<a href="https://qworld.net" target="_blank" align="left"></a>
$$
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\abs}[1]{\left\lvert#1\right\rvert}
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\newcommand{\inner}[2]{\left\langle#1,#2\right\rangle}
\newcommand{\bra}[1]{\left\langle#1\right|}
\newcommand{\ket}[1]{\left|#1\right\rangle}
\newcommand{\braket}[2]{\left\langle#1|#2\right\rangle}
\newcommand{\ketbra}[2]{\left|#1\right\rangle\left\langle#2\right|}
\newcommand{\angleset}[1]{\left\langle#1\right\rangle}
\newcommand{\expected}[1]{\left\langle#1\right\rangle}
\newcommand{\dv}[2]{\frac{d#1}{d#2}}
\newcommand{\real}[0]{\mathfrak{Re}}
$$

# Worked Example

_prepared by Israel Gelover_

Let

\begin{equation*}
    \begin{split}
        B &= \set{\ket{0}, \ket{1}, \ket{2}} \enspace \text{(orthonormal base of } {}_\mathbb{C}\mathbb{C}^3)\\
        \hat{H} &= \ketbra{0}{0} + i\ketbra{1}{2} - i\ketbra{2}{1} \enspace \text{(hamiltonian)} \\
        \hat{A} &= \ketbra{0}{0} + \ketbra{1}{1} + 3\ketbra{2}{2} - 2\ketbra{0}{1} - 2\ketbra{1}{0} \enspace \text{(observable)} \\
        \ket{\psi(0)} &= N(\ket{0} - 2\ket{1} + i\ket{2}) \enspace \text{(initial state)}
    \end{split}
\end{equation*}

where an observable corresponds to a measurable physical magnitude of the system being modeled and $N$ is the normalization coefficient. These observables are self-adjoint operators and the Hamiltonian is one of those observables that corresponds to the energy of the system.

Here, we want to know

1. What is the value of $N$?
2. What are the possible values that we can get from measuring $\hat{A}$ and with what probability can we get them at time $t=0$?
3. What is the value of $\ket{\psi(t)}$ at an arbitrary time $t$?
4. What is the probability $P(a_i)_t$ of measuring each eigenvalue $a_i$ of $\hat{A}$ at time $t$?
5. If when measuring $\hat{A}$ at time $t=0$ we get the eigenvalue $a_2$ (or $a_3$), what is the status of the system immediately after the measurement?

### <a name="1">Normalization constant</a>

1. What is the value of $N$?

Let us recall that by <a href="./Postulates.ipynb#definition_3_1">Postulate I</a> the states of the system are unit vectors and furthermore they are invariant with respcet to global phases. Thus

\begin{equation*}
    \begin{split}
        1 &= \braket{\psi(0)}{\psi(0)} \\
        &= N^*(\bra{0} - 2\bra{1} - i\bra{2}) N(\ket{0} - 2\ket{1} + i\ket{2}) \\
        &= N^*N(\braket{0}{0} + 4\braket{1}{1} + \braket{2}{2}) \\
        &= \abs{N}^2(1+4+1) \\
        &= 6\abs{N}^2 \\
        \implies \abs{N}^2 &= \frac{1}{6} \\
        \implies N &= \frac{1}{\sqrt{6}}
    \end{split}
\end{equation*}

Since by the global phase invariance we can think that $N \in \mathbb{R}$. Therefore

\begin{equation}\label{fi_t0}
    \ket{\psi(0)} = \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2})
\end{equation}

### <a name="2">Observables and their probabilities at the initial state</a>

2. What are the possible values that we can get from measuring $\hat{A}$ and with what probability can we get them at time $t=0$?

Recall that by <a href="./Postulates.ipynb#definition_3_1">Postulate IV</a> the observables are self-adjoint operators and the results of measurements of that operator correspond to eigenvalues of the operator.

