I Momentum operator acting to the left

[DuckAmuck]

TL;DR Summary

Change direction of Operator

Is the following true if the momentum operator changes the direction in which it acts?
\langle \phi | p_\mu | \psi \rangle = -\langle \phi |\overleftarrow{p}_\mu| \psi \rangle

My reasoning:
\langle \phi | p_\mu | \psi \rangle = -i\hbar \langle \phi | \partial_\mu | \psi \rangle
\langle \phi | \partial_\mu | \psi \rangle = \int \phi^\dagger \partial_\mu \psi dx = \phi^\dagger \psi |_\text{bound} - \int \psi \partial_\mu \phi^\dagger dx
= - \int \psi \partial_\mu \phi^\dagger dx = - \langle \phi | \overleftarrow{\partial_\mu} |\psi \rangle

is this correct? if not, why?

 

[anuttarasammyak]

What you say seems same as
p_{\mu} ^\dagger = i \hbar \frac{\partial}{\partial x^\mu}=-p_{\mu}
where $\dagger$ means Hermitian conjugate.

 

[vanhees71]

You are mixing up different notions of the momentum operator, and that's usually doomed to fail.

On the level you wrote the question, you work in the Dirac formalism not specifying a "representation", i.e., you work in a basis free way. It's very confusing when you introduce operators to act to the left.

By definition all operators that represent observables are self-adjoint operators and thus you have for any vectors $|\phi \rangle$ and $|\psi \rangle$
$$\langle \phi|\hat{p}_j \psi \rangle = \langle \hat{p}_j \phi|\psi \rangle.$$
Here the operators always act to the right.

Now you can introduce a basis and write the operators wrt. this basis. If you choose the (generalized) position eigenbasis you deal with position-space wave functions, $\psi(\vec{x})=\langle \vec{x} |\psi \rangle$ and all you need is the completeness relation
$$\int_{\mathbb{R}^3} \mathrm{d}^3 x |\vec{x} \rangle \langle \vec{x}|=\hat{1}.$$
Then from the Heisenberg algebra,
$$[\hat{x}_j,\hat{x}_k]=[\hat{p}_j,\hat{p}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \hbar \delta_{jk} \hat{1},$$
you get
$$\hat{p}_j \psi(\vec{x}):=\langle \vec{x}|\hat{p} \psi \rangle=-\mathrm{i} \hbar \partial_j \langle \vec{x}|\psi \rangle = -\mathrm{i} \hbar \partial_j \psi(\vec{x}).$$
The scalar product reads
$$\langle \phi|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \langle \phi|\vec{x} \rangle \langle \vec{x}|\psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \phi^*(\vec{x}) \psi(\vec{x}).$$
Then the self-adjointness is easy to show using the momentum operator in position-space representation,
$$\langle \phi|\hat{p}_j \psi \rangle=\int_{\mathbb{R}^3} \mathrm{d}^3 x \phi^*(\vec{x}) (-\mathrm{i} \hbar \partial_j \psi(\vec{x}) = \int_{\mathbb{R}^3} \mathrm{d}^3 x [-\mathrm{i} \hbar \partial_j \phi(\vec{x})]^* \psi(\vec{x}),$$
which you prove by integration by parts.

 

[DuckAmuck]

What you say seems same as
p_{\mu} ^\dagger = i \hbar \frac{\partial}{\partial x^\mu}=-p_{\mu}
where $\dagger$ means Hermitian conjugate.

Right. I get that but I’m not sure where I go wrong here.

 

[vanhees71]

Have you understood what I wrote above?

 

[anuttarasammyak]

Right. I get that but I’m not sure where I go wrong here.

(<\psi_2|p|\psi_1>)^\dagger=<\psi_1|p^\dagger|\psi_2>
so
-i\hbar \int \psi_2^*(x)\frac{\partial \psi_1(x)}{\partial x} dx=<\psi_2|p|\psi_1>=(<\psi_1|p^\dagger|\psi_2>)^\dagger=(i\hbar \int \psi_1^*(x)\frac{\partial \psi_2(x)}{\partial x} dx)^*=-i\hbar \int \psi_1(x)\frac{\partial \psi_2^*(x)}{\partial x} dx
The rightest side is the way operator works on the original left neighbor as you say $\leftarrow$.

 

[atyy]

Summary:: Change direction of Operator

Is the following true if the momentum operator changes the direction in which it acts?

As pointed out by @anuttarasammyak and @vanhees71 above, it is better to rephrase your question as asking about the Hermitian conjugate of the momentum operator: https://quantummechanics.ucsd.edu/ph130a/130_notes/node133.html

The Hermitian conjugate of the derivative operator is negative of itself, while the Hermitian conjugate of the momentum operator is itself: https://quantummechanics.ucsd.edu/ph130a/130_notes/node144.html#example:Hconj2