Finding a density matrix for a distribution of pure states

Question

Let $\theta$ be a Gaussian variable with mean 0 and variance 1. Then for $t>0$, the variable $\theta \sqrt{t}$ is also Gaussian with mean $0$ and variance $t$. Let $|\psi_0\rangle$ be an arbitrary state on a Bloch sphere. Define $$|\psi(t)\rangle := R_x(\theta \sqrt{t}) |\psi_0\rangle$$ where $R_x(w) := \cos(w)I -i \sin(w)X$ is a rotation around $x$-axis of a Bloch sphere.

Question: I want to find the density matrix $\rho(t)$ that describes the distribution of all $|\psi(t)\rangle$.

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My initial thought was that the correct approach would involve computing the expectation $$\mathbb{E}_{\theta} [|\psi(t)\rangle \langle \psi(t)|],$$ but this becomes cumbersome as I need to compute expectations of $\cos^2$ and $\sin^2$.

Instead, I considered a more straightforward, though perhaps incorrect, approach. First, compute the expectation of $|\psi(t)\rangle$ and then use that result to construct the density matrix.

To this end, the averaged $|\psi(t)\rangle$ over all $\theta \sqrt{t} \sim N(0, t)$ is \begin{align} \mathbb{E}[|\psi(t)\rangle] &= \mathbb{E}\left[\cos \left(\theta \sqrt{t}\right) \right] |\psi_0\rangle - i \mathbb{E} \left[\sin \left (\theta \sqrt{t} \right) \right]X|\psi_0\rangle \\ &= e^{-\frac{t}{2}}|\psi_0\rangle - i 0 X |\psi_0\rangle \\ &= e^{-\frac{t}{2}}|\psi_0\rangle. \end{align} So, on average, $|\psi(t)\rangle$ is $e^{-t/2}|\psi_0\rangle$. The derivation for $\mathbb{E}[\cos{\theta \sqrt{t}}]$ can be found here, and $\mathbb{E}[\sin(\theta \sqrt{t})]=0$ because $\sin(x)$ is odd symmetric.

To make it a density operator, I could try writing $$\rho(t) = e^{-t}|\psi_0\rangle \langle \psi_0|,$$ but then $\text{Tr}[\rho(t)] \neq 1$ for $t>0$. A sensible guess would be $$\rho(t) = e^{-t}|\psi_0\rangle\langle \psi_0| + (1- e^{-t})X |\psi_0\rangle\langle\psi_0|X. \tag{1}$$

The issue is that this is just a conjecture. I don’t have a formal proof (or a sound argument) for (1), nor am I sure this approach is valid.

Answer 1

It's a bit tedious, but still doable.

First of all, without loss of generality, we can write: $$\newcommand\ket[1]{\left|#1\right\rangle}\ket{\psi_0}=\begin{pmatrix}\alpha\\\beta\mathrm{e}^{\mathrm{i}\varphi}\end{pmatrix}$$ with $\alpha\geqslant0$, $\beta\geqslant0$ and $\varphi\in[0,2\pi)$. Let us fix $t$. For a given $\theta$, we have: $$\ket{\psi(t)}=\begin{pmatrix}\cos(\theta\sqrt{t})&-\mathrm{i}\sin(\theta\sqrt{t})\\-\mathrm{i}\sin(\theta\sqrt{t})&\cos(\theta\sqrt{t})\end{pmatrix}\begin{pmatrix}\alpha\\\beta\mathrm{e}^{\mathrm{i}\varphi}\end{pmatrix}=\begin{pmatrix}\alpha\cos(\theta\sqrt{t})-\mathrm{i}\beta\sin(\theta\sqrt{t})\mathrm{e}^{\mathrm{i}\varphi}\\\beta\cos(\theta\sqrt{t})\mathrm{e}^{\mathrm{i}\varphi}-\mathrm{i}\alpha\sin(\theta\sqrt{t})\end{pmatrix}\,.$$ Using the fact that we want to compute a density matrix, the $|1\rangle\!\langle0|$ coefficient is the complex conjugate of the $|0\rangle\!\langle1|$ coefficient, and the $|1\rangle\!\langle1|$ coefficient is equal to $1$ minus the $|0\rangle\!\langle0|$ coefficient. Thus, we only have to care about these two coefficients to determine the density matrix.