Since $\hat{A} = \ketbra{0} + \ketbra{1} + 3\ketbra{2} - 2\ketbra{0}{1} - 2\ketbra{1}{0}$, we can view it in matrix form to calculate its eigenvalues more easily, using the method described <a href="../2-math/SpectralTheory.ipynb#examples">here</a>

\begin{equation*}
    \hat{A} = \begin{pmatrix}
        1 & -2 & 0 \\
        -2 & 1 & 0 \\
        0 & 0 & 3
    \end{pmatrix}
\end{equation*}

It is clear that $\hat{A}$ is self-adjoint since $A = A^\dagger = (A^\top)^*$, so the associated eigenvalues and eigenvectors are:

\begin{equation*}
    \begin{split}
        a_1 = -1 \qquad&\qquad \ket{a_1} = \frac{1}{\sqrt{2}}(\ket{0} + \ket{1}) \\
        a_2 = 3 \qquad&\qquad \ket{a_2} = \frac{1}{\sqrt{2}}(\ket{0} - \ket{1}) \\
        a_3 = 3 \qquad&\qquad \ket{a_3} = \ket{2}
    \end{split}
\end{equation*}

From this, the possible values that we can get by measuring $\hat{A}$ are $-1$ and $3$. Now, by <a href="./Postulates.ipynb#definition_3_1">Postulate V</a>:

- Since $a_1$ has multiplicity $1$, from the calculation for the normalization constant we have that

\begin{equation}\label{prob_a1}
    \begin{split}
        P(a_1)_{t=0} &= \abs{\braket{a_1}{\psi(0)}}^2 \\
        &= \abs{\frac{1}{\sqrt{2}}(\bra{0} + \bra{1}) \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2})}^2 \\
        &= \abs{\frac{1}{\sqrt{12}}(\braket{0}{0} - 2\braket{1}{1})}^2 \\
        &= \abs{\frac{-1}{\sqrt{12}}}^2 = \frac{1}{12}
    \end{split}
\end{equation}

- Since eigenvalue $3$ has multiplicity $2$ we have

\begin{equation}\label{prob_a2}
    \begin{split}
        P(a_2)_{t=0} = P(a_3)_{t=0} &= \abs{\braket{a_2}{\psi(0)}}^2 + \abs{\braket{a_3}{\psi(0)}}^2 \\
        &= \abs{\frac{1}{\sqrt{2}}(\bra{0} - \bra{1})\frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2})}^2 \\
        &+ \abs{\bra{2} \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2})}^2 \\
        &= \abs{\frac{1}{\sqrt{12}}(\braket{0}{0} + 2\braket{1}{1})}^2 + \abs{\frac{1}{\sqrt{6}}i\braket{2}{2}}^2 \\
        &= \abs{\frac{3}{\sqrt{12}}}^2 + \abs{\frac{i}{\sqrt{6}}}^2 \\
        &= \frac{9}{12} + \frac{1}{6} = \frac{11}{12}
    \end{split}
\end{equation}

Note that since we only have two different eigenvalues: $P(a_2)_{t=0} = P(a_3)_{t=0} = 1 - P(a_1)_{t=0}$.

### <a name="3">Time evolution</a>

3. What is the value of $\ket{\psi(t)}$ at an arbitrary time $t$?

Let us remember that by <a href="./Postulates.ipynb#definition_3_1">Postulate VII</a> and <a href="./TimeEvolution.ipynb#remark_3_4">Remark 3.4</a>, we need to apply the time evolution operator. Therefore we can write $\ket{\psi(t)}$ at time $t=0$ as

\begin{equation*}
    \ket{\psi(t)} = e^{-\frac{i}{\hbar}t\hat{H}}\ket{\psi(0)}
\end{equation*}

We also know that

\begin{equation*}
    \hat{H} = \ketbra{0} + i\ketbra{1}{2} - i\ketbra{2}{1}
\end{equation*}

Let's write $\hat{H}$ in its matrix form to calculate its eigenvalues and eigenvectors

\begin{equation*}
    \begin{pmatrix}
        1 & 0 & 0 \\
        0 & 0 & i \\
        0 &-i & 0
    \end{pmatrix}
\end{equation*}

