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I have some doubts about an exercise. It asks me the expectation value of the momentum of a wavefunction in a two dimensional box. So I have to square the momentum and then operate with that on the wavefunction and integrate but still have this doubt about how to square it.

Is it true that the square of $\hat{P}$ operator in two dimension is $$\hat{P}^2=(ih)^2\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)$$

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  • $\begingroup$ In position space, the momentum operator in 1D is $\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}$. Squaring this would be $\hat{p}^2=\frac{\hbar^2}{i^2}\frac{\partial^2}{\partial x^2}$. Using the superposition principle, in which x and y components of the momentum are uncorrelated, we arrive at the same result you obtained. $\endgroup$
    – DomDoe
    Commented May 7, 2018 at 13:44
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    $\begingroup$ @DomDoe I'm uncertain about your line of reasoning. What would you say the operator $\hat{p}^4$ is in two dimensions? $\endgroup$
    – gj255
    Commented May 7, 2018 at 14:03
  • $\begingroup$ @gj255 $\hat{p}^4 = \hat{p}^2 \hat{p}^2 =\left( \vec{\hat{p}} \cdot \vec{\hat{p}} \right)^2 $ i would say. But i think understand what you're aiming at. $\endgroup$
    – DomDoe
    Commented May 7, 2018 at 14:56

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If the exercise is asking for momentum expectation value, then you don't have to square the operator. Namely the momentum operator is $\hat{\vec{P}} = (\hat{P}_x,\hat{P}_y)$ in your two-dimensional problem. So you are asked to compute for a given wave-function $\psi$ the quantity: $$\langle\psi|\hat{\vec{P}}|\psi\rangle = \bigg(\langle\psi|\hat{P}_x|\psi\rangle\; , \,\langle\psi|\hat{P}_y|\psi\rangle\bigg)$$ Notice the answer is a vector containing the expectation values for each direction.

EDIT

$$P_x = -i\hslash \frac{\partial}{\partial x}$$ So if you compute the square of $\hat{\vec{P}}$ as defined above: $$\hat{P}^2 = \hat{\vec{P}}\cdot \hat{\vec{P}} = -\hslash^2\left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\right)$$

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  • $\begingroup$ Careful! $<\hat P_x>$ is probably zero, as it's as likely to be positive as negative... $\endgroup$
    – RogerJBarlow
    Commented May 7, 2018 at 14:22
  • $\begingroup$ Yes if there is no special potential breaking the translation symmetry in $x$ (or $y$) then it should be 0, that is exactly what the OP has to find out, don't give away the answers ;). $\endgroup$
    – ohneVal
    Commented May 7, 2018 at 16:28

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