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I know that the operator form of $1/P_x$ ($P_x$ is the $x$-component of momentum operator $\mathbf{P}$) should have an integral form like: $$\frac{i}{\hbar}\int\,dx,$$ but I'm not sure about the limits of integral. Any help would be appreciated.

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2 Answers 2

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The operator $$\frac{1}{P_x}$$ produces a divergent result if it acts on a wave function with $P_x=0$ (the eigenstate with this eigenvalue looks like a constant function of $x$). The superpositions of a $P_x=0$ states with the eigenstates associated with nonzero values of $P_x$ are still singular.

That's why a well-defined result of $(1/P_x)|\psi\rangle$ is only obtained if $|\psi\rangle$ contains no $P_x=0$ admixture in the superposition. But that's equivalent to $$ \int_{-\infty}^{+\infty} dx \,\psi(x) = 0$$ The left hand side is nothing else than the $P_x=0$ Fourier component. For that reason, the prescriptions $$[\frac{1}{P_x} \psi] (x) \leftrightarrow \frac{i}{\hbar}\int_{-\infty}^x dx'\,\psi(x') $$ and $$[\frac{1}{P_x}\psi](x) \leftrightarrow -\frac{i}{\hbar}\int_{x}^\infty dx'\,\psi(x') $$ produce the same result.

If you wanted to generalize the action of an operator $1/P_x$ on a wave function that does contain a $P_x=0$ piece, then you would have to decide whether you mean the "principal value" of $1/P_x$ or $1/(P_x+i\epsilon)$ or something else. This extra refinement could make the operator well-defined for a broader set of test functions $\psi(x)$.

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    $\begingroup$ It should also be noted that the domain of definition of $1/P_x$, as defined in this answer, is substantially different to that of $P_x$, though it would be interesting to see a comparison between that domain and the image of $P_x$. $\endgroup$ Dec 5, 2013 at 13:54
  • $\begingroup$ Dear Emilio, well, $P_x$ and $1/P_x$ are different operators - inverse to each other - so of course that their definitions are different. One of them is proportional to the derivative; the other to the integration. You could be comparing the domain of one with the image of the other but the answer would depend on exactly "how pathological or generalized" functions (or distributions) you are willing to consider. Morally, they're inverse operators to each other so the image of one is the domain of the other. $\endgroup$ Dec 6, 2013 at 7:38
  • $\begingroup$ Thanks dear Lubos. I didn't get how did you choose x for the limit of integral? $\endgroup$
    – Abolfazl
    Dec 7, 2013 at 20:26
  • $\begingroup$ I showed your answer to my professor. He also said theat P is symetric in space and the interval for the intergral should be symetric but P=0 is a problem yet. $\endgroup$
    – Abolfazl
    Dec 19, 2013 at 8:31
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One way to treat Quantum mechanics is that writing the operators in terms of matrices.

One can think of operators $\hat{X}$, $\hat{P_x}$ and $\hat{P_x}^{-1}$ in terms of matrix quantum mechanics. In the $|x \rangle$ basis (where $\hat{X} |x \rangle=x|x \rangle$), we can discretize the space up to some overall constant as a unit of spacing, and write the operators as

$$\hat{X}=a{\begin{pmatrix} \ddots & 0 & 0 & 0 & 0 & 0&0\\ 0 & -2 & 0 & 0 & 0 & 0&0\\ 0 & 0 & -1 & 0 & 0 & 0&0\\ 0 & 0 & 0 & 0 & 0 & 0&0\\ 0 & 0 & 0 & 0 & 1 & 0&0 \\ 0 & 0 & 0 & 0 & 0 & 2&0 \\ 0 & 0 & 0 & 0 & 0 & 0& \ddots \\ \end{pmatrix}}$$

$$\hat{P_x}=-i \hbar \partial_X=i \hbar a^{-1}{\begin{pmatrix} \ddots & 0 & 0 & 0 & 0 & 0&0\\ 1 & 0 & -1 & 0 & 0 & 0&0\\ 0 & 1 & 0 & -1 & 0 & 0&0\\ 0 & 0 & 1& 0 & -1 & 0&0\\ 0 & 0 & 0 & 1 & 0 & -1&0 \\ 0 & 0 & 0 & 0 & 1 & 0&-1 \\ 0 & 0 & 0 & 0 & 0 & 0& \ddots \\ \end{pmatrix}}$$

Easy to check $$[\hat{X},\hat{P_x}]=i\hbar$$

Thus, $$\frac{1}{\hat{P_x}}=\hat{P_x}^{-1}$$

Thus, $\frac{1}{\hat{P_x}}=\hat{P_x}^{-1}$ is just the inverse matrix of $\hat{P_x}$. One may need to consider two cases: periodic boundary conditions on a 1D circle $S^1$ or an infinite size 1D system. Treated two cases separately.

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