\theta)=e^{\frac{iZ\theta}{2}}=\left[\begin{array}{cc} e^{\frac{i\theta}{2}} & 0\\ 0 & e^{-\frac{i\theta}{2}} \end{array}\right] \end{split}\]

and

\[ e^{-i*theta/2} \]

on the diagonal, zeros off the diagonal. And we can see immediately that x and y rotations on their own aren’t going to give us this z-rotation.

So we can write down x theta and y theta, which are just rotations around these axes by angles theta.

\[E^{ix*theta/2}, e^{i*y*theta/2}\]

And this guy is equal to cosine of theta over two minus i $sin(\frac{\theta}{2}) -i\sin(\frac{\theta}{2})$

\[\begin{split} X(\theta)=e^{\frac{iX\theta}{2}}=\left[\begin{array}{cc} \cos(\frac{\theta}{2}) & -i\sin(\frac{\theta}{2})\\ -i\sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2}) \end{array}\right] \end{split}\]

\[ E^{i*y*theta/2} \]

is quite similar. Cosine theta over two, sine theta over two, minus sine theta over two, cosine theta over two. Now, there’s a product of these rotations. A y an x and a y, that will excecute one of these sets.

\[\begin{split} Y(\theta)=e^{\frac{iY\theta}{2}}=\left[\begin{array}{cc} \cos(\frac{\theta}{2}) & -\sin(\frac{\theta}{2})\\ -\sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2}) \end{array}\right] \end{split}\]

Lets look at those matrix rotations first. What I am going to do is set theta, well, equal to pi over two for a y rotation. And that’s going to give me one over root two, one one, minus one, one. Y theta equal to minus pi over 2 will give me operations that look like one, minus one, one, one. You might recognize these to be quite similar to the Hadamard matrix that changes from the computational basis to the plus minus basis, and that is by design.

\[\begin{split} Y(\theta=\frac{\pi}{2})=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right] \end{split}\]

\[\begin{split} Y(\theta=-\frac{\pi}{2})=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ -1 & 1 \end{array}\right] \end{split}\]

Now lets take a look at what happens if we sandwich an x rotations around theta with these two matrices, which we lovingly call y 90 and y -90. Because they are 90 degree rotations. So, we’re going to get a, well, y pi by 2 x theta y minus pi 2. Is equal to 1 over root two one, minus one, one,one, cosine theta over two, minus i sine theta over two, minus i sine theta over two, cosine of theta over two. Then we have another one over root two, one, minus one, one, one. We multiply these through. So, first off I can take these two factors of one over square root of two, make them a factor of a half. One minus one one one. Then if I multiply these here, I get cos theta over two plus I sine theta over two, because these two minus signs cancel. So that’s cos theta over two plus I sine theta over two.

\[\begin{split} Y_{\frac{\pi}{2}}X_{\theta}Y_{-\frac{\pi}{2}} =\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right]\cdot\left[\begin{array}{cc} \cos(\frac{\theta}{2}) & -i\sin(\frac{\theta}{2})\\ -i\sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2}) \end{array}\right]\cdot\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\\ -1 & 1 \end{array}\right] =\frac{1}{2}\left[\begin{array}{cc} 1 & -1\\ 1 & 1 \end{array}\right]\cdot\left[\begin{array}{cc} \cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2}) & \cos(\frac{\theta}{2})-i\sin(\frac{\theta}{2})\\ -(\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})) & \cos(\frac{\theta}{2})-i\sin(\frac{\theta}{2}) \end{array}\right] =\frac{1}{2}\left[\begin{array}{cc} 2\left(\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})\right) & 0\\ 0 & 2?\left(\cos(\frac{\theta}{2})-i\sin(\frac{\theta}{2})\right) \end{array}\right] \end{split}\]

Here I’ll get cosine theta over two minus I sine theta over two. Here I’ll get minus cosine, minus I sine. So that’s minus I cosine theta plus I sine theta over two. And then here I’ll get cosine minus I sine. So, cosine theta over two minus I sine theta over two. Continuing on with all this math I can notice that these two elements are equal and opposite. So If I take one and minus one, I’ll get cosine plus I sine, minus negative cosine plus I sine. Which is simply two cosine theta over two plus I sine theta over two. Here I take a number and it’s opposite, I get zero. Here I can take these two numbers which are equal and add them. That gives me theta over two minus I sine theta over two. And if I take one and minus one and these two I get zero as well.

That’s a lot of matrix multiplication. Now, let’s also use Eulers identity. We know that this is equal to e^{itheta/2} and that this is equal to e^{-iTheta/2}.

\[ \begin{align}\begin{aligned} e^{i\theta}=\cos(\frac{\theta}{2})+i\sin(\frac{\theta}{2})\\e^{-i\theta}=\cos(\frac{\theta}{2})-i\sin(\frac{\theta}{2})\\From arithmetic using complex numbers. And so we have managed to synthesize this z rotation around theta just by using x rotations and y rotations.\\\begin{split}=\frac{1}{2}\left[\begin{array}{cc} 2e^{i\theta} & 0\\ 0 & 2?e^{-i\theta} \end{array}\right] \end{split}\end{aligned}\end{align} \]

So by using a little bit of ket notation and some trigonometric identities, we can show that we can obtain universal control using only the tools that are available to the experimentalists. This is nice because it means we don’t have to design new hardware. We can just put together operations that we already have in order to obtain universal control.

