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I already derived a QM expectation value for ordinary momentum which is:

$$ \langle p \rangle= \int\limits_{-\infty}^{\infty} \overline{\Psi} \left(- i\hbar\frac{d}{dx}\right) \Psi \, d x $$

And I can read clearly that operator for momentum equals $\widehat{p}=- i\hbar\frac{d}{dx}$. Is there an easy way to derive an expectation value for $\langle p^2 \rangle$ and its QM operator $\widehat{p^2}$?

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  • $\begingroup$ If you had to guess as to what the square of the momentum operator looked like, what would you guess? $\endgroup$
    – kηives
    Mar 22, 2013 at 20:01
  • $\begingroup$ How were you able to derive this relationship for the expectation value? How is that approach not applicable to $\hat p^2$? $\endgroup$
    – Muphrid
    Mar 22, 2013 at 20:16
  • $\begingroup$ I derived it from $d\langle x \rangle/dt$ using 2 partial derivation, some u-substitutions and derivation of a product. It took me 2 days to calculate it. I did it like Griffith in his book... $\endgroup$
    – 71GA
    Mar 22, 2013 at 22:54

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Well, $\widehat{p^2} = \hat{p}^2= \hat{p} \hat{p}$.

So, in the position basis it is $-\hbar^2 \frac{d^2}{dx^2}$, and $\langle p^2 \rangle = \int_{-\infty}^\infty \bar{\Psi}\left(-\hbar^2 \frac{d^2}{dx^2} \right)\Psi dx$.

Note: $\hat{p}$ is technically not equal to $-i\hbar d/dx$, but rather in the position basis $\langle x | \hat{p}| x' \rangle = -i\hbar d/dx \delta(x-x')$.

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  • $\begingroup$ Are you sure i can do this $\widehat{p^2} = \hat{p}^2= \hat{p} \hat{p}$? Afterall i am not alowed to do it with expectation values as $\langle p^2 \rangle \neq {\langle p \rangle} ^2$. Could you point me to any proof? $\endgroup$
    – 71GA
    Mar 22, 2013 at 20:14
  • $\begingroup$ One way to look at it is to start from the classical theory and then quantize it - anytime you see a $x$ you change it to $\hat{x}$ and anything you see a $p$ you write a $\hat{p}$. Then you impose the commutation relation $[x,p] = i\hbar$. There are some ordering issues so you need to choose an ordering too. Now to quantize the classical $p^2$ term, $p^2 = p p \to \hat{p} \hat{p} = \hat{p}^2$. So the quantum version of $p^2$, which we call $\widehat{p^2}$, is equal to $\hat{p}^2$. The expectation values not being equal as you have mentioned follow from this. $\endgroup$
    – nervxxx
    Mar 22, 2013 at 20:35
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    $\begingroup$ @71GA that's the definition of what it means to square an operator. $\endgroup$
    – David Z
    Mar 22, 2013 at 23:29
  • $\begingroup$ Where can i read more about this? Is this somehow connected to operator theory??? $\endgroup$
    – 71GA
    Mar 23, 2013 at 16:04
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    $\begingroup$ @71GA if you mean my note... well. Griffiths, from the outset, uses the position basis to teach QM. There's nothing wrong with that, but it obscures the fact that the fundamental object is a ket vector $| \psi \rangle>$ in an abstract Hilbert space, and the wavefunction $\psi(x)$ is just a representation of that ket vector in real space: $\langle x | \psi \rangle$. After you're done with Griffiths, Shankar will be a good 2nd read and Sakurai a good 3rd read. $\endgroup$
    – nervxxx
    Mar 23, 2013 at 21:22
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The expectation value for some operator $A$ is given by

$$\langle A\rangle = \int \Psi^*A\Psi.$$

If we set $A=p$ then we get the expression you've written above. Now just set $A = p^2$ to get what you want.

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Between any two vectors $| \psi_1 \rangle$, $| \psi_2 \rangle \in \mathcal H$ $$\langle \psi_1 | \hat p^2 | \psi_2 \rangle = \langle \psi_1 | \hat p | \psi_2' \rangle = \langle \hat p \psi_1 | \psi_2' \rangle = \langle \hat p \psi_1 | \hat p \psi_2 \rangle.$$ In the first equality, $|\psi_2'\rangle = \hat p | \psi_2 \rangle \in \mathcal H$ by definition of a linear operator acting on a vector space. In the second, we used that $\hat p$ is hermitian. In particular, for an expectation, $$\langle \psi | \hat p^2 | \psi \rangle = \langle \hat p \psi | \hat p \psi \rangle = \int_{\mathcal D} ( \hat p \psi)^* (\hat p \psi)\,\mathrm{d}x,$$ so one need only really compute $\hat p \psi(x)$.

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