HW 2 - Starting QC Calculations

2.18

Theorem

If $⟨ u | = \sum_{i} b_{i} ⟨ i |$ and $| v ⟩ = \sum_{i} a_{i} | i ⟩$ then:

⟨ u | v ⟩ = \sum_{i} a_{i} b_{i}

\begin{proof}

\begin{aligned} ⟨ u | v ⟩ & = (\sum_{i} b_{i} ⟨ i |) (\sum_{i} a_{i} | i ⟩) \\ = \sum_{i} \sum_{j} b_{i} ⟨ i | a_{j} | j ⟩ \\ = \sum_{i} \sum_{j} b_{i} a_{j} ⟨ i | j ⟩ \\ = \sum_{i} \sum_{j} a_{j} b_{i} χ (i = j) \\ = \sum_{i} a_{i} b_{i} \end{aligned}

\end{proof}

3.5

Theorem

The Hamiltonian of a quantum state $| ψ ⟩ = [1, 2, 1]^{*}$ is given by $H = [\begin{matrix} 2 & 0 & 3 \\ 0 & 1 & 0 \\ 3 & 0 & 2 \end{matrix}]$ . Supposing, if we measure the quantum state in an arbitrary basis, which eigenvalue is the most probable? Also, calculate the expectation value of that state.

\begin{proof}
We find the eigenvalues of $H$ . First, finding all the eigenstates implies finding all the eigenvectors for $H$ :

\begin{aligned} p (λ) & = (2 - λ) (1 - λ) (2 - λ) - 3 (- 3) \\ = (2 - λ)^{2} (1 - λ) + 9 \\ = (4 - 4 λ + λ^{2}) (1 - λ) + 9 \\ = \dots \end{aligned}

λ_{1} = 5, λ_{2} = - 1, λ_{3} = 1

with corresponding e-vectors:

v_{1} = (1, 0, 1), v_{2} = (- 1, 0, 1), v_{3} = (0, 1, 0)

this is our eigenbasis. We need to compose $| ψ ⟩$ out of these:

| ψ ⟩ = v_{1} + 0 v_{2} + 2 v_{3}

Therefore, when we measure we'll (most likely) get $λ_{3} = 1$ since the number in front of $v_{3}$ is the largest magnitude.

For expectation value, we just do:

⟨ \hat{H} ⟩ = ⟨ ψ | \hat{H} | ψ ⟩ = (1, 2, 1) H (1, 2, 1)^{*} = (1, 2, 1) (5, 2, 5)^{*} = 5 + 4 + 5 = 14

Note that if we had to do it with $v_{1}, v_{2}$ then we'd need to normalize, then find the expectation value.
\end{proof}

3.8

Theorem

The basis vectors for a qutrit are $| 0 ⟩, | 1 ⟩, | 2 ⟩$ . In matrix form, the basis states are defined as:

| 0 ⟩ = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}], | 1 ⟩ = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}], | 2 ⟩ = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}]

If the state of the system is defined as $| ψ ⟩ = \frac{2 | 0 ⟩ + 3 | 1 ⟩ + 5 | 2 ⟩}{\sqrt{38}}$ , and the measurement operator $[\begin{matrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & - 6 \end{matrix}]$ , what are the possible measurements? What are their probabilities?

\begin{proof}
What helps is to find the eigenvalues $λ_{i}$ and eigenvectors $| e_{i} ⟩$ for our measurement operator. Notice that it's a diagonal matrix, so we can shortcut and say that $| 0 ⟩, | 1 ⟩, | 2 ⟩$ are the eigenstates, with corresponding eigenvalues of $2, 3, - 6$ respectively. These are our measurements.

To get their probabilities, we just take the magnitude of each ket:

\begin{aligned} P (| 0 ⟩) & = {| \frac{2}{\sqrt{38}} |}^{2} \approx 10.53 % \\ P (| 1 ⟩) & = {| \frac{3}{\sqrt{38}} |}^{2} \approx 23.68 % \\ P (| 2 ⟩) & = {| \frac{5}{\sqrt{38}} |}^{2} \approx 65.79 % \end{aligned}

and the measurements would be the expectation values of $| 0 ⟩, | 1 ⟩, | 2 ⟩$ respectively, which are $λ_{1} = 2, λ_{2} = 3, λ_{3} = - 6$ respectively.
\end{proof}

