So I had found two question based on the title one was talking about momentum operator in bound state and the other was a more general. Where in the first bound state calculation they had related $\langle p \rangle \sim \langle[H,x]\rangle $ and then further proved $$\langle [H,x] \rangle = \langle n | [H,x] | n \rangle = \langle n | Hx-xH | n \rangle = E_n ( \langle n| x | n \rangle - \langle n| x | n \rangle) =0.$$

Here I did not understand why bound state was a necessary condition since they never used any potential term or let alone expand any operators. Moreover in the general solution, they didn't get 0 but were stuck at $$ \int\left(i \hbar \frac{\partial}{\partial x} \psi^*\right) \psi d x $$ this integral in the end which did not give zero.

Could someone explain why the solutions work differently and why the first solution can't be true for any 0 potential system?

Note: Both the solution make sense and seem logical but I'm not able to understand why it's different because of the conditions.


1 Answer 1


It depends on your definition of bounded state. If a bounded state is just a smooth, rapidly vanishing for $|x|\to \infty$, wavefunction, the the claim is generally false.

Indeed, consider a smooth rapidly vanishing wavefunction $\psi$, mathematically a function of the Schwartz space ${\cal S}(\mathbb{R}^n)$. The Fourier transform send this function to a function of the same space $\hat{\psi}$.

The Fourier transform (I assume $\hbar=1$) however moves the theory form the position picture to the momentum picture, where the momentum operator acts as a multiplication. Here things become evident:

$$\langle \psi | P_j \psi\rangle = \int_{\mathbb{R}^n} \overline{\hat{\psi}(p)}p_j \hat{\psi}(p) d^np= \int_{\mathbb{R}^n} |\hat{\psi}(p)|^2 p_j d^np$$ It is easy to construct Schwartz functions such that the integral above does not vanish: If it vanishes for a certain function, by changing the function in a small neighborhood of a point far from the origin (thus remaining in the Schwartz space) the integral ceases to vanish.

Also considering the case $n=1$ a particle on the line the integral generally does not vanish, unless for instance the function is symmetric (up to a phase) under $p\to -p$. The eigenstates of the 1D harmonic oscillator are a special example.

If the wavfunction is a normalized eigenvector of the Hamiltonian operator with special symmetries, then the thesis is true as it happens for the harmonic oscillator. However, just considering superpositions of these eigenvectors (preserving the fact that the final wavefunction rapidly vanishes) the thesis becomes false.

  • $\begingroup$ In the linked posts, the bound state under consideration are normalized eigenstates of the Hamiltonian. $\endgroup$ Sep 25 at 11:14
  • $\begingroup$ @SolubleFisch Indeed, my post above refers to the general case. I know the other answers. and yours in particular concernin eigevectors of the Hamiltonian in 1D. The above issue, conversely, seems to refer to the general case... Or, at least, it was my interpretation. $\endgroup$ Sep 25 at 12:02

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