In this way, it is clear that $\hat{H}$ is self-adjoint since $\hat{H} = (\hat{H}^\top)^*$, so the associated eigenvalues and eigenvectors are:

\begin{equation*}
    \begin{split}
        \varepsilon_1 = 1 \qquad&\qquad \ket{\varepsilon_1} = \ket{0} \\
        \varepsilon_2 = 1 \qquad&\qquad \ket{\varepsilon_2} = \frac{1}{\sqrt{2}}(\ket{1} + i\ket{2}) \\
        \varepsilon_3 = -1 \qquad&\qquad \ket{\varepsilon_3} = \frac{1}{\sqrt{2}}(\ket{1} - i\ket{2})
    \end{split}
\end{equation*}

Note that $E=\set{\ket{\varepsilon_1}, \ket{\varepsilon_2}, \ket{\varepsilon_3}}$ is also an orthonormal basis for the space since $\hat{H}$ is self-adjoint.

Now, in order to apply the time evolution operator we need to express $\ket{\psi(0)}$ in terms of the base $E$, that is, in terms of the eigenvectors of $\hat{H}$. For this, let's consider that by <a href="../2-math/SpectralTheory.ipynb#definition_2_21">Definition 2.21</a> we have that

\begin{equation*}
    \sum_{j=1}^{3}\ketbra{\varepsilon_j}{\varepsilon_j} = I
\end{equation*}

So, if we apply this form of the identity operator to $\ket{\psi(0)}$ we have that

\begin{equation*}
    \ket{\psi(0)} = \sum_{j=1}^{3}\ket{\varepsilon_j}\braket{\varepsilon_j}{\psi(0)}
\end{equation*}

Thus, we can express $\ket{\psi(0)}$ in terms of the base $E$

\begin{equation*}
    \begin{split}
        \braket{\varepsilon_1}{\psi(0)} &= \bra{0} \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2}) \\
        &= \frac{1}{\sqrt{6}}(\braket{0}{0}) = \frac{1}{\sqrt{6}} \\
        \braket{\varepsilon_2}{\psi(0)} &= \frac{1}{\sqrt{2}}(\bra{1} + i\bra{2}) \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2}) \\
        &= \frac{1}{\sqrt{12}}(-2\braket{1}{1} - \braket{2}{2}) = \frac{-3}{\sqrt{12}} \\
        \braket{\varepsilon_3}{\psi(0)} &= \frac{1}{\sqrt{2}}(\bra{1} - i\bra{2}) \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2}) \\
        &= \frac{1}{\sqrt{12}}(-2\braket{1}{1} + \braket{2}{2}) = \frac{-1}{\sqrt{12}} \\
        \implies \ket{\psi(0)} &= \frac{1}{\sqrt{6}}\ket{\varepsilon_1} - \frac{3}{\sqrt{12}}\ket{\varepsilon_2} - \frac{1}{\sqrt{12}}\ket{\varepsilon_3}
    \end{split}
\end{equation*}

From this expression we can see that $\ket{\psi(0)}$ is unitary, and with $\ket{\psi(0)}$ expressed as a linear combination of the eigenvectors of $\hat{H}$ we can apply the time evolution operator as we obtained it in <a href="./TimeEvolution.ipynb#remark_3_4">Remark 3.4</a> and in this way we have

\begin{equation*}
    \begin{split}
        \ket{\psi(t)} &= e^{-\frac{i}{\hbar}t\hat{H}}\left(\frac{1}{\sqrt{6}}\ket{\varepsilon_1} - \frac{3}{\sqrt{12}}\ket{\varepsilon_2} - \frac{1}{\sqrt{12}}\ket{\varepsilon_3}\right) \\
        &= \frac{1}{\sqrt{6}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{\varepsilon_1} - \frac{3}{\sqrt{12}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{\varepsilon_2} - \frac{1}{\sqrt{12}}e^{-\frac{i}{\hbar}t\hat{H}}\ket{\varepsilon_3}
    \end{split}
\end{equation*}

