Does the momentum operator applied to a position state vanish?

Question

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.

Answer 1

$i\hbar\,\partial/\partial x$ 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 $$\hat{p}=\int{\rm d}x\,|x\rangle\left(-i\hbar\frac{\partial}{\partial x}\right)\langle x|$$

Answer 2

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 : \begin{equation} \hat p |p\rangle = p|p\rangle\tag{1}\end{equation}
• $|x\rangle$ 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.