But this is not that simple as it could be. There is a far easier way to see this, which we can do using the Bloch sphere. The action of this sequence of unitary operations. Y 90, x around some angle theta and then y minus 90 can be expressed in terms of matrices. But it’s much easier to see using the Bloch sphere. If we write down the Bloch sphere with the x axis pointing out of the board, we can see the effect of a y 90, a rotation around 90 degrees of the y axis, is just turning the sphere sort of like a steering wheel, so that the x axis now faces down and the z axis is where the x axis used to be. If we now apply a rotation around the x axis, that is giving us a rotation around our old z axis. And once we have that done, all we have to do is reverse the y 90 with a y minus 90 to restore the orientation of our original axes, having picked up a phase on the z. And that’s how we can synthesize a diagonal unitary. e^{itheta/2}, e^{-i*theta/2} out of the operations to which the experimentalists have access. A little bit of notations saves you a lot of hardware design.

8.2. State decomposition

Often in quantum computing , it is necessary to express one state in terms of a basis of some other states. I am going to go ahead and call this state decomposition, although you can see it under a variety of names and usually it’s clear from the context that they mean expressing one state in terms of a basis that is more familiar or convenient to. Now the rule of state decomposition is that any state phi can be expressed as a sum over k terms, where each term consists of an inner product of phi - the trial state - with one of the basis states phi_k, and the states phi_k. Number times vector. And this is the same as decomposing a vector in a basis if you are already familiar with linear algebra.

\[ \mid\phi\rangle=\sum_{k}\underbrace{\langle\psi_{k}\mid\phi}*{\text{number}}\mid\underbrace{\psi*{k}}_{\text{vector}}\rangle \]

As an example of this, let’s take a look at a trial wave function of zero, and a basis which is the Hadamard basis, the plus minus basis, that we reviewed in the lecture. The sum over psi_k, phi, psi_k is then the inner product of plus and zero times plus and the inner product of minus and zero times minus. Now here I figured out those inner product. The inner product of plus zero is one over root two times the dot product of these two vectors, which is just one over root two. And the calculation for minus zero is very similar, there is a minus sign here, but it gets multiplied by zero, making no difference, and we end up with one over root two again. Therefore, zero can be decomposed as one over root two, plus + minus, just as the plus state can be decomposed as one over root two, zero + one.

Example:

\[ \begin{align}\begin{aligned} \mid\phi\rangle=\mid0\rangle,\,\,\,\mid\psi_{0}\rangle=\mid+\rangle,\,\,\,\mid\psi_{1}\rangle=\mid-1\rangle\\\mid\phi\rangle=\sum_{k}\langle\psi_{k}\mid\phi\rangle\mid\psi_{k}\rangle=\langle+\mid0\rangle\mid+\rangle+\langle-\mid0\rangle\mid-\rangle\\\begin{split}\langle+\mid0\rangle=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & 1\end{array}\right]\left[\begin{array}{c} 1\\ 0 \end{array}\right]=\frac{1}{\sqrt{2}}\end{split}\\\begin{split}\langle-\mid0\rangle=\frac{1}{\sqrt{2}}\left[\begin{array}{cc} 1 & -1\end{array}\right]\left[\begin{array}{c} 1\\ 0 \end{array}\right]=\frac{1}{\sqrt{2}}\end{split}\\\therefore\mid0\rangle=\frac{1}{\sqrt{2}}\left(\mid+\rangle+\mid-\rangle\right)=\frac{1}{\sqrt{2}} \end{aligned}\end{align} \]

8.3. Performing arbitrary measurements

It could be that your measurement device is only capable of measuring in one basis. Ket notations helps us to show that it will not be difficult to let this simple measurement device measure in any basis. In this example Ben will show you how this works.

I’d like to talk today about how a little bit of ket-notation can save us a big experimental headache, when it comes time to try measure a qubit in an arbitrary basis. First, we have to review a few of the important facts.

Arbitrary measutremnent

The conjugate transpose of a matrix or a vector is given by first taking the transpose of this 2-by-2 matrix that leaves the diagonal elements in the same place, and exchanges the elements being b and c here, but also taking the complex conjugate of each element in the matrix.

\[\begin{split} \mid fU=\frac{1}{2}\left[\begin{array}{cc} a & b\\ c & d \end{array}\right],\,\,\,U^{\dagger}=\left[\begin{array}{cc} a^{*} & b^{*}\\ c^{*} & d^{*} \end{array}\right] \end{split}\]

A second important fact, which you can prove yourself using this definition, and I encourage you to do so, is that the dagger or complex conjugate transpose of U- psi (ket) is just the bra for psi times U-dagger. Now, that follows from linear algebra, but it is easy to check for yourself.

\[ \begin{align}\begin{aligned} \langle U\mid\psi\rangle\rangle^{\dagger}=\langle\psi\mid U^{\dagger}\\\begin{split}\mid\psi\rangle=\left[\begin{array}{c} \alpha\\ \beta \end{array}\right],\,\,\,\mid\psi_{\bot}\rangle=\left[\begin{array}{c} \beta^{*}\\ -\alpha^{*} \end{array}\right], \end{split}\end{aligned}\end{align} \]

There’s also another construction that I’d like to use, which is that for every state psi, there is another state which I call psi-perp(endicular), which consists of these coefficients. So, if psi is alpha-beta, psi-perp is the conjugate of beta and minus the conjugate of alpha. And these states are by construction orthogonal, which you can also see just by taking the inner product and seeing that it’s always zero regardless of what alpha and beta are.

