# QM - calculating expectation value for velocity of an electron

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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 $\sigma_v^2=\langle v^2 \rangle-\langle v \rangle^2$.

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### 71GA

Updated on August 08, 2020

• 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 years
Well 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 years
Never mind I found it in this PDF - page 32 - Now give me some downvotes for being lazy :D
• 71GA almost 9 years
I 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 years
You can't derive a vector ($\vec v$) from its square ($T\sim v^2$)
• 71GA almost 9 years
So 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 years
Well if you want just magnitude of velocity, then it's possible.
• 71GA almost 9 years
I will be satisfied with only a magnitude so far ... don-t 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 years
I 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 years
After 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 Ali-s anwser.
• Ruslan almost 9 years
We 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 --- ex-moderator kitten almost 3 years
While 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 years
Regarding 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 years
Oops, that's a mistake.