Furthermore, because of linearity we can afford to consider each coefficient separately and compute its expectation value.

The $|0\rangle\!\langle0|$ coefficient is the squared modulus of the $\ket{0}$ coefficient of the statevector, that is: $$\begin{align*}&\left(\alpha\cos(\theta\sqrt{t})+\beta\sin(\theta\sqrt{t})\sin(\varphi)\right)^2+\beta^2\sin^2(\theta\sqrt{t})\cos^2(\varphi)\\={}&\alpha^2\cos^2(\theta\sqrt{t})+\beta^2\sin^2(\theta\sqrt{t})+\alpha\beta\sin(\varphi)\sin(2\theta\sqrt{t})\,.\end{align*}$$ We now want to compute the expectation value over $\theta$ of this expression. As you mentioned, the rightmost term is going to be nil, since $\sin$ is an odd function. Furthermore, since $\cos^2+\sin^2=1$, finding the expectation value of $\cos^2(\theta\sqrt{t})$ is enough to find that of $\sin^2(\theta\sqrt{t})$. That is, we want to compute: $$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\cos^2(\theta\sqrt{t})\mathrm{e}^{-\frac{\theta^2}{2}}\,\mathrm{d}\theta$$ Since I don't want to deal with this monstrosity (I mean, who would?), let us leave Wolfram Alpha do its thing: $$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\cos^2(\theta\sqrt{t})\mathrm{e}^{-\frac{\theta^2}{2}}\,\mathrm{d}\theta=\mathrm{e}^{-t}\cosh(t)=\frac{1+\mathrm{e}^{-2t}}{2}$$ from which we deduce: $$\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}\sin^2(\theta\sqrt{t})\mathrm{e}^{-\frac{\theta^2}{2}}\,\mathrm{d}\theta=\frac{1-\mathrm{e}^{-2t}}{2}\,.$$ All in all, the $|0\rangle\!\langle0|$ term of the density matrix is: $$\frac12+\frac{\alpha^2-\beta^2}{2}\mathrm{e}^{-2t}\,.$$ We now move on to the $|0\rangle\!\langle1|$ coefficient. Its expression is: $$\begin{align*} &\left(\alpha\cos(\theta\sqrt{t})-\mathrm{i}\beta\sin(\theta\sqrt{t})\mathrm{e}^{\mathrm{i}\varphi}\right)\left(\beta\cos(\theta\sqrt{t})\mathrm{e}^{-\mathrm{i}\varphi}+\mathrm{i}\alpha\sin(\theta\sqrt{t})\right)\\ ={}&\alpha\beta\cos^2(\theta\sqrt{t})\mathrm{e}^{-\mathrm{i}\varphi}+\mathrm{i}\alpha^2\frac{\sin(2\theta\sqrt{t})}{2}-\mathrm{i}\beta^2\frac{\sin(2\theta\sqrt{t})}{2}+\alpha\beta\sin^2(\theta\sqrt{t})\mathrm{e}^{\mathrm{i}\varphi} \end{align*}$$ We can reuse the computations we just made for $|0\rangle\!\langle0|$, which gives us an expectation value of: $$\alpha\beta\left(\cos(\varphi)-\mathrm{i}\mathrm{e}^{-2t}\sin(\varphi)\right)\,.$$ All in all, the density matrix you're looking for is: $$\rho(t)=\begin{pmatrix}\frac12+\frac{\alpha^2-\beta^2}{2}\mathrm{e}^{-2t}&\alpha\beta\left(\cos(\varphi)-\mathrm{i}\mathrm{e}^{-2t}\sin(\varphi)\right)\\\alpha\beta\left(\cos(\varphi)+\mathrm{i}\mathrm{e}^{-2t}\sin(\varphi)\right)&\frac12+\frac{\beta^2-\alpha^2}{2}\mathrm{e}^{-2t}\end{pmatrix}\,.$$

Alternatively, a much simpler proof is using the matrix representation from the beginning as you've started. Using the same arguments, we find that the final expression is in fact: $$\frac{1+\mathrm{e}^{-2t}}{2}\left|\psi_0\right\rangle\!\left\langle\psi_0\right|+\frac{1-\mathrm{e}^{-2t}}{2}X\left|\psi_0\right\rangle\!\left\langle\psi_0\right|X\,.$$