And with those three ingredients, we can solve this question.

´+ So, experimentalists usually measure a single operator: the Pauli-Z operator, that was perhaps discussed earlier. But we’d like to be able to measure in whatever basis we want.

\[ \{\mid0\rangle,\mid1\rangle\rangle,\,\,\,\text{basis}(Z)} \]

So that if we have a question: “Is this state this or that?”, we can answer it without having to restrict ourselves to the 0-1 basis. So, we have a two-step plan to solve this. First, we’re going to apply some operator U, and then we’re going to measure in the 0-1 basis.

+How to measure in other basis

Apply U
Measure in {\mid0\rangle,\mid1\rangle\rangle

Now, if you’ve got a controllable qubit, you can apply an operator U as you see fit, and as discussed in the statement of the question, we can measure in the 0-1 basis. We can measure the Z-operator. So, the expectation value of this measurement, if we first apply U on psi, that replaces psi with psi-U-dagger in the bra and U-psi in the ket, and then we take the familiar sandwich product to determine the expectation value of the operator, we end up with psi-U-dagger-Z-U-psi.

Expectation value;

\[ =\left(\langle\psi\mid U^{\dagger}\right)Z\left(U\mid\psi\rangle\right) =\langle\psi\mid\left(U^{\dagger}ZU\right)\mid\psi\rangle \]

\[ U^{\dagger}ZU=\underbrace{U^{\dagger}\mid0\rangle}*{\mid\psi\rangle}\underbrace{\langle0\mid U}*{\langle\psi\mid}-\underbrace{U^{\dagger}\mid1\rangle}*{\mid\psi*{\bot}\rangle}\underbrace{\langle1\mid U}*{\langle\psi*{\bot}\mid} \]

So, it’s as if we had our original state psi, and instead of measuring Z, we measured U-dagger-Z-U. Now, U-dagger-Z-U we can write out like so. And if we label the state U-dagger-0 as psi, then U-dagger-Z-U is equal to psi-psi minus psi-perp-psi-perp. This is a measurement that will return 1 if the state is psi, and -1 if the state is psi-perp, giving us an arbitrary basis to measure in.

9. A fact about maximally entangled states

Ket notation is not always the best approach to find an answer or solve an equation. Ben will give you an example about entangled states where it would be a lot of work to use ket notation.

Often when doing quantum information, ket-notation can fail you. This happens a lot when you have a densely packed vector full of a bunch of different coefficients and there’s no obvious structure. Now, hopefully when that happens to you, you’re only going to deal with a few qubits. Otherwise, you are going to writing out coefficients for a very long time.

\[ \frac{1}{\sqrt{2}}\left(\mid\psi^{*}\psi\rangle+\mid\psi_{\bot}^{*}\psi_{\bot}\rangle\right)=\frac{1}{\sqrt{2}}\left(\mid00\rangle+\mid11\rangle\right)\qquad\text{always} \]

Here, we can see a not too bad example with two qubits where we’re going to prove that for any maximally entangled state of the form psi-star-psi plus psi-perp-star-psi-perp, it’s always equal to the Bell state that’s fully correlated that we know and love: 1 over root 2 (times) 0-0 plus 1-1.

\[ \begin{align}\begin{aligned}\begin{split} \mid\psi\rangle=\left[\begin{array}{c} \alpha\\ \beta \end{array}\right],\,\,\,\mid\psi^{*}\rangle=\left[\begin{array}{c} \alpha^{*}\\ \beta^{*} \end{array}\right],\end{split}\\\begin{split}\mid\psi_{\bot}\rangle=\left[\begin{array}{c} \alpha\\ \beta \end{array}\right],\,\,\,\mid\psi_{\bot}^{*}\rangle=\left[\begin{array}{c} \beta\\ -\alpha \end{array}\right], \end{split}\end{aligned}\end{align} \]

And this is regardless of what psi is. So, if we take an arbitrary wavevector psi which is two complex coefficients alpha and beta, we can define psi-star, which is just the complex conjugate of that state, which is also a valid state, psi-perp, which is some state which is orthogonal to psi, which I encourage you to check for yourself by taking the inner product, and psi-perp-star, which is the complex conjugate of the orthogonal state. And if we write out all of our tensor products using the formula that we learned earlier, we obtain a pair of densely packed vectors full of those coefficients, which would be very awkward in ket-notation.

\[\begin{split} \frac{1}{\sqrt{2}}\left(\mid\psi^{*}\psi\rangle+\mid\psi_{\bot}^{*}\psi_{\bot}\rangle\right) =\frac{1}{\sqrt{2}}\left[\left[\begin{array}{c} \alpha^{*}\alpha\\ \alpha^{*}\beta\\ \beta^{*}\alpha\\ \beta^{*}\beta \end{array}\right]+\left[\begin{array}{c} \beta^{*}\beta\\ -\beta\alpha^{*}\\ -\alpha\beta^{*}\\ \alpha\alpha^{*} \end{array}\right]\right] =\frac{1}{\sqrt{2}}\left[\left[\begin{array}{c} \mid\alpha\mid^{2}+\mid\beta\mid^{2}\\ 0\\ 0\\ \mid\alpha\mid^{2}+\mid\beta\mid^{2} \end{array}\right]\right] =\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right] \end{split}\]

