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## Homework Statement

Find the spectrum of the Momentum operator in the Hilbert Space defined by L^2([-L,L]), consisting of all square integrable functions ψ(x) in the range -L, to L

## Homework Equations

We can get the resolvent set containting all λ in ℂ such that

you can always find a vector ψ(x) in the Domain of [itex]\hat{p}[/itex] that satisfies the following equation:

(λI - [itex]\hat{p}[/itex]) ψ(x) = [itex]\Phi[/itex](x)

where [itex]\Phi[/itex](x) is any state in the Hilbert Space.

The Spectrum of the operator is the complement of the resolvent set with respect to the complex plane

The momentum operator, in position representation, is given by [itex]\hat{p}[/itex] =[itex]\frac{\hbar}{i}\frac{d}{dx}[/itex] with domain consisting of all functions in the Hilbert Space with square integrable derivatives and whose values at -L and L are equal.

## The Attempt at a Solution

I apply the operators on the state ψ(x) to get the following expression for the LHS:

λψ(x) - [itex]\frac{\hbar}{i}[/itex]ψ'(x)

I then move the derivative term to the RHS and divide by lamda to get an expression for ψ(x):

[itex]\frac{\hbar}{iλ}[/itex]ψ'(x) + ([itex]\Phi[/itex](x))/λ.

Now here's the problem. If I take the absolute square of both sides and integrate, lamda just becomes a constant outside the integral. Now both integrals will converge, the second term due to it being part of the Hilbert Space and the first term being a prerequisite for being in the Hilbert Space. So the only value of λ that is not in the resolvent set (and hence in the spectrum) is 0. And even that is suspect, as that would mean we divided by zero in the initial steps. Even if I differentiate both sides before integrating (to test the second condition for being in the domain of p_hat), lamda still falls out, which instead leads to the entire complex plane being in the spectrum, as the derivative of [itex]\Phi[/itex] is not necessarily square integrable.

I can't make sense of both answers, so surely I've done something terribly wrong. What is it?