# Confusion about average value of $v^2$

I know that expected value of velocity means $$\langle v\rangle=\frac{d\langle x\rangle}{dt}$$ and since I know how to calculate $$\langle x\rangle$$ namely $$\langle x\rangle=\int_{-\infty}^{\infty}x|\Psi|^2dx,$$ I can compute $$\langle v\rangle$$.

Now my question is what is meant by $$\langle v^2\rangle$$? Does it mean $$\frac{d\langle x^2\rangle}{dt}$$?

• Quick comment (not quite an answer): velocities are rarely (if ever) worked with in QM. I suppose one could define a "velocity operator" $\hat{v}=\hat{p}/m$ but this isn't done very often. Commented Apr 4, 2019 at 11:35

In the context of (non-relativistic) quantum mechanics there is no velocity operator. There is instead the momentum operator whose powers you want to calculate the expectation value of, on a state $$|\alpha\rangle$$.

Thus one has, by definition: $$\langle \hat{p}^2\rangle = \langle \alpha|\hat{p}^2|\alpha\rangle$$ In order to explicitly calculate the above quantity one must know how the momentum operator acts on the state $$|\alpha\rangle$$. Usually you can express $$|\alpha\rangle = \sum_{a}c_a|\phi_{a}\rangle$$ with $$|\phi_{a}\rangle$$ being a complete set of eigenstates of the Hamiltonian, and maybe you know how the momentum operator acts on the energy eigenstates.

Otherwise, you can insert the identity operator in the bra-ket scalar product and re-write everything in terms of integrals and derivatives of the wave function, in case you have an explicit expression thereof.

No it is not the same as $$\frac{d\langle x^2\rangle}{dt}$$. This is because the derivative will become $$2\langle x \rangle \frac{d\langle x\rangle}{dt}$$ (I've basically used chain rule/ implicit differentiation). For this one, it'll be easier to work out $$\langle p^2 \rangle$$ first using the momentum operator :)

• Does it mean $$\frac{d \langle \hat{X}^2 \rangle}{dt}$$?

No. Suppose we have $$\hat{H}$$ and an observable $$\hat{A}$$ both independent of time, from the Schrodinger equation $$\frac{d}{dt}|\psi(t)\rangle=\hat{H}|\psi(t)\rangle$$ we have

$$\frac{d}{dt} \langle \psi(t)| \hat{A} |\psi(t)\rangle = \frac{\text{i}}{\hbar}\langle \psi(t)| [\hat{H},\hat{A}] |\psi(t)\rangle$$

If $$\hat{A}=\hat{X}$$ then we recover your first identity. But if $$\hat{A}=\hat{X^2}$$, you can easily compute the commutator and realize it is not $$\hat{V}^2$$.

• What is meant by $$\langle \hat{V}^2 \rangle$$?

The mathematical definition of $$\langle \hat{V}^2 \rangle$$ is given on the first line provided $$|\psi(t)\rangle$$ is normalized, and can be computed, for example, in the momentum basis as the last two lines \begin{align*} \langle \hat{V}^2 \rangle &=\langle\psi(t)| \frac{\hat{P}^2 }{m^2} |\psi(t) \rangle \\ &=\int_{-\infty}^{\infty} dp \langle\psi(t)|p\rangle \langle p| \frac{\hat{P}^2 }{m^2}|\psi(t)\rangle\\ &=\int_{-\infty}^{\infty} dp |\langle p|\psi(t)\rangle|^2 \frac{p^2}{m^2} \end{align*}

The last expression tells you what it means: If you assume a large collection ($$N \longrightarrow \infty$$) of systems each one in the state $$|\psi(t)\rangle$$ and measure the momentum (or the velocity) you can get its distribution, $$|\langle p|\psi(t)\rangle|^2$$, which then allows you to get the average of squared velocity.

Additionally this average is related to the average kinetic energy since \begin{align*} \langle \hat{T}\rangle = \frac{1}{2}m \langle \psi(t)|\hat{V}^2 |\psi(t)\rangle \end{align*} and satisfies the energy conservation. For $$\hat{H}=\frac{\hat{P}^2}{2m}+ V(\hat{X})$$ we have the average

$$$$\langle \psi(t)|\hat{H}|\psi(t)\rangle=E=\langle \psi(t)|\frac{\hat{P}^2}{2m}|\psi(t)\rangle+\langle \psi(t)|V(\hat{X})|\psi(t)\rangle$$$$