2
$\begingroup$

I'm trying to calculate uncertainty in momentum, and I know that

$$\langle\hat P^2\rangle=\int^{\infty}_{-\infty}\hat P^2|\Psi(x)|^2\,\text dx$$

But I'm confused by what that symbol means. Does it mean I perform the operator on $|\Psi(x)|^2$ twice, or does it mean:

$$\langle\hat P^2\rangle=\int^{\infty}_{-\infty}\langle\Psi(x)|\hat P^2|\Psi(x)\rangle \,\text dx$$ where I just perform the operator on $\Psi$ twice? Or are they equivalent?

$\endgroup$
7
$\begingroup$

You seem a bit confused about how to use Dirac notation, so I'll derive the result from scratch. For any operator $O$, the definition of the expectation value is $$\langle O \rangle = \langle \psi | O | \psi \rangle.$$ In order to write this as an integral, just note that $$1 = \int dx \, | x \rangle \langle x |.$$ By putting in two "factors of $1$", we get $$\langle O \rangle = \int dx \, dx ' \, \langle \psi | x \rangle \langle x | O | x' \rangle \langle x' | \psi \rangle.$$ The definition of the wavefunction is $$\psi(x) = \langle x | \psi \rangle.$$ In particular, it doesn't make sense to write $|\psi(x) \rangle$ in Dirac notation, since $\psi(x)$ is just a number, not a ket. Anyway, using the definition of the wavefunction, we have $$\langle O \rangle = \int dx \, dx ' \, \psi^*(x) \langle x | O | x' \rangle \psi(x').$$ This is how you calculate the expectation value of any operator in the position basis. In your case, $$\langle x | p^2 | x ' \rangle = \left( \frac{\hbar}{i} \right)^2 \delta''(x - x')$$ essentially by the definition of $p$, so $$\langle p^2 \rangle = - \hbar^2 \int dx \, dx' \, \psi^*(x) \delta''(x - x') \psi(x').$$ Now integrate by parts with respect to $x'$ twice, to get $$\langle p^2 \rangle = - \hbar^2 \int dx \, dx' \, \psi^*(x) \delta(x - x') \psi''(x') = - \hbar^2 \int dx \, \psi^*(x) \psi''(x).$$ That's your answer.

tl;dr: Neither of your expressions are right. Instead, differentiate $\psi(x)$ twice.

$\endgroup$
  • $\begingroup$ $1 = \int dx \, | x \rangle \langle x |$ — doesn't unnormalizability of $|x\rangle$ make this false, whatever normalization you choose? $\endgroup$ – Ruslan Feb 28 at 11:37
  • $\begingroup$ No, it does not. They are normalised on $\langle x|x' \rangle=\delta(x-x')$ and thus the relation holds. It follows directly from the spectral theorem for continuous eigenvalues. $\endgroup$ – Jan2103 Feb 28 at 14:47
2
$\begingroup$

By definition \begin{align} \langle O\rangle := \int dx \psi^*(x) \hat O\psi(x) \end{align} so in your case \begin{align} \langle P^2\rangle = \int dx\psi^*(x) \left(-i\hbar \frac{d}{dx}\right)\left(-i\hbar \frac{d}{dx}\right)\psi^*(x)=-\hbar^2 \int dx\,\psi^*(x)\psi^{\prime\prime}(x) \end{align}

$\endgroup$
0
$\begingroup$

Your second equation is right (I don't understand why knzhou doesn't like it). It's the quantum mechanical expression for $\left<\hat P^2 \right>$. But your first equation is wrong - because this is quantum mechanics.

If $\hat P$ were just a function, say $f(x)$, you could trivially rewrite $\Psi^*(x) f^2(x) \Psi(x)$ as $f^2(x) \Psi^*(x) \Psi(x)=f^2(x) |\Psi(x)^2|$ and get your first equation, which is the standard statistical expression for $\left<\hat P^2 \right>$ and you've probably got that from a textbook on statistics or statistical mechanics.

And in that context it's fine, but Quantum Mechanics breaks it. Because $\hat P$, being momentum, involves not just multiplication but differentiation and you can't do that trivial rewrite. This is one of those points where previous knowledge and formulae get disturbed and extended by Quantum Mechanics.

$\endgroup$
  • 1
    $\begingroup$ The problem with the second expression in the question is that $|\psi(x)\rangle$ does not make sense (or, at best, is confusing). It is a common mistake for beginners to confuse the vector $|\psi\rangle$ with its position representation $\psi(x) = \langle x | \psi \rangle$, see e.g. physics.stackexchange.com/questions/120015/… $\endgroup$ – Noiralef Feb 28 at 12:32
  • $\begingroup$ OK, yes, thanks, Should be $|\Psi>$ not $|\Psi(x)>$. I missed that. $\endgroup$ – RogerJBarlow Feb 28 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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