But we have here alpha-star-alpha, alpha-star-beta, beta-star-alpha and beta-star-beta. And we’re going to add to that: beta-beta-star, minus beta-alpha-star, minus alpha-beta-star and alpha-alpha-star. Now, beta-star-beta plus alpha-alpha-star is just the magnitude of alpha squared plus the magnitude of beta squared, which is 1. And we see that same thing in the top term here, alpha-star-alpha plus beta-beta-star, that’s 1. And then these inner terms cancel. Because you have alpha-star-beta minus beta-alpha-star, that has just been flipped here. And a beta-star-alpha minus alpha-beta-star, if I flipped these two, it becomes obvious that they are equal and opposite, so they cancel. This implies for example, that for example if Alice and Bob share a Bell state, and Alice measure in a 0-1 basis, she can get a state 0 or 1 and she can tell that Bob has the same state. Now she measures instead in the basis psi-star psi-star-perp, she can tell that Bob has the state psi or psi-perp depending on her measurement result.

10. Experimental and theoretical measurements

In this section we will explain the difference between measurements that occur in a laboratory and measurements that are described in theory. What are the differences and how are those similar?

Let’s take this opportunity to clarify something that confuses a lot when first learning about quantum mechanics, which is the difference between measurements that occur in a laboratory and measurements that we describe in theory. And also point out hopefully the ways in which these things are similar. So, you may have already seen a picture like this one. But, in the laboratories we can measure some currents in nanoamperes, and if there’s a spin in the down state pointing parallel to the field, there will be no bump in the current. And if it’s in the up state, there will be a bump. And this is a large macroscopic classical signal that they can detect. And this relates to a measurement operator. You can imagine that we take the total area under this curve; the total amount of current, in the case that there’s no bump, and that corresponds to the spin down state. And the total current in the case that there is a bump will be slightly higher, and that corresponds to the spin up state. So, that real number: the amount of current times the projector on to the down state plus the total amount of current in the case that there is bump times the projector on to the up state gives us our measurement operator. Now, the important thing about these two signals being distinguishable, but there being a large difference between them, is that if they’re equal, we end up with some number times down-down plus some number which is identical, times up-up. So, the measurement operator becomes the identity, which doesn’t discern between the up and down states. So as useless is the measurement. But if these numbers are discernably different, then down-down and up-up receive different measurement outcomes, and out measurement operator is \mathbf{1}more like a Z, which is one we could use to accurately distinguish between the computational basis states in quantum computing.

10.1. Practice Quiz 3

10.1.1. Question 1:

Suppose we are given a pure, two-qubit state $$\psi\rangle_{AB}=\alpha_{00}|00\rangle+\alpha_{01}|01\rangle+\alpha_{10}|10\rangle+\alpha_{11}|11\rangle, where \left|\alpha_{00}\right|^{2}+\left|\alpha_{01}\right|^{2}+\left|\alpha_{10}\right|^{2}+\left|\alpha_{11}\right|^{2}=1 $$. We can compute the amount of entanglement between the two qubits, as quantified by the entanglement measure known as the concurrence, using the following expression: $$C\left(|\psi\rangle_{AB}\right)=2\left|\alpha_{00}\alpha_{11}-\alpha_{01}\alpha_{10}\right\mid$$. If the two qubits are maximally entangled, then $$C(|\psi\rangle)=1$$; and if the qubits are not entangled at all, or are in a product state, then $$C(|\psi\rangle)=0$$. Use the above expression to compute the amount of entanglement between the qubits in each of the following states:

how much entanglement is in that state?

[] $\mid\psi\rangle_{AB}=\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)=\mathbf{1}$
[] $\psi\rangle_{AB}=\frac{1}{\sqrt{2}}(|01\rangle+|10\rangle)=\mathbf{1}$
[] $\psi\rangle_{AB}=|00\rangle=\mathbf{0}$
$\psi\rangle_{AB}=\sqrt{\frac{48}{100}}(|00\rangle+\sqrt{\frac{48}{100}}|11\rangle)+\sqrt{\frac{4}{100}}\mid10\rangle=\mathbf{0.96}$

Explanation

Without a programming language, every input into the computer would need to be an exact instruction on the CPU. This would make using computers very hard and would take a lot of time.

0 points possible (ungraded)

10.2. Quiz 3: Ket notation

10.2.1. Question 1: Expectation values

The expectation value $\langle A\rangle$ of an operator $A$ upon a state $\mid\psi\rangle$ is, as mentioned, the average of very many outcomes of measurements of that operator. Mathematically speaking, it is the inner product of the state with itself, with the operator ‘sandwiched’ inbetween: $\langle\psi|A|\psi\rangle$.

