Testing if the momentum operator is Hermitian

[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

EDIT: sorry, I found the answer https://www.physicsforums.com/showthread.php?t=138552"

 

[jtbell]

Now how do I deal with the first term? Does it reduce to 0 somehow?

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

 

[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.

 

[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)^{*}

 

[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.