Now, if we consider $f(x) = e^x$, from <a href="./TimeEvolution.ipynb#proposition_3_5">Proposition 3.5</a> we have that $e^{\hat{H}}\ket{\varepsilon_j} = e^{\varepsilon_j}\ket{\varepsilon_j}, j \in \set{1,2,3}$. Therefore

\begin{equation}\label{estado_t}
    \ket{\psi(t)} = \frac{1}{\sqrt{6}}e^{-\frac{i}{\hbar}t}\ket{\varepsilon_1} - \frac{3}{\sqrt{12}}e^{-\frac{i}{\hbar}t}\ket{\varepsilon_2} - \frac{1}{\sqrt{12}}e^{\frac{i}{\hbar}t}\ket{\varepsilon_3}
\end{equation}

If necessary, we can rewrite $\ket{\psi(t)}$ in terms of base $B$, but this expression that we obtained already answers the question asked.

### <a name="4">Observables and their probabilities at an arbitrary time</a>

4. What is the probability $P(a_i)_t$ of measuring each eigenvalue $a_i$ of $\hat{A}$ at time $t$?

By <a href="./Postulates.ipynb#definition_3_1">Postulate V</a> and the calculation we just did for $\ket{\psi(t)}$ we know that

\begin{equation*}
    \begin{split}
        P(a_1)_t &= \abs{\braket{a_1}{\psi(t)}}^2 \\
        &= \abs{\bra{a_1}\left(\frac{1}{\sqrt{6}}e^{-\frac{i}{\hbar}t}\ket{\varepsilon_1} - \frac{3}{\sqrt{12}}e^{-\frac{i}{\hbar}t}\ket{\varepsilon_2} - \frac{1}{\sqrt{12}}e^{\frac{i}{\hbar}t}\ket{\varepsilon_3}\right)}^2 \\
        &= \abs{\frac{1}{\sqrt{6}}e^{-\frac{i}{\hbar}t}\braket{a_1}{\varepsilon_1} - \frac{3}{\sqrt{12}}e^{-\frac{i}{\hbar}t}\braket{a_1}{\varepsilon_2} - \frac{1}{\sqrt{12}}e^{\frac{i}{\hbar}t}\braket{a_1}{\varepsilon_3}}^2 \\
    \end{split}
\end{equation*}

Calculating the inner products

\begin{equation*}
    \begin{split}
        \braket{a_1}{\varepsilon_1} &= \frac{1}{\sqrt{2}}(\bra{0} + \bra{1}) \ket{0} = \frac{1}{\sqrt{2}}(\braket{0}{0}) = \frac{1}{\sqrt{2}} \\
        \braket{a_1}{\varepsilon_2} &= \frac{1}{\sqrt{2}}(\bra{0} + \bra{1}) \frac{1}{\sqrt{2}}(\ket{1} + i\ket{2}) = \frac{1}{2}\braket{1}{1} = \frac{1}{2} \\
        \braket{a_1}{\varepsilon_3} &= \frac{1}{\sqrt{2}}(\bra{0} + \bra{1}) \frac{1}{\sqrt{2}}(\ket{1} - i\ket{2}) = \frac{1}{2}\braket{1}{1} = \frac{1}{2}
    \end{split}
\end{equation*}
$\implies$
\begin{equation*}
    \begin{split}
        P(a_1)_t &= \abs{\frac{1}{\sqrt{6}}\frac{1}{\sqrt{2}}e^{-\frac{i}{\hbar}t} - \frac{3}{\sqrt{12}}\frac{1}{2}e^{-\frac{i}{\hbar}t} - \frac{1}{\sqrt{12}}\frac{1}{2}e^{\frac{i}{\hbar}t})}^2 \\
        &= \abs{\frac{1}{\sqrt{12}}e^{-\frac{i}{\hbar}t} - \frac{3}{2\sqrt{12}}e^{-\frac{i}{\hbar}t} - \frac{1}{2\sqrt{12}}e^{\frac{i}{\hbar}t})}^2 \\
        &= \abs{\underbrace{-\frac{1}{2\sqrt{12}}e^{-\frac{i}{\hbar}t}}_{z} \underbrace{-\frac{1}{2\sqrt{12}}e^{\frac{i}{\hbar}t}}_{z^*}}^2 = \abs{\underbrace{-\frac{1}{\sqrt{12}}\cos\left(\frac{t}{\hbar}\right)}_{2\real(z)}}^2 \\
        &= \frac{1}{12}\cos^2\left(\frac{t}{\hbar}\right)
    \end{split}
\end{equation*}