The easiest way to solve this is using the linear algebra representation: if, for instance, $$ A = \left[\begin{array}{ccc} 1 & 0\\ 0 & -1 \end{array}\right], and |\psi\rangle equals \left[\begin{array}{l} 1\\ 0 \end{array}\right], then: \langle\psi|A|\psi\rangle=\left[\begin{array}{ll} 1 & 0\end{array}\right]\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{l} 1\\ 0 \end{array}\right] For the state \sqrt{\frac{1}{3}}|0\rangle+\sqrt{\frac{2}{3}}i|1\rangle$$

calculate the expectation value of the following operators:

10.2.1.1. Hadamard matrix:

\[ \begin{align}\begin{aligned}\begin{split} H=\sqrt{\frac{1}{2}}\left[\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right],\end{split}\\\mid\psi\rangle=\sqrt{\frac{1}{3}}|0\rangle+\sqrt{\frac{2}{3}}i|1\rangle\\\begin{split}\mid\psi\rangle=\sqrt{\frac{1}{3}}\left[\begin{array}{l} 1\\ 0 \end{array}\right]+\sqrt{\frac{2}{3}}i\left[\begin{array}{l} 0\\ 1 \end{array}\right]\end{split}\\\begin{split}\mid\psi\rangle=\sqrt{\frac{1}{3}}|0\rangle+\sqrt{\frac{2}{3}}i|1\rangle=\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ 0 \end{array}\right]+\left[\begin{array}{c} 0\\ \sqrt{\frac{2}{3}}i \end{array}\right]=\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ \sqrt{\frac{2}{3}}i \end{array}\right] \end{split}\end{aligned}\end{align} \]

Complex conjugate transpose: $$ \langle\psi\mid=\mid\psi*\rangle=\sqrt{\frac{1}{3}}|0\rangle-\sqrt{\frac{2}{3}}i|1\rangle=\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right] $$

\[\begin{split} \langle\psi\mid H\mid\psi\rangle=\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\sqrt{\frac{1}{2}}\left[\begin{array}{cc} 1 & 1\\ 1 & -1 \end{array}\right]\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ \sqrt{\frac{2}{3}}i \end{array}\right] \end{split}\]

\[\begin{split} \left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\sqrt{\frac{1}{2}}\left[\begin{array}{c} \sqrt{\frac{1}{3}}+\sqrt{\frac{2}{3}}i\\ \sqrt{\frac{1}{3}}-\sqrt{\frac{2}{3}}i \end{array}\right] =\sqrt{\frac{1}{2}}\left(\sqrt{\frac{1}{3}}\left(\sqrt{\frac{1}{3}}+\sqrt{\frac{2}{3}}i\right)-\sqrt{\frac{2}{3}}i\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{2}{3}}i\right)\right) =\sqrt{\frac{1}{2}}\left(\left(\sqrt{\frac{1}{3}}\sqrt{\frac{1}{3}}\right)+\left(\sqrt{\frac{1}{3}}\sqrt{\frac{2}{3}}i\right)-\left(\sqrt{\frac{2}{3}}i\sqrt{\frac{1}{3}}\right)+\left(\sqrt{\frac{2}{3}}i\sqrt{\frac{2}{3}}i\right)\right) =\sqrt{\frac{1}{2}}\left(\frac{1}{3}+\frac{\sqrt{2}}{3}i-\frac{\sqrt{2}}{3}i-\frac{2}{3}\right) =\sqrt{\frac{1}{2}}\left(\frac{1}{3}-\frac{2}{3}\right)=-\frac{1}{3\sqrt{2}} \end{split}\]

10.2.1.2. Pauli X

\[\begin{split} X=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right] \end{split}\]

\[\begin{split} \langle\psi \mid X\mid\psi\rangle=\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ \sqrt{\frac{2}{3}}i \end{array}\right] =\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\left[\begin{array}{c} \sqrt{\frac{2}{3}}i\\ \sqrt{\frac{1}{3}} \end{array}\right] =\begin{array}{cc} \sqrt{\frac{1}{3}}\sqrt{\frac{2}{3}}i & -\sqrt{\frac{2}{3}}i\sqrt{\frac{1}{3}}\end{array} =0 \end{split}\]

10.2.1.3. Pauli Y matrix

\[\begin{split} Y=\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right] \end{split}\]

\[\begin{split} \langle\psi \mid X\mid\psi\rangle=\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ \sqrt{\frac{2}{3}}i \end{array}\right] =\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\left[\begin{array}{c} \sqrt{\frac{2}{3}}\\ \sqrt{\frac{1}{3}}i \end{array}\right] =\sqrt{\frac{1}{3}}\sqrt{\frac{2}{3}}-\sqrt{\frac{2}{3}}i\sqrt{\frac{1}{3}}i =\frac{\sqrt{2}}{3}+\frac{\sqrt{2}}{3} =\frac{2\sqrt{2}}{3}\approx0.4714 \end{split}\]

10.2.1.4. Pauli Z matrix:

\[\begin{split} Z=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \end{split}\]

\[\begin{split} \langle\psi \mid X\mid\psi\rangle=\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\sqrt{\frac{1}{2}}\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ \sqrt{\frac{2}{3}}i \end{array}\right] =\left[\begin{array}{cc} \sqrt{\frac{1}{3}} & -\sqrt{\frac{2}{3}}i\end{array}\right]\left[\begin{array}{c} \sqrt{\frac{1}{3}}\\ -\sqrt{\frac{2}{3}}i \end{array}\right] =\frac{1}{3}-\frac{2}{3}=-\frac{1}{3} \end{split}\]

10.2.1.5. Question 2: Entanglement between qubits

An entangled state is a multi-qubit quantum state that can not be written as a Kronecker product of single-qubit states.