5.2

Theorem

The state of a three-qubit system is described as $\frac{1}{\sqrt{2^{3}}} (| 0 ⟩ + | 0 ⟩ + | 1 ⟩)$ . Verify whether this is a valid quantum state. (Hint: the norm of a valid quantum state is 1.)

\begin{proof}
As the hint implies, we should make sure that $| ψ ⟩$ is of unit length. As such:

| ψ ⟩ = \frac{2 | 0 ⟩ + | 1 ⟩}{\sqrt{2^{3}}}

Here, notice that if we just have the top part:

| ψ^{'} ⟩ = 2 | 0 ⟩ + | 1 ⟩ \to \frac{| ψ^{'} ⟩}{‖ ψ^{'} ‖} = \frac{| ψ^{'} ⟩}{\sqrt{2^{2} + 1^{2}}} = \frac{| ψ^{'} ⟩}{\sqrt{5}} \neq | ψ ⟩

Notice that our $| ψ ⟩$ is not the same size as our norm of the same vector $| ψ^{'} ⟩$ , which point in the same direction while $| ψ^{'} ⟩$ is of unit length. As such, $| ψ ⟩$ is not a valid quantum state.

Note

We can use this to show that HW 2 - Starting QC Calculations#^f307aa's $| ψ ⟩$ is not a valid state there. However, the answer there is still correct, assuming a normalized $| ψ ⟩$ .

\end{proof}

5.3

Theorem

The state of a qubit is described by $| ψ ⟩ = a | 0 ⟩ + b | 1 ⟩$ , where $a = \frac{1}{3} (1 + i)$ and $b = \frac{\sqrt{3.5}}{3} (1 + i)$ . Check whether the wavefunction is normalized, and if it is, evaluate the probability of measuring the qubit in a state of $| 0 ⟩$ or $| 1 ⟩$ .

\begin{proof}
We can determine the norm of the vector:

‖ | ψ ⟩ ‖ = | a |^{2} + | b |^{2} = {| \frac{1}{3} (1 + i) |}^{2} + {| \frac{\sqrt{3.5}}{3} (1 + i) |}^{2} = \frac{1}{9} {\sqrt{2}}^{2} + \frac{3.5}{9} {\sqrt{2}}^{2} = \frac{2 + 7}{9} = 1

thus the norm is 1, so $| ψ ⟩$ is correctly normalized. To evaluate the probability of measuring the qubit in a state of $| 0 ⟩$ or $| 1 ⟩$ :

\begin{aligned} P (| 0 ⟩) & = {| \frac{1}{3} (1 + i) |}^{2} = \frac{2}{9} \\ P (| 1 ⟩) & = {| \frac{\sqrt{3.5}}{3} (1 + i) |}^{2} = \frac{7}{9} \end{aligned}

Thus it is more likely to measure the qubit in a state of $| 1 ⟩$ than $| 0 ⟩$ .
\end{proof}

5.4

Theorem

When we discussed global and relative phases, it became clear that $| 1 ⟩$ and $- | 1 ⟩$ are the same state, excepting up to a phase, which cannot be measured. Explain why the same argument does not hold good for the states $\frac{1}{\sqrt{2}} (| 0 ⟩ + | 1 ⟩)$ and $\frac{1}{\sqrt{2}} (| 0 ⟩ - | 1 ⟩)$ .

\begin{proof}
Looking back at comparing $| 1 ⟩$ and $- | 1 ⟩$ , the reason that these were the same state, barring phase, was that their probabilities are equivalent:

| ⟨ 1 | \hat{H} | - 1 ⟩ |^{2} = H^{2}

thus why we can't measure the 180 degree phase difference in their $a_{i}$ 's. Furthermore, when we measure them, we can do so relative to an eigenbasis where $| 1 ⟩$ is a basis vector, so the $- 1$ factor still means that these $| 1 ⟩$ 's are relating to the same set of eigenvectors (as any eigenvector is a scalar multiple of some normed on, which $| 1 ⟩$ and $- | 1 ⟩$ are just scalar multiples of each other).

But for this new case, notice that the negative effect doesn't occur to the whole vector, just the $| 1 ⟩$ part. If we had a flipped sign on $| 0 ⟩$ then the same argument applies. To make sure this is correct, one can consider the evolution between states:

\begin{array}{r} {| [\begin{array}{c} 2^{- 1 / 2} & 2^{- 1 / 2} \end{array}] \hat{H} [\begin{array}{c} 2^{- 1 / 2} \\ - 2^{- 1 / 2} \end{array}] |}^{2} \neq H^{2} \end{array}

\end{proof}