QM  calculating expectation value for velocity of an electron
Calculating the energy eigenvalue, will give you $\langle v^2 \rangle$. This is how it's done:
$$\langle v^2 \rangle = \frac{2}{m}\langle T \rangle=\frac{2}{m}\langle \psi\hat T\psi\rangle=\frac{2}{m}\int\psi^*(x)\hat T \psi(x)dx$$
Where you should write $T$(the kinematic energy) as an operator. This can be done by writing it as a function of $x$ and $p$, and then replacing $p$ with its operator.
However this is not what we call the expectation value of speed. To calculate the expectation value of speed, we calculate the expectation value of its momentum:
$$\langle v \rangle = \frac{\langle p\rangle}{m}=\frac{1}{m}\int \psi^*(x)p\psi(x)dx$$
Which can be calculated either by transforming $\psi$ to the momentum space, or replacing $p$ with its operator $\mathfrak i \hbar\frac{\partial}{\partial x}$.
An important thing to note is: $\langle v^2 \rangle \ne \langle v\rangle^2$ in general. E.g. consider the symmetric harmonic potential where $\langle v \rangle=0$ but $\langle v^2 \rangle > 0$. The difference is called variance $\sigma_v^2=\langle v^2 \rangle\langle v \rangle^2$.
71GA
Updated on August 08, 2020Comments

71GA almost 2 years
How do we calculate the expectation value for speed? I have heard that we must first calculate the expectaion value for kinetic energy. Someone please explain a bit what options do we have.

71GA almost 9 yearsWell i asked this because I cant find an anwser in my book which only describes operators of kinetic energy, potential energy, momentum, position and position square... If i write expectation value for speed, Google returns mostly these same operators...

71GA almost 9 yearsNever mind I found it in this PDF  page 32  Now give me some downvotes for being lazy :D

71GA almost 9 yearsI have discovered that in this PDF they diferentiate the wavefunction over time but i have a stationary state! Do i have to multiply it with $\exp\frac{i}{\hbar E t}$ and then diferentiate it? Is there any easier way if i already calculated expectation value for kinetic energy $\langle T \rangle$?

Ruslan almost 9 yearsYou can't derive a vector ($\vec v$) from its square ($T\sim v^2$)

71GA almost 9 yearsSo this is not possible? $\langle E_k\rangle = \tfrac{1}{2}m\langle v \rangle^2 \longrightarrow \langle v\rangle = \sqrt{2 \langle E_k \rangle / m}$

Ruslan almost 9 yearsWell if you want just magnitude of velocity, then it's possible.

71GA almost 9 yearsI will be satisfied with only a magnitude so far ... dont know enough of QM yet to think about QM vectors :). Do zou think this classical approximation is good enough for a particle in a box. I mean i calculated $\langle E_k \rangle=338.79eV$.

Ruslan almost 9 yearsI think this still isn't quite good (compare $\sqrt{\left<v^2\right>}$ with $\left<v\right>$ for some random set of $v$), you'd better compute it from velocity operator not from kinetic energy.

71GA almost 9 yearsAfter looking at @Ali's anwser I believe that you meant "momentum operator" instead of the "velocity operator". Thanks for pointing this out as i wasn't sure if i d get $\langle v \rangle^2$ or $\rangle v^2 \langle$ using the kinetic energy method  i guess it is the later huh  can i ask you, how do we know this? Why do we know that by using the kinetic energy method we get $\langle v^2\rangle$ and not $\langle v\rangle^ 2$? For the solution i will take Alis anwser.

Ruslan almost 9 yearsWe know that we get $\left<v^2\right>$ because we find expectation value of $\frac{\hat p^2}{m^2}$ not $\frac{\hat{\vec p}}{m}$.

dmckee  exmoderator kitten almost 3 yearsWhile both the question and the selected answer seem to be conflating them, the expectation values of velocity and speed are not the same thing.For any sufficiently symmetric bound system $\langle v \rangle = 0$ while $\langle v \rangle > 0$.


71GA almost 9 yearsRegarding your 1st paragraph. Why did you use the energy $\langle E \rangle$ and an operator $\hat{H}$ instead of $\langle E_k \rangle$ and an operator $\langle \hat{T}\rangle$?

Ali almost 9 yearsOops, that's a mistake.