For example, if we take the tensor product of two arbitrary states |\psi\rangle=\left[\begin{array}{l}

\[\begin{split} \alpha\\ \beta \end{array}\right] and |\phi\rangle=\left[\begin{array}{l} \gamma\\ \delta \end{array}\right], we obtain |\psi\rangle\otimes|\phi\rangle=\left[\begin{array}{c} \alpha\gamma\\ \alpha\delta\\ \beta\gamma\\ \beta\delta \end{array}\right]. \end{split}\]

If we are given a vector, such as

\[\begin{split} \left[\begin{array}{l} 1\\ 0\\ 0\\ 0 \end{array}\right] \end{split}\]

and we want to show that it is separable (unentangled), we need to find a set of numbers $\{\alpha,\beta,\gamma,\delta\}$ such that $\alpha\gamma=1,\alpha\delta=0,\beta\gamma=0$, and $\beta\delta=0$. The assignment $\alpha=\gamma=1$, $\beta=\delta=0$ accomplishes this, and we can say that the state is $\mid0\rangle\otimes|0\rangle$, and therefore separable. If such an assignment does not exist, then the state is entangled. Mark all states below that are entangled:

how much entanglement is in that state?

[] $\mid\psi\rangle=\sqrt{\frac{1}{4}}\left[\begin{array}{l} 1\\ 1\\ 1\\ 1\end{array}\right]$
$\mid\psi\rangle=\sqrt{\frac{1}{4}}\left[\begin{array}{l} 1\\ 1\\ 1\\ 1 \end{array}\right]$
[] $\mid\psi\rangle=\sqrt{\frac{1}{2}}\left[\begin{array}{l} 0\\ 0\\ 0\\ 0\\ 1\\ 0\\ 0\\ 1 \end{array}\right]v$
$\mid$
[] \mid\psi\rangle=\sqrt{\frac{1}{2}}\left[\begin{array}{l} 0\ 0\ 1\ 0\ 0\ 0\ 1\ 0 \end{array}\right]
[] \mid\psi\rangle=\frac{\mid00\rangle+\mid10\rangle}{\sqrt{2}}
\mid\psi\rangle=\frac{\mid00\rangle+\mid11\rangle}{\sqrt{2}}
\mid\psi\rangle=\frac{\mid++\rangle+\mid–\rangle}{\sqrt{2}}

Explanation

Following the method from the hints, the state $$\sqrt{\frac{1}{4}}\left[\begin{array}{l} 1\\ 1\\ 1\\ 1 \end{array}\right] has coefficients a_{00}=a_{01}=a_{10}=a_{11}=\frac{1}{\sqrt{4}}$$

So if we write out the kronecker product, we see that

$\alpha^{*}\gamma,\alpha^{*}\delta,\beta^{*}\gamma$ and $\beta^{*}\delta$ should all be equal to $\frac{1}{\sqrt{2}}$.

This is easily the case for $\alpha=\beta=\gamma=\delta=\frac{1}{\sqrt{2}}$. We can conclude that we can write this state as $(\alpha|0\rangle+\beta|1\rangle)\otimes(\gamma|0\rangle+\delta|1\rangle)$ with $\alpha=\beta=\gamma=\delta=\frac{1}{\sqrt{2}}$, so the state is not entangled. If we try to write $$\mid\psi\rangle=\sqrt{\frac{1}{3}}\left[\begin{array}{l} 1\\ 0\\ 1\\ 1 \end{array}\right$$ as a kronecker product, we see that $\alpha^{}\delta=0$, but that $\alpha^{}\gamma=\beta^{*}\delta=\frac{1}{\sqrt{3}}$. This is a direct qubit states, and is thus entangled.

10.2.1.6. Question 3: Two-qubit operations

2 qubit operations are operations that involve, as expected, 2 qubits. Just as 1 qubit operations they can be written in matrix form. However, since 2 qubits are now involved, these matrices are now 4\times4 unitaries. For the following 2 qubit operations, click the appropiate matrix that corresponds with the operation. - The 2 qubit operation that applies a bit flip on the second qubit, conditioned on the first qubit being in the $\mid1\rangle$ state. So, a bit flip is applied to the second qubit if the first qubit is in the $\mid1\rangle$ state, and no flip is applied if the first qubit is in the $\mid0\rangle$ state. A bit flip (or X ) turns a 0 into a 1 and a 1 into a 0 . so

