# Testing if the momentum operator is Hermitian

1. Sep 8, 2008

### kkan2243

Hi. I'm not too good at maths and I'm having some trouble figuring out the basics of what to do with complex conjugates of functions.

Our lecturer has set a couple problems requiring us to test if a few operators are Hermitian. Before I can get to those I thought I'd test the basic momentum operator: $$(\frac{\hbar}{i} \frac{\partial}{\partial x})$$.

Using integration by parts:

$$\int_{-\infty}^\infty \Psi^\ast(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi(x,t) dx = \frac{\hbar}{i} [\Psi^\ast(x,t)\Psi(x,t)]^\infty_{-\infty} - \int_{-\infty}^\infty \Psi(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi^\ast(x,t) dx$$

Now how do I deal with the first term? Does it reduce to 0 somehow? I recongnise that the second term should be equal to the LHS for the momentum operator to be shown as Hermitian.

Cheers. Kaan

Last edited by a moderator: Apr 23, 2017
2. Sep 8, 2008

### Staff: Mentor

$\Psi$ must be normalizable, therefore $\Psi^* \Psi$ must go to zero as x goes to infinity in either direction.

3. Sep 8, 2008

### h0dgey84bc

Yes first term on RHS goes to zero because of the normalization condition on the wavefunction $$\Psi(x,t) ->0$$ as $$x-> \infty$$ ( or $$-\infty$$ ).

So you're left with:

$$\int_{-\infty}^\infty \Psi^\ast(x,t) (\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi(x,t) dx = \int_{-\infty}^\infty \Psi(x,t) (-\frac{\hbar}{i} \frac{\partial}{\partial x}) \Psi^\ast(x,t) dx$$

Thus the Hermitian condition, $$\int_{-\infty}^\infty \Psi^\ast(x,t) \hat{P} \Psi(x,t) dx = \int_{-\infty}^\infty \Psi(x,t) \hat{P^\ast} \Psi^\ast(x,t) dx$$

is satisified since the conjugate of the momentum operator is just minus the momentum operator.

4. Nov 5, 2009

### Bill Foster

Is this a valid proof?

Given an arbitrary ket $$|a\rangle$$:

$$\langle a|\hat{\textbf{p}}a\rangle$$

$$=\langle \hat{\textbf{p}}^{\dagger}a|a\rangle$$

If $\hat{\textbf{p}}$ is Hermitian, then

$$\hat{\textbf{p}}^{\dagger}=\hat{\textbf{p}}$$

So

$$\langle \hat{\textbf{p}}^{\dagger}a|a\rangle = \langle \hat{\textbf{p}}a|a\rangle$$

$$=\left(\langle a|\hat{\textbf{p}}a\rangle\right)^{*}$$

5. Nov 5, 2009

### dextercioby

You can't assume P is hermitean, you have to prove such thing ! You can't prove it in the abstract Dirac language. You must choose a Hilbert space. Define the operators and only then check for properties such as continuity/boundedness, hermiticity, unitarity, etc.