Since $z + z^* = 2\real(z) \forall z \in \mathbb{C}$, that is

\begin{equation*}
    z = -\frac{1}{2\sqrt{12}}e^{-i\frac{t}{\hbar}} = -\frac{1}{2\sqrt{12}}\left(\cos\left(-\frac{t}{\hbar}\right) + i\sin\left(-\frac{t}{\hbar}\right)\right) \implies 2\real(z) = -\frac{1}{\sqrt{12}}\cos\left(\frac{t}{\hbar}\right)
\end{equation*}

since cosine is an even function. Therefore

\begin{equation}\label{pa1}
    P(a_1)_t = \frac{1}{12}\cos^2\left(\frac{t}{\hbar}\right)
\end{equation}

Which makes sense with what we find in <a href="#2">Observables and their probabilities at the initial state</a>, because if in this last equation we substitute $t=0$ we have

\begin{equation*}
    P(a_1)_{t=0} = \frac{1}{12}\cos^2(0) = \frac{1}{12}
\end{equation*}

We can also observe that for certain values of $t$ this probability will be $0$ and the maximum value that it will have is precisely $\frac{1}{12}$.

On the other hand, since we are calculating the probability of two different eigenvalues

\begin{equation*}
    P(a_2)_t = P(a_3)_t = 1-P(a_1)_t = 1 - \frac{1}{12}cos^2\left(\frac{t}{\hbar}\right) = \frac{11}{12} + \frac{1}{12}\sin^2\left(\frac{t}{\hbar}\right)
\end{equation*}

Which is also congruent with what we found in <a href="#2">Observables and their probabilities at the initial state</a>, because if we substitute $t=0$ we have

\begin{equation*}
    P(a_2)_{t=0} = \frac{11}{12} + \frac{1}{12}\sin^2(0) = \frac{11}{12}
\end{equation*}

Note that this is the minimum value that it can take and for certain values of $t$ this probability can be $1$.

### <a name="5">Wave function collapse</a>

5. If when measuring $\hat{A}$ at time $t=0$ we get the eigenvalue $a_2$ (or $a_3$), what is the status of the system immediately after the measurement?

By <a href="./Postulates.ipynb#definition_3_1">Postulate VI</a> we must calculate the normalized projection to the subspace generated by the eigenvectors associated with the eigenvalue. From the calculation of the <a href="#1">Normalization constant</a>, at time $t=0$ we know that

\begin{equation*}
    \ket{\psi(0)} = \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2})
\end{equation*}

We have to project this vector to the subspace generated by $\set{\ket{a_2}, \ket{a_3}}$ using the projection operator

\begin{equation*}
    \hat{P} = \ketbra{a_2}{a_2} + \ketbra{a_3}{a_3}
\end{equation*}

Which is a particular case of the <a href="../2-math/SpectralTheory.ipynb#definition_2_21">Definition 2.21</a>, since in the Completeness Relation we project over the entire space or rather, over the vectors that make up an orthonormal basis of the space, hence obtain the identity, in this case we are interested in doing it only on the eigenvectors associated with the measured eigenvalue.
     After that, we need to renormalize the resulting vector since the projection can change the size of the vector. In short, what we need to do is
     
\begin{equation*}
    \ket{\psi(0)} \xrightarrow{\text{after measurement}} \frac{\hat{P}\ket{\psi(0)}}{\sqrt{P(a_2)_{t=0}}}
\end{equation*}

That is

\begin{equation*}
    \hat{P}\ket{\psi(0)} = \ket{a_2}\braket{a_2}{\psi(0)} + \ket{a_3}\braket{a_3}{\psi(0)}
\end{equation*}