\[ \begin{align}\begin{aligned}X(\alpha|0\rangle+\beta|1\rangle)=\beta|0\rangle+\alpha|1\rangle$$. \\ What we see in this question is a controlled bit flip, or a CNOT gate, with the first qubit as control qubit, and the second qubit as target qubit. We only do a bitflip on the second qubit, if the first one is 1. \\\begin{split}```{admonition} how much entanglement is in that state? - [x] $\left[\begin{array}{llll} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]$ - [] $\left[\begin{array}{llll} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{array}\right]$ - [] $\left[\begin{array}{llll} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]$ - [] $\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1$ - [] $\right] \fullmoon\qquad\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1$ - [x] $\right]\newmoon\qquad\left[\begin{array}{llll} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0$ - [] $\right] \fullmoon\qquad\left[\begin{array}{llll} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]$ ```\end{split}\\```{admonition} Explanation - The 2 qubit operation that applies a phase flip on the second qubit, conditioned on the first qubit being in the |-\rangle state. A phase flip (or Z ) is the equivelant of a bit flip. But in the Hadamard basis. It turns a + intoa - and a - into +. You could think of this as changing the sign of the |0\rangle component; Z(\alpha|0\rangle+\beta|1\rangle)=\alpha|0\rangle-\beta|1\rangle. Here we ask for a controlled phase gate, or a CPHASE gate. But in a different basis; We only do a phase flip, if the first qubit is |-\rangle \fullmoon\qquad\\Explanation The first operation is exactly the CNOT. You should be able to recognize this one! The second operation looks very much like the CPHASE, but it has one important difference: the condition is on the |-\rangle state and not on the |1\rangle state. The Hadamard transform these to eachother, so that we need the CPHASE by that matrix that transforms the first qubit, and does nothing on the second qubit: H\otimes I. We need to map, perform the CPHASE, and map back, so the total operation becomes: (H\otimes I)CPHASE(H\otimes I). In terms of 4\times4 matrices this is\\\begin{split}$$\left[\begin{array}{cccc} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & -1 & 0\\ 0 & 1 & 0 & -1 \end{array}\right]\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{cccc} 1 & 0 & 1 & 0\\ 0 & 1 & 0 & 1\\ 1 & 0 & -1 & 0\\ 0 & 1 & 0 & -1 \end{array}\right]\end{split}\end{aligned}\end{align} \]

Calculating this product results in the desired matrix.

### Question 4: Unitaries, or mapping one basis to another

Let the basis 

$$\{\mathbf{v}\} be the set of vectors \left\{ \left[\begin{array}{c}
\cos\theta_{1}\\
\sin\theta_{1}
\end{array}\right],\left[\begin{array}{c}
-\sin\theta_{1}\\
\cos\theta_{1}
\end{array}\right]\right\}  
$$

And let the basis 

$$\{\mathbf{b}\} be the set of vectors \left\{ \left[\begin{array}{c}
\cos\theta_{2}\\
\sin\theta_{2}
\end{array}\right],\left[\begin{array}{c}
-\sin\theta_{2}\\
\cos\theta_{2}
\end{array}\right]\right\}  
$$
Calculate the matrix that maps $\{\mathbf{b}\}$ to $\{\mathbf{v}\}$. To input the questions, fill every 4 entries in seperately below. Simplify answers as good as possible by using these identities.

$$
\begin{array}{|l|}
\hline \operatorname{Product-to-sum}\\
\hline \cos\theta\cos\varphi=\frac{\cos(\theta-\varphi)+\cos(\theta+\varphi)}{2}\\
\sin\theta\sin\varphi=\frac{\cos(\theta-\varphi)-\cos(\theta+\varphi)}{2}\\
\sin\theta\cos\varphi=\frac{\sin(\theta+\varphi)+\sin(\theta-\varphi)}{2}\\
\cos\theta\sin\varphi=\frac{\sin(\theta+\varphi)-\sin(\theta-\varphi)}{2}\\
\tan\theta\tan\varphi=\frac{\cos(\theta-\varphi)-\cos(\theta+\varphi)}{\cos(\theta-\varphi)+\cos(\theta+\varphi)}\\
\hline \prod_{k=1}^{n}\cos\theta_{k}=\frac{1}{2^{n}}\sum_{e\in S}\cos\left(e_{1}\theta_{1}+\cdots+e_{n}\theta_{n}\right)\\
\text{ where }e=\left(e_{1},\cdots,e_{n}\right)\in S=\{1,-1\}^{n}\\
\prod_{k=1}^{n}\sin\theta_{k}=\frac{(-1)^{\left\lfloor \frac{n}{2}\right\rfloor }}{2^{n}}\left\{ \sum_{e\in S}\cos\left(e_{1}\theta_{1}+\cdots+e_{n}\theta_{n}\right)\prod_{j=1}^{n}e_{j}\text{ if }n\right.\text{ is even, }
\\\hline \end{array}
$$
$$
\begin{aligned} & \text{ Sum-to-product }{}^{[28]}\\
 & \sin\theta\pm\sin\varphi=2\sin\left(\frac{\theta\pm\varphi}{2}\right)\cos\left(\frac{\theta\mp\varphi}{2}\right)\\
 & \cos\theta+\cos\varphi=2\cos\left(\frac{\theta+\varphi}{2}\right)\cos\left(\frac{\theta-\varphi}{2}\right)\\
 & \cos\theta-\cos\varphi=-2\sin\left(\frac{\theta+\varphi}{2}\right)\sin\left(\frac{\theta-\varphi}{2}\right)\\
 & \tan\theta\pm\tan\varphi=\frac{\sin(\theta\pm\varphi)}{\cos\theta\cos\varphi}
\end{aligned}
$$

