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I'm dealing with some basic calculations with plane waves and I'm having some trouble with an idea.

It has been said in another question that if you take to momenta like, for example $\boldsymbol{p}=(p_{1},p_{2},p_{3})$ and $\boldsymbol{p}'=(p_{1}',p_{2}',p_{3}')$, and calculate the transition probability for the position operator the result is

\begin{equation} ⟨\boldsymbol p'|\hat{X}|\boldsymbol p⟩=-i\hbar\frac{\partial}{\partial p_{x}}\delta(\boldsymbol{p}'-\boldsymbol{p}),\text{ where } ⟨\boldsymbol r|\boldsymbol p⟩=e^{i \boldsymbol{p}\cdot\boldsymbol{r}/\hbar}, \end{equation}

where the differentiation is taken with respect to the variable $\boldsymbol p$ because the $\hat{X}$ operator is acting to the right, but if you take the operator acting to the left then you get a similar result with the sign changed and the derivation is not taken with respect to $\boldsymbol{p}'$ that is

\begin{equation} (⟨\boldsymbol p'|\hat{X})|\boldsymbol p⟩=i\hbar\frac{\partial}{\partial p_{x}'}\delta(\boldsymbol{p}'-\boldsymbol{p}) \end{equation}

And, from my knowledge, both results must be the same. But what happens when you have that $⟨\boldsymbol r|\boldsymbol p⟩= f(\boldsymbol p) e^{i\boldsymbol{p} \cdot \boldsymbol{r}/\hbar}$ and $⟨\boldsymbol p'|\boldsymbol r⟩=g^{*}(p')e^{-i\boldsymbol{p}\cdot\boldsymbol{r}/\hbar}$? Then these two results,

\begin{align} &⟨\boldsymbol p'|(\hat{X}|\boldsymbol{p}⟩)=-i\hbar g^{*}(\boldsymbol p')f(\boldsymbol p)\frac{\partial}{\partial p_{x}}\delta(\boldsymbol{p}'-\boldsymbol{p})=i\hbar g^{*}(\boldsymbol p')\frac{\partial}{\partial p_{x}}(f(\boldsymbol p))\delta(\boldsymbol{p}'-\boldsymbol{p}) \\ & (⟨\boldsymbol p'|\hat{X})|\boldsymbol p⟩=i\hbar g^{*}(\boldsymbol p')f(\boldsymbol p)\frac{\partial}{\partial p_{x}'}\delta(\boldsymbol{p}'-\boldsymbol{p})=-i\hbar \frac{\partial}{\partial p'_{x}}(g^{*}(\boldsymbol p'))f(\boldsymbol p)\delta(\boldsymbol{p}'-\boldsymbol{p}) \end{align}

are clearly not the same. Can you help me understanding this?

Qmechanic
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Tomas Libutti
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1 Answers1

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First of all, I would encourage you to think of position of acting on momentum states to the left, that is, to commute them with the bra: $$ ⟨\mathbf p|\hat x=-i\hbar \frac{\partial}{\partial p_x}⟨\mathbf p|, $$ where the differentiation is over anything to the right of it, so for instance $$⟨\mathbf p|\hat x|\psi⟩=-i\hbar \frac{\partial}{\partial p_x}⟨\mathbf p|\psi⟩=-i\hbar \frac{\partial\,\tilde \psi(\mathbf p)}{\partial p_x}.$$

To answer your question, the simple fact is that states that are not plane waves, like the wavefunctions you are considering, cannot be momentum eigenstates. You have discovered one way to show this, but the deep reason is that they are incompatible with the canonical commutation relation $[\hat x_i,\hat p_j]=i\hbar \delta_{ij}$, because von Neumann's unicity theorem demands that any system with that commutation relationship have $\hat x$ and $\hat p$ eigenstates whose wavefunctions are plane waves in the other representation.

Emilio Pisanty
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