# Does the momentum operator applied to a position state vanish?

In quantum mechanics we have $$\begin{equation*} \langle x|p\rangle=C\exp\left(\frac{ipx}{\hbar}\right) \end{equation*}$$ where $$C$$ is a normalization constant.

It follows that $$\begin{equation*} -i\hbar\frac{\partial}{\partial x}\bigl(\langle x|p\rangle\bigr) =-i\hbar C\frac{\partial}{\partial x}\exp\left(\frac{ipx}{\hbar}\right) =p\langle x|p\rangle \tag{1} \end{equation*}$$

However by the product rule we can also write $$\begin{equation*} -i\hbar\frac{\partial}{\partial x}\bigl(\langle x|p\rangle\bigr) =-i\hbar \left[ \left(\frac{\partial}{\partial x}\langle x|\right)|p\rangle +\langle x|\left(\frac{\partial}{\partial x}|p\rangle\right) \right] \tag{1.1} \end{equation*}$$

Rewrite as $$\begin{equation*} -i\hbar\frac{\partial}{\partial x}\bigl(\langle x|p\rangle\bigr) =-i\hbar\left(\frac{\partial}{\partial x}\langle x|\right) |p\rangle +p\langle x|p\rangle \tag{2} \end{equation*}$$

Then by equivalence of (1) and (2) we have $$\begin{equation*} -i\hbar\left(\frac{\partial}{\partial x}\langle x|\right) |p\rangle=0 \tag{3} \end{equation*}$$

Is this correct?

EDIT: No, it is not correct. The mistake is in equation (2). As pointed out in the accepted answer, the correct form of (2) is $$\begin{equation*} -i\hbar\frac{\partial}{\partial x}\bigl(\langle x|p\rangle\bigr) =-i\hbar\left(\frac{\partial}{\partial x}\langle x|\right) |p\rangle \tag{2'} \end{equation*}$$ because $$\begin{equation*} \langle x|\left(\frac{\partial}{\partial x}|p\rangle\right)=0 \end{equation*}$$

It follows from (2') that $$\begin{equation*} -i\hbar\frac{\partial}{\partial x}\bigl(\langle x|p\rangle\bigr) =p\langle x|p\rangle \end{equation*}$$ which is equivalent to equation (1) as required.

• Your notation is very confusing (which may be the cause of your trouble). Commented Sep 9, 2023 at 17:01
• Careful when taking the derivative of a ket as you wrote it: this is like writing $\frac{\partial}{\partial x}\hat x$, which in itself makes no sense. After all, $\vert x\rangle$ is a unit vector. Commented Sep 9, 2023 at 17:13

$$i\hbar\,\partial/\partial x$$ is the momentum operator in the position representation. It is not the momentum operator that acts in the Hilbert space of abstract kets like $$|p\rangle$$, so your eq. $$(2)$$ doesn't follow. There is simply no way to make sense of such an expression.

Rather, starting from $$\hat{p}|p\rangle=p|p\rangle$$, we have that

\begin{align*} \hat{p}|p\rangle=p\int{\rm d}x\,e^{ipx/\hbar}|x\rangle=\int{\rm d}x\,|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)e^{ipx/\hbar}, \end{align*} so, from $$e^{ipx/\hbar}=\langle x|p\rangle$$, it follows that $$\hat{p}=\int{\rm d}x\,|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|$$

• @TheLittleDogLaughed my suggestion would be to go back to the definition of the derivative: ${\rm d}f/{\rm d}x=\lim_{h\to 0}(f(x+h)-f(x))/h$. Can you recover the product rule when the "product" is $f(x)=\langle x|p\rangle$? (spoiler: no, you cannot, because it is not a product of functions of $x$). Commented Sep 9, 2023 at 17:54
• I believe you are technically out by a factor of $1/\sqrt{2\pi \hbar}$ in converting $\langle x\vert p\rangle$ to the exponential. Commented Sep 9, 2023 at 19:05
• @ZeroTheHero I always forget which factors go where so you could very well be correct. I think in this case if you choose $\langle p|p^\prime\rangle=2\pi\hbar\,\delta(p-p^\prime)$ then it comes out as I wrote, while yours corresponds to the choice $\langle p|p^\prime\rangle=\delta(p-p^\prime)$. Commented Sep 11, 2023 at 2:42
• The natural choice is $\langle x\vert p\rangle=e^{ipx/\hbar}/\sqrt{2\pi \hbar}$ so that $\langle p\vert x\rangle=\langle x\vert p\rangle^*$ so you never have to worry about $2\pi$ factors as you do. Commented Sep 11, 2023 at 13:05

Your notations are confusing and not at all consistent with the conventional Dirac notation :

• an operator $$\hat {\mathcal O}$$ should be applied to a ket $$|\psi\rangle$$, forming an expression like $$\hat{\mathcal O}|\psi\rangle$$. For example, the fact that $$|p\rangle$$ is an eigenvector of the operator $$\hat p$$ is written : $$$$\hat p |p\rangle = p|p\rangle\tag{1}$$$$
• $$|x\rangle$$ and $$|p\rangle$$ are ket-valued functions of $$x$$ and $$p$$ respectively. In other words for each value of $$x$$ there is a ket $$|x\rangle$$ representing a particle at the position $$x$$. Consequently, by taking the hermitian product $$\langle x|p\rangle$$, we obtain a complex-valued function of $$x$$ and $$p$$, and specifically : $$\langle x|p\rangle = C\exp\left(\frac{ipx}{\hbar}\right) \tag 2$$
• the operator $$\hat p$$ is $$-i\hbar \partial_x$$ in the position representation. This can be written in several different ways, one of which is that for any ket $$|\psi\rangle$$, we have : $$\langle x |\hat p|\psi\rangle = -i\hbar \partial_x (\langle x|\psi\rangle) \tag 3$$ Everything makes sense in the RHS, as $$\langle x|\psi\rangle$$ is a complex-valued function which we can differentiate easily.

Now, it might seem that equations $$(1),(2)$$ and $$(3)$$ allow us to compute $$\langle x |\hat p|p\rangle$$ in two different ways, so we might wan to check that everything is consistent. On one hand we have : $$\langle x |\hat p|p\rangle = \langle x |\big( p|p\rangle\big) = p\langle x |p\rangle$$ while on the other hand we have : \begin{align} \langle x |\hat p|p\rangle &= -i\hbar \partial_x ( \langle x |p\rangle) \\ &= -i\hbar \Big[ \partial_x \big(\langle x|\big) |p\rangle + \langle x| \partial_x \big(|p\rangle\big)\Big] \end{align} Here, we have applied the product rule to the two ket-valued functions $$|x\rangle$$ and $$|p\rangle$$. However, $$|p\rangle$$ does not depend on $$x$$, so its derivative with respect to $$x$$ vanishes and we are left with : $$\langle x |\hat p|p\rangle = -i\hbar \partial_x \big(\langle x|\big) |p\rangle \tag 4$$

This equation holds for any value of $$p$$. As the kets $$|p\rangle$$ form a basis of the Hilbert space, equation $$(4)$$ is equivalent to : $$\langle x |\hat p = -i\hbar \partial_x \big(\langle x|\big)$$ or, taking the hermitian conjugate, to : $$\hat p |x\rangle = i\hbar \partial_x |x\rangle \tag{4'}$$ We see that what we are left is just an equivalent way of writing down $$\hat p$$ in the position representation.