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How would I find the time evolution of the standard deviation of an operator? For example, how might I find the time evolution $\sigma_x (t)$ of the standard deviation $\sigma_x = \sqrt{ \langle \hat{x}^2 \rangle - \langle \hat{x} \rangle^2}$ of the position operator $\hat{x}$ given a state $| 0 \rangle$ representing a particle in the ground state of a harmonic oscillator?

Can I multiply the initial value $\sigma_x (0) = \sqrt{ \dfrac{\hbar}{2 m \omega} }$ by the generator of time translation $\hat{U} = \large e^{\frac{-i \hat{H} t}{\hbar}}$? I've also tried switching to the Heisenberg picture and applying the Heisenberg equations of motion but have been unable to reach a conclusion. Any help would be greatly appreciated, thanks.

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3 Answers 3

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Adding to @Vadim answer's: note that when a system is in an eigenstate of the Hamiltonian, it is stationary. If an operator does not have an explicit time dependence, its expectation value will be constant. This is true regardless of the details of the Hamiltonian or the operator.

To see this, you can note that if the state is an eigenstate $H|\psi\rangle = E |\psi\rangle$ then $|\psi(t)\rangle = \exp(-iEt/\hbar) |\psi\rangle$, meaning that for any observable $$ \langle \hat{O}(t) \rangle = \langle \psi(t) | \hat{O} | \psi(t) \rangle = \langle \psi | \hat{O} | \psi \rangle$$ independent of time.

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In case you're planning to actually compute it, here's how:

Derivatives commute with operator averages in the Heisenberg picture (since your state is constant; alternatively in the Schrodinger picture, you can take derivatives of the state), so you can do something like this:

$$\frac{d}{dt}\sqrt{\langle x^2\rangle-\langle x\rangle^2}=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\frac{d}{dt}(\langle x^2\rangle-\langle x\rangle^2)$$ $$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\left(\langle \frac{d}{dt} x^2\rangle-\frac{d}{dt}\langle x\rangle^2\right)$$ $$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\left(\langle \frac{d}{dt} x^2\rangle-2\langle x\rangle\frac{d}{dt}\langle x\rangle\right)$$

Notice the product rule for operator derivatives is not commutative

$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\left(\langle \frac{dx}{dt} x\rangle+\langle x\frac{dx}{dt}\rangle-2\langle x\rangle\langle \frac{dx}{dt}\rangle\right).$$

Then plug in Heisenberg's equation $\frac{dx}{dt}=\frac{i}{\hbar}[H,x]$,

$$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\frac{i}{\hbar}\left(\langle [H,x] x\rangle+\langle x[H,x]\rangle-2\langle x\rangle\langle[H,x]\rangle\right)$$ $$=\frac{1}{2\sqrt{\langle x^2\rangle-\langle x\rangle^2}}\frac{i}{\hbar}\left(\langle [H,x^2]\rangle-2\langle x\rangle\langle[H,x]\rangle\right)$$

And some further operator bashing will give you the answer. Alternatively, you can do this in the Schrodinger picture directly. Here, for example, you can evaluate derivatives like so:

$$i\hbar\frac{d}{dt}\langle\psi\rvert x^2\lvert \psi\rangle=i\hbar\left(\frac{d}{dt}\langle\psi\rvert\right) x^2\lvert \psi\rangle+i\hbar\langle\psi\rvert x^2\frac{d}{dt}\lvert \psi\rangle=i\hbar\left(\frac{d}{dt}\lvert \psi\rangle\right)^\dagger x^2\lvert \psi\rangle+\langle\psi\rvert x^2H\lvert \psi\rangle$$ $$=-\langle \psi\lvert H x^2\lvert \psi\rangle+\langle\psi\rvert x^2H\lvert \psi\rangle=-\langle \psi\lvert [H, x^2]\lvert \psi\rangle$$

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The standard deviation in question is not an operator. One could find the time evolution of $\langle\hat{x}\rangle$ and $\langle\hat{x}^2\rangle$ and then calculate the time-dependent standard deviation. How one calculates them depends on whether one uses the Schrödinger or the Heisenberg picture.

In Schrödinger picture the wave function carries time dependence, determined by the Schrödinger equation. Then the time-dependent average of an operator can be calculated as: $$\langle \hat{O}\rangle(t) = \int dx \Psi^*(x,t)\hat{O}\Psi(x,t).$$

In the Heisenberg picture the wave function is time-independent, whereas the time evolution of the operators can be found from the Heisenberg equations of motion: $$\frac{d}{dt}\hat{O}(t) = \frac{1}{i\hbar}\left[\hat{O}(t), \hat{H}\right],$$ which are identical with the Hamiltonian equations of motion and can be easily solved for simple cases, e.g., for a harmonic oscillator (keeping in mind that the constants of integration are the operators). The time-dependent averages are then found as $$\langle \hat{O}(t)\rangle = \int dx \Psi^*(x)\hat{O}(t)\Psi(x).$$

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