# Action of momentum operator on wavefunction in momentum space

In a previous question How to get the position operator in the momentum representation from knowing the momentum operator in the position representation? it was mentioned that

\begin{align} \langle p|[\hat x, \hat p]|\psi\rangle &= \langle p|\hat x\hat p-\hat p\hat x|\psi\rangle \\ &= \langle p|\hat x\hat p|\psi\rangle - \langle p|\hat p \hat x|\psi\rangle \\ &= \langle p|\hat x \hat p|\psi\rangle - p\langle p|\hat x|\psi\rangle. \end{align}

1. In the above expressions, the $$p$$ is a wavefunction in momentum space but $$\hat p$$ is an operator in $$x$$ i.e $$\frac{\hbar}{i}\frac{\partial}{\partial x}$$, so can it act on momentum space wavefunction too?

2. Also, is $$\langle p|\psi \rangle$$ defined? If so, what is its value?

3. Lastly, how do we go from $$\langle p|\hat p \hat x|\psi\rangle \$$ to $$p\langle p|\hat x|\psi\rangle$$?

In the above expressions, the $$p$$ is a wavefunction in momentum space but $$\hat p$$ is an operator in $$x$$ i.e $$\frac{\hbar}{i}\frac{\partial}{\partial x}$$, so can it act on momentum space wavefunction too?

$$p$$ is not a wavefunction. The wavefunction is expressed in the basis of $$|p\rangle$$ as $$\langle p | \psi \rangle \equiv \psi(p)$$. In the momentum space representation, $$\hat p$$ is still the same operator, it just looks different because we've changed our basis. In this case, the action of $$\hat p$$ on $$\psi(p)$$ is just multiplication by $$p$$. The operator $$\hat p$$ is not "in $$x$$" as if it is only defined in the position space. Rather, the representation of $$\hat p$$ in the position basis is defined to be $$\frac{\hbar}{i}\frac{\partial}{\partial x}$$; in different representations, it will be different.

Also, is $$\langle p | \psi \rangle$$ defined? If so, what is its value?

Yes. It's defined to be $$\psi(p)$$ in the same way $$\langle x | \psi \rangle \equiv \psi(x)$$.

Lastly, how do we go from $$\langle p | \hat p \hat x |\psi \rangle$$ to $$p\langle p | \hat x |\psi \rangle$$

$$|p\rangle$$ is an eigenfunction of $$\hat p$$ with eigenvalue $$p$$, hence we can take $$\big[\langle p | \hat p \big]\hat x |\psi \rangle = \big[\langle p | p \big]\hat x |\psi \rangle$$, where I've dropped the hat on $$\hat p$$. Then we just pull $$p$$ out.

$$\langle p | \psi \rangle$$ is the wavefunction expressed in the momentum basis: $$\psi (p)$$. Nothing more can be said about it unless more context is provided.

Importantly, the operator $$\hat{p}$$ can be expressed in any basis, position space, as you defined it in your answer is only one possible basis.

Lastly, the momentum operator acting on the wavefunction expressed in momentum space will return the momentum of the particle (I'm assuming it's a one-particle wavefunction), and won't change the wavefunction because it's an eigenfunction of the momentum operator:

$$\hat{p} \space \psi (p) = p \space \psi (p)$$

There are two elements to the answer. The first is to understand what is $$\psi(p)$$ and the other the action of $$\hat p$$ on such functions.

To understand $$\psi(p)$$, the simplest is to refer to this question and adapt the answer to think of $$\psi(p)=\langle p\vert \psi\rangle$$. Much like $$\vert x_0\rangle$$ is such that $$\hat x\vert x_0\rangle=x_0\vert x_0\rangle$$, we have $$\vert p_0\rangle$$ such that $$\hat p\vert p_0\rangle=p_0\vert p_0\rangle$$, i.e. the states $$\vert p’\rangle$$ are eigenstates of the operator $$\hat p$$ with eigenvalue $$p’$$.

Now, $$\hat p\psi(p)$$ is by definition $$\hat p\psi(p)\equiv \langle p\vert \hat p\vert\psi\rangle\, .$$ The rest is just manipulations: \begin{align} \langle p\vert p\vert\psi\rangle =\langle \psi\vert \hat p^\dagger \vert p\rangle^* &=\langle \psi\vert\hat p\vert p\rangle^* \qquad \qquad \hbox{since \hat p is hermitian}\, ,\\ &=p\langle \psi\vert p\rangle^* \qquad \qquad \hbox{since eigenvalues of \hat p are real}\, ,\\ &=p\langle p\vert\psi\rangle = p\psi(p)\, . \end{align} In other words, just like $$\hat x$$ is multiplication by $$x$$ on functions $$\psi(x)$$ of $$x$$, $$\hat p$$ is multiplication by $$p$$ on functions $$\psi(p)$$.

