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This is an extension to the previous post.

We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $\mathbb{C}^2$. We denote by $\left . \left | 0 \right \rangle\right .$ and $\left . \left | 1 \right \rangle\right .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by $$ H=\begin{pmatrix} 0 &-i \\ -i &0 \end{pmatrix} $$ and assume that for $t=0$ the state of the system is just given by $\psi(t=0)=\left . \left | 0 \right \rangle\right .$. In the following, we also assume natural units in which $\hbar=1$.

Problems:

i) Compute the expectation value of a $Z$-measurement at time $t$: $\left \langle \sigma_z \right \rangle_{\psi(t)}=\left \langle \psi(t)\mid \sigma_z\psi(t) \right \rangle$, where $$ \sigma_z=\begin{pmatrix} 1 &0 \\ 0 &-1 \end{pmatrix} $$

ii) Instead of evolving the quantum states in time, we can alternatively evolve the observables according to $\sigma_z(t)=e^{iHt}\sigma_ze^{-iHt}$, called the Heisenberg evolution of $\sigma_z$. At which time should we perform our measurement in order to maximize the expectation value of $\sigma_z$?

In the previous post, I've concluded that $$ \psi(t)=\begin{pmatrix} \frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\ -\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} \end{pmatrix} $$ I am not really sure how to calculate $\left \langle \psi(t)\mid \sigma_z\psi(t) \right \rangle$. Can you help me with this part?

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Assuming your $\psi(t)$ is correct, you need to perform this: \begin{align*} \langle\psi(t)|\sigma_z\,\psi(t)\rangle &=\left\langle\left[\begin{matrix}\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\-\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} \end{matrix}\right]\Bigg|\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]\left[\begin{matrix}\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\-\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} \end{matrix}\right]\right\rangle \\ &=\left[\begin{matrix}\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}&-\frac{1}{2}e^{t}+\frac{1}{2}e^{-t} \end{matrix}\right]\left[\begin{matrix}\frac{1}{2}e^{t}+\frac{1}{2}e^{-t}\\\frac{1}{2}e^{t}-\frac{1}{2}e^{-t} \end{matrix}\right], \end{align*} and so on. You know, this would likely be a bit easier using hyperbolic trig functions: \begin{align*} \langle\psi(t)|\sigma_z\,\psi(t)\rangle &=\left\langle\left[\begin{matrix}\cosh(t)\\-\sinh(t) \end{matrix}\right]\Bigg|\left[\begin{matrix}1&0\\0&-1\end{matrix}\right]\left[\begin{matrix}\cosh(t)\\-\sinh(t) \end{matrix}\right]\right\rangle \\ &=\left[\begin{matrix}\cosh(t)&-\sinh(t) \end{matrix}\right]\left[\begin{matrix}\cosh(t)\\\sinh(t)\end{matrix}\right] \\ &=\cosh^2(t)-\sinh^2(t)\\ &=1, \end{align*} somewhat miraculously.

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  • $\begingroup$ May I know which definition of $<.|.>$ you used to calculate it? $\endgroup$ – UnknownW Jan 31 at 21:04
  • $\begingroup$ In QM, the usual inner product is that for vectors $|\psi\rangle, \; |\varphi\rangle,$ you have $\langle \varphi|\psi\rangle=|\varphi\rangle^{\dagger}\cdot|\psi\rangle$. So you take the complex conjugate transpose of the first vector, and dot it into the second. $\endgroup$ – Adrian Keister Jan 31 at 21:07
  • $\begingroup$ Thank you. Do you have time to help me with the last problem? $\endgroup$ – UnknownW Jan 31 at 21:33
  • $\begingroup$ I'm a little puzzled by the second part. The Heisenberg picture and the Schrödinger picture are equivalent, and we just calculated the expected value of $\sigma_z$ to be a constant: $1$. If it's not varying in time, I don't see how the time matters. I would just answer that the time doesn't matter. $\endgroup$ – Adrian Keister Jan 31 at 21:43
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First calculate

$$ \sigma_z |\psi(t)\rangle = \pmatrix{1 & 0 \\ 0 & -1} \pmatrix{\cosh t \\ -\sinh t } = \pmatrix{\cosh t \\ \sinh t } $$

And then

$$ \langle \psi(t)|\sigma_z |\psi(t)\rangle = \pmatrix{\cosh t & -\sinh t} \pmatrix{\cosh t \\ \sinh t } = \cosh^2 t - \sinh^2t = 1 $$

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