I'm a newbie reading quantum mechanics from "Inroduction to Quantum Meachanics" by Griffiths and in the early pages of the book the author defines:

$$\langle x\rangle =\int_{-\infty}^{\infty} x|\Psi(x,t)|\,dx = \int_{-\infty}^{\infty} \Psi^* (x)\Psi \,dx,$$

$$\langle v\rangle = \frac{d}{dt}\left(\langle x\rangle\right)= -\frac{i\hbar}{m}\int_{-\infty}^{\infty} \Psi^*\frac{\partial\Psi}{\partial x} \,dx,$$

$$\langle p\rangle = m\langle v\rangle= -i\hbar\int_{-\infty}^{\infty} \Psi^*\frac{\partial\Psi}{\partial x} \,dx,$$

so to me the author seems to be working out with expectations, which made perfect sense to me. I then googled the expression for kinetic energy and I was expecting to find out that:

$$\langle T\rangle=\frac{\langle p \rangle^2}{2m},$$

but instead, it seems that

$$\langle T\rangle=\frac{\langle p^2 \rangle}{2m}.$$

Why is this? I don't understand what happened in the case of kinetic energy. Why isn't the author now working with expected momentum in the case of expected kinetic energy? Can you perhaps show me a derivation of $\langle T\rangle $ and more importantly, explanation on why it is done like that? In the book, the author says that generally:

$$\langle Q(x, p)\rangle = \int \Psi^*Q(x, \frac{\hbar}{i}\frac{\partial}{\partial x})\Psi\,dx,$$

with advising that every $p$ should be replaced with $\frac{\hbar}{i}\frac{\partial}{\partial x}$ when calculating the expectation of interest. The why-part for this was however a bit non-existing.

  • 1
    $\begingroup$ The first answer here might help you: physics.stackexchange.com/questions/424800/… $\endgroup$
    – Martin C.
    Commented Nov 1, 2018 at 13:11
  • $\begingroup$ In the mean time, I will explain with intuition rather than math. Momentum has a direction, kinetic energy does not. You can have a mean momentum of $0$ but a non-zero mean kinetic energy. Performing the average of the square of momentum fixes this. $\endgroup$ Commented Nov 1, 2018 at 13:13
  • $\begingroup$ @MartinC. Thank you, I actually checked that answer before but it didn't really open up to me unfortunately :/ $\endgroup$
    – jjepsuomi
    Commented Nov 1, 2018 at 13:16
  • $\begingroup$ @AaronStevens thank you, the intuition already helped :) Of course the details are still in the mist for me. Any other book recommendation where this might be explicitly derived? $\endgroup$
    – jjepsuomi
    Commented Nov 1, 2018 at 13:19
  • 1
    $\begingroup$ If $A = B$, then $\langle A \rangle = \langle B \rangle$, because we can do the same thing to both sides of an equation. That's absolutely all there is to it. If you think $H = p^2/2m$, then $\langle H \rangle = \langle p^2 / 2m \rangle$. $\endgroup$
    – knzhou
    Commented Nov 1, 2018 at 14:08

2 Answers 2


Why is this?

For concreteness, let's look at a specific example for which $\langle T \rangle \ne \frac{\langle P \rangle^2}{2m}$

Consider the case that we have a particle with state vector (working in 1D for simplicity)

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|+p\rangle + |-p\rangle\right)$$

where $p \ne 0$ and $P\,|\pm p\rangle = \pm p\,|\pm p\rangle$ (these are eigenkets of the momentum operator).

Clearly, the expectation value of momentum is

$$\langle P\rangle = \langle\psi|P|\psi\rangle = \frac{1}{2}\left(+p -p\right) = 0$$

This is because the momentum measurement has equal chance of yielding $+p$ and $-p$.

However, a kinetic energy measurement measurement can only yield

$$T = \frac{(\pm p)^2}{2m} = \frac{p^2}{2m}$$

and so

$$\langle T \rangle = \frac{p^2}{2m} \ne \frac{\langle P \rangle^2}{2m} = 0$$


If you think about it, this really doesn't come down to QM and just depends on how you take averages. QM only comes into play if you actually want to calculate those averages given the state vector of the system.

We know that $T=\frac{p^2}{2m}$, so the average of this is then $$\langle T\rangle=\left\langle\frac{p^2}{2m}\right\rangle=\frac{\langle p^2\rangle}{2m}$$

Since, in general, $\langle p^2\rangle\neq\langle p \rangle^2$, this is where we end up.

If you want to find this value using the position basis, then we invoke QM: $$\langle T\rangle=-\frac{\hbar^2}{2m}\int\Psi^*\frac{\partial^2}{\partial x^2}\Psi\ dx$$

This is because in the position basis, the $P^2$ operator is $-\hbar^2\frac{\partial^2}{\partial x^2}$.

  • $\begingroup$ Thank you, it helped me a bit, but I still 100% don't get it. The reason why I think it confuses me, is because in the case of velocity, we used expected position. In the case of momentum, we used expected velocity. But in the case of kinetic energy we did NOT use expected momentum. Do you happen to know any books perhaps where this is explained in detail? :) $\endgroup$
    – jjepsuomi
    Commented Nov 1, 2018 at 13:57
  • $\begingroup$ @jjepsuomi Which part? How averages work with functions of other variables, or how you calculate those averages? Keep in mind that the heart of your question (why $\langle p^2\rangle$ instead of $\langle p \rangle^2$) isn't about QM $\endgroup$ Commented Nov 1, 2018 at 13:59
  • 2
    $\begingroup$ @jjepsuomi The kinetic energy uses the expectation value of the momentum squared. The answer explicitly shows how this works: $\langle T\rangle=\left\langle\frac{p^2}{2m}\right\rangle=\frac{\langle p^2\rangle}{2m}$. $\endgroup$
    – Martin C.
    Commented Nov 1, 2018 at 14:11
  • $\begingroup$ Thank you, I think I now got it, huh it was that simple then x) $\endgroup$
    – jjepsuomi
    Commented Nov 1, 2018 at 14:14
  • $\begingroup$ @jjepsuomi Sorry, I didn't realize you just wanted an example worked through showing that $\langle p\rangle^2\neq\langle p^2\rangle$ for a specific case. $\endgroup$ Commented Nov 1, 2018 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.