A nice video that explains transformations of bases can be found here. Fill in 

$$\cos\theta_{1} as \cos\left(\right. theta_1) \cos\left(\theta_{1}+\theta_{2}\right)$$ as 
$$\cos (theta_1 + theta_2), \cos\theta_{1}\cos\theta_{2} as \cos\left(\right. theta_1)$$ 

$$\cos ($ theta_ 2$), \cos \theta_{1}+\cos \theta_{2}$$ 
as 
$$\cos ($ theta_1 $+$ $\cos$ (theta_2)$$, etc. Also, answers provided in the most compact form: $$\sin \left(\theta_{1}+\theta_{2}\right)$$ or similar are preferred as others might raise problems in being detected as correct even if they are. Remember and use these identities:

$$\sin\left(\theta_{1}\pm\theta_{2}\right)=\sin\left(\theta_{1}\right)\cos\left(\theta_{2}\right)\pm\cos\left(\theta_{1}\right)\sin\left(\theta_{2}\right)$$

$$\cos\left(\theta_{1}\pm\theta_{2}\right)=\cos\left(\theta_{1}\right)\cos\left(\theta_{2}\right)\mp\sin\left(\theta_{1}\right)\sin\left(\theta_{2}\right)$$


```{admonition} Explanation
Hint (1 of 3): Since it is a mapping from one basis to another, the matrix should be unitary.

Hint (2 of 3): You can use a geometrical argument to solve this

Hint (3 of 3): The top left entry is the inner product between \boldsymbol{v}*{1}and \boldsymbol{b*{1}:}\langle\boldsymbol{v_{1}}\mid\boldsymbol{b_{1}}\rangle

The top left entry in the matrix is:

- [x] $cos(theta_{1}-theta_{2})$$

$$ cos(theta_1 - theta_2) or cos(theta_2 - theta_1) or cos(theta_1)cos(theta_2) + sin(theta_1)sin(theta_2) or cos(theta_1-theta_2) or cos(theta_2-theta_1) or cos(theta_1)cos(theta_2)+sin(theta_1)sin(theta_2) or cos(theta_2)cos(theta_1) + sin(theta_2)sin(theta_1) or cos(theta_2)cos(theta_1)+sin(theta_2)sin(theta_1)$$

The top right entry in the matrix is:

- [x] $sin(theta_{1}-theta_{2})$ 

$$ sin(theta_1 - theta_2) or sin(theta_2 - theta_1) or cos(theta_1)sin(theta_2) - sin(theta_1)cos(theta_2) or -sin(theta_1-theta_2) or sin(theta_2-theta_1)$$ or 
$$cos(theta_1)sin(theta_2)-sin(theta_1)cos(theta_2) or sin(theta_2)cos(theta_1)-cos(theta_2)sin(theta_1) or sin(theta_2)cos(theta_1) - cos(theta_2)sin(theta_1)$$

The bottom left entry in the matrix is:

- [x] sin(theta_{1}-theta_{2})

sin(theta_1 - theta_2) or -sin(theta_2 - theta_1) or sin(theta_1)cos(theta_2) - cos(theta_1)sin(theta_2) or sin(theta_1-theta_2) or -sin(theta_2-theta_1) or sin(theta_1)cos(theta_2)-cos(theta_1)sin(theta_2) or cos(theta_2)sin(theta_1) - sin(theta_2)cos(theta_1) or cos(theta_2)sin(theta_1)-sin(theta_2)cos(theta_1)

The bottom right entry in the matrix is:

- [x] cos(theta_{1}-theta_{2})

$$ cos(theta_1 - theta_2) or cos(theta_2 - theta_1) or cos(theta_1 - theta_2) or cos(theta_2 - theta_1) or cos(theta_2-theta_1) or cos(theta_1-theta_2)$$

**Explanation:** The inner product between \boldsymbol{v_{1}\,}\text{and}\,\boldsymbol{b_{1}}\rangleis cos(𝜃1)cos(𝜃2)+sin(𝜃1)sin(𝜃2). Using the identities in the link, this can be written as cos(𝜃1−𝜃2). The rest of the inner products can be calculated the same way. Note that the answer only depends on the difference between $\theta_{1}$ and $\theta_{1}$!

4. The Hardware of Quantum Computers

4.1. Quiz 1

4.1.1. Question 1: Designing the quantum computer

4.1.2. Question 2: Quantum Algorithms: beyond a speedup

4.1.3. Question 3: A programming language for a computer

4.1.4. Question 4: The compilery

4.1.5. Question 5: The current bottleneck

5. Quantum materials

5.1. Main takeaways

5.1.1. Material Criteria test quizz

5.1.2. Question 1: The usefulness of Cooper pairs

5.1.3. Question 2

5.1.4. Question 3: Mobility

6. Introduction to Ket notation

6.1. Operations & Observables

6.1.1. Unitary operations

6.1.2. The Bloch Sphere

7. Multi-Qubit States & Operations

7.1. Example: Bell State Preparation

7.2. Example: Bell State Preparation

7.2.1. Main takeaways

8. Advantages and dissadvantages of Ket notation

8.1. Synthesising rotations

8.2. State decomposition

8.3. Performing arbitrary measurements

9. A fact about maximally entangled states

10. Experimental and theoretical measurements

10.1. Practice Quiz 3

10.1.1. Question 1:

10.2. Quiz 3: Ket notation

10.2.1. Question 1: Expectation values

10.2.1.1. Hadamard matrix:

10.2.1.2. Pauli X

10.2.1.3. Pauli Y matrix

10.2.1.4. Pauli Z matrix:

10.2.1.5. Question 2: Entanglement between qubits

10.2.1.6. Question 3: Two-qubit operations