Eigenfunction of momentum and operators

[Philip Land]

Homework Statement

Homework Equations

$\hat{P}= -ih d/dx$

The Attempt at a Solution

To actually obtain $\psi_{p_0}$ I guess one can apply the momentum operator on the spatial wavefunction. If we consider a free particle (V=0) we can easily get obtain $\psi = e^{\pm i kx}$, where $k= \sqrt{2mE/ \hbar}$.

By now applying momentum operator we get $\hat{P} \psi= -ih \cdot \pm i \sqrt{2mE/ \hbar}$. This is as far as I get, how do I actually get the eigenfunction of momentum on the required form?

And 2:
How do I show that the completeness relation is satisfied for that 'specific' expression? I know that we can express $|x> = \int dp |p><p|x>,$take $-i \hbar \frac{d}{d_x} f_p(x) = p f_p(x)$ and turn into a Fourier transformation (including k in the kernel) and get $p' \tilde{f_p(p')}$ which can be solved picking p=p' and any function of p can be converted into an eigenfunction if we transform back to x s.t $f_p(x) = \int dp e^{ipx \hbar} \tilde{f_p(p)}$, but I have no idea how to use this to confirm the completeness relation for the above expression...

 

[Orodruin]

1) You do not need to derive the eigenfunction. Just show that the given function is the eigenfunction.

2) You need to show that $\int |p\rangle \langle p| dp$ is the identity operator, i.e., that acting on any state with it gives back the same state.

 

[Philip Land]

Ahh okey, that made the first part very easy.

2) You need to show that ∫|p⟩⟨p|dp∫|p⟩⟨p|dp\int |p\rangle \langle p| dp is the identity operator, i.e., that acting on any state with it gives back the same state.

And on 2) what exactly is p in that equation?

 

[Orodruin]

It is just a parameter labelling the momentum eigenstates.

 

[Philip Land]

It is just a parameter labelling the momentum eigenstates.

Okey, so I'm a bit confused about the state vs $\psi_{p_0}$ which is a function. Could I rewrite the function as a state somehow?

 

[Philip Land]

Okey, so I'm a bit confused about the state vs $\psi_{p_0}$ which is a function. Could I rewrite the function as a state somehow?

Because plugging in $\psi$ in the equation returns $p_0^2 / \hbar \cdot \int dp.$

 

[Orodruin]

The wave function $\psi(x)$ for any state $|\psi\rangle$ is defined as $\psi(x) = \langle x|\psi\rangle$.

 

[Philip Land]

Oki I see. Are there way to get p for this specific case which I ultimately can plug into the completeness theorem?

 

[Orodruin]

There is no specific p. The $p$ in the completeness relation is an integration variable.

 

[Philip Land]

There is no specific p. The $p$ in the completeness relation is an integration variable.

If there's no specific, p, how can I check it for a specific case?

 

[Orodruin]

You should not be checking it for a specific case. You should show that it is always the identity operator.

 

[Philip Land]

You should not be checking it for a specific case. You should show that it is always the identity operator.

"Show that the expression above satisfies it"

 

[Orodruin]

"Show that the expression above satisfies it"

Yes? You are still not understanding the question correctly. The wavefunction you are given is the wavefunction corresponding to the state $|p_0\rangle$. You need to show that the states with this type of wavefunction lead to the states satisfying the completeness relation.

 

[Philip Land]

Yes? You are still not understanding the question correctly. The wavefunction you are given is the wavefunction corresponding to the state $|p_0\rangle$. You need to show that the states with this type of wavefunction lead to the states satisfying the completeness relation.

Cheers. Yes the reason I ask is that I obviously don't understand it.

 

[Orodruin]

That the wavefunction of the momentum states is $\psi_{p_0}(x) = \frac{1}{\sqrt{\hbar}} e^{ip_0x/\hbar}$ by definition means that
$$
\langle x|p_0\rangle = \frac{1}{\sqrt{\hbar}} e^{ip_0x/\hbar}.
$$
From this and the completeness relation for $|x\rangle$, i.e.,
$$
\int |x\rangle \langle x| dx = 1,
$$
you should be able to show the states $|p\rangle$ with the above wavefunction satisfy the completeness relation. You can do this by showing that
$$
\langle x|\psi\rangle = \int \langle x|p \rangle \langle p |\psi \rangle dp
$$
for any state $|\psi\rangle$.