The interesting part is to obtain the action of $$\hat x$$ on functions $$\psi(p)$$. This is done by converting from the $$p$$-representation to the $$x$$-representation using the basic expression $$\langle x\vert p\rangle = \frac{1}{\sqrt{2\pi}}e^{i x p/\hbar}$$ which follows because states of definite momentum $$\vert p\rangle$$ are expressed as plane wave in space, $$e^{i x p/\hbar}$$. The factor $$1/\sqrt{2\pi}$$ is there for convenience as the plane waves cannot be normalized. With this: $$\hat x\psi(p)\equiv \langle p\vert \hat x \vert \psi\rangle \tag{1}$$ Inserting now the unit operator as a sum over all the $$\vert x\rangle$$ projections: $$\hat 1=\int dx \vert x\rangle \langle x\vert$$ transforms (1) into \begin{align} X\psi(p)&=\int dx’ \langle p\vert \hat x \vert x’\rangle \langle x’\vert\psi\rangle\, ,\\ &=\int dx’ x’ \langle p\vert x’\rangle\langle \psi\rangle \, ,\\ &= \int dx’ x’ \frac{1}{\sqrt{2\pi}} e^{-i x’ p/\hbar} \psi(x)\, ,\\ &=i\hbar\frac{\partial}{\partial p} \int dx’ \langle p\vert x’\rangle\langle x’\vert\psi\rangle\, ,\\ &=i\hbar \frac{\partial}{\partial p} \langle p\vert \psi\rangle \, ,\tag{2}\\ &=i\hbar \frac{\partial}{\partial p}\psi(p)\, . \tag{3} \end{align} There is a number of mathematically loose steps in there, but that’s the basic idea. In (2) I’ve used $$\int dx’ \vert x’\rangle\langle x’\vert=\hat 1$$ to get rid of the integral, and I’ve also assume that one can “pull out” the derivative w/r to $$p$$ out of the integral since I’m integrating over $$x’$$ but taking a derivative w/r to $$p$$.

Note the important sign difference between (3) and the more common action of $$\hat p$$ on $$\psi(x)$$.

You final query follows by using $$\langle p\vert \hat p \hat x\vert\psi\rangle = \langle \psi \hat x\hat p\vert p\rangle ^* = p\langle p\vert \hat x\vert\psi\rangle$$ where I’ve use hermiticity of $$\hat x$$ and $$\hat p$$, and the reality of the eigenvalue $$p$$.

The Dirac notation $$|p\rangle$$ means that $$|p\rangle$$ is an eigenstate of the momentum operator $$\hat p$$ with eigenvalue $$p$$: $$\hat p |p\rangle = p |p\rangle .$$ If $$|\psi\rangle$$ is an eigenstate of $$\hat p$$ with eigenvalue $$p^{\prime}$$ (i.e. if it's the plane wave $$\propto exp(i p^{\prime} x/\hbar)$$), $$\langle p|\psi\rangle = \delta(p-p^{\prime})$$. If $$|\psi\rangle$$ is not a momentum eigenstate, it can still be expanded in momentum eigenstates: $$|\psi\rangle = \sum_p |p\rangle \cdot \langle p|\psi \rangle .$$ (The above equation uses the fact that $$\sum_p |p\rangle\langle p|$$ is identity.) $$\langle p|\psi\rangle$$ is the projection of $$|\psi\rangle$$ onto the momentum eigenstate $$|p\rangle$$ (commonly simply written as $$\psi(p)$$).

Since $$|p\rangle$$ is an eigenstate (more precisely, an eigenket) of $$\hat p$$ with eigenvalue $$p$$, so is the Hermitian conjugate $$\langle p|$$ ($$\langle p|$$ is the corresponding eigenbra in Dirac notation). Going from $$\langle p|\hat p \hat x|\psi\rangle$$ to $$p\langle p|\hat x|\psi\rangle$$ is using that $$\langle p|$$ is an eigenbra of the momentum operator $$\hat p$$, and one lets $$\hat p$$ act to the left ($$\langle p| \hat p = p \langle p|$$), so that one can pull the eigenvalue $$p$$ out of the scalar product.