Calculating

\begin{equation*}
    \begin{split}
        \braket{a_2}{\psi(0)} &= \frac{1}{\sqrt{2}}(\bra{0} - \bra{1}) \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2}) \\
        &= \frac{1}{\sqrt{12}}(\braket{0}{0} + 2\braket{1}{1}) = \frac{3}{\sqrt{12}} \\
        \braket{a_3}{\psi(0)} &= \bra{2} \frac{1}{\sqrt{6}}(\ket{0} - 2\ket{1} + i\ket{2}) \\
        &= \frac{1}{\sqrt{6}}(i\braket{2}{2}) = \frac{i}{\sqrt{6}}
    \end{split}
\end{equation*}

Where

\begin{equation*}
    \hat{P}\ket{\psi(0)} = \frac{3}{\sqrt{12}}\ket{a_2} + \frac{i}{\sqrt{6}}\ket{a_3}
\end{equation*}

Renormalizing

\begin{equation*}
    \begin{split}
        \frac{\hat{P}\ket{\psi(0)}}{\sqrt{P(a_2)_{t=0}}} &= \frac{\hat{P}\ket{\psi(0)}}{\sqrt{\frac{11}{12}}} \\
        &= \sqrt{\frac{12}{11}}\left(\frac{3}{\sqrt{12}}\ket{a_2} + \frac{i}{\sqrt{6}}\ket{a_3}\right) \\
        &= \frac{\sqrt{12}}{\sqrt{11}}\frac{3}{\sqrt{12}}\ket{a_2} + \frac{\sqrt{2}\sqrt{6}}{\sqrt{11}}\frac{i}{\sqrt{6}}\ket{a_3} \\
        &= \frac{3}{\sqrt{11}}\ket{a_2} + \frac{i\sqrt{2}}{\sqrt{11}}\ket{a_3}
        \end{split}
\end{equation*}

Therefore the state of the system immediately after having measured the eigenvalue $a_2$ (or $a_3$) is

\begin{equation*}
    \frac{3}{\sqrt{11}}\ket{a_2} + \frac{i\sqrt{2}}{\sqrt{11}}\ket{a_3}
\end{equation*}

### Task 1.

A physical system is described at time $t=0$ by the state

\begin{equation*}
    \ket{\psi(0)} = \frac{1}{\sqrt{2}}(\ket{01} + i\ket{10}) \in \mathbb{B}^{\otimes 2}
\end{equation*}

The evolution of the system is given by the Hamiltonian

\begin{equation*}
    \mathcal{H} = \varepsilon(\sigma_x \otimes \sigma_z), \quad \varepsilon \in \mathbb{R}
\end{equation*}

Let us consider the observable

\begin{equation*}
    \mathcal{A} = a(\sigma_y \otimes \sigma_x), \quad a \in \mathbb{R}^+
\end{equation*}

What are the possible values that can be obtained by measuring $\mathcal{A}$?

<a href="./WorkedExample_Solutions.ipynb#task_1">Click here for solution</a>

### Task 2.

Find the probability of measuring  at time $t=0$ each of the values found in **Task 1**.

<a href="./WorkedExample_Solutions.ipynb#task_2">Click here for solution</a>

### Task 3.

Find the probability of measuring  at an arbitrary time $t$ each of the values found in **Task 1**.

<a href="./WorkedExample_Solutions.ipynb#task_3">Click here for solution</a>

### Task 4.

If at time $t=0 $ the observable $\mathcal{A}$ is measured and the positive value is obtained, what is the state of the system immediately after the measurement?

<a href="./WorkedExample_Solutions.ipynb#task_4">Click here for solution</a>

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ipynb
Jupyter Notebook
3-qm/WorkedExample.ipynb
igelover/math-intro-to-qc
4ad535402cb6ccd619c4e4fd5c257ae8c6ce76d3
MIT
2022-01-18T00:26:27.000Z
2022-01-18T00:26:27.000Z
3-qm/WorkedExample.ipynb
igelover/math-intro-to-qc
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MIT
3-qm/WorkedExample.ipynb
igelover/math-intro-to-qc
4ad535402cb6ccd619c4e4fd5c257ae8c6ce76d3
MIT
Qwen/Qwen-72B
1. YES
2. YES