# Inner Product/Wavefunction Proof

Tags:
1. Oct 28, 2015

### MrPhoton

1. The problem statement, all variables and given/known data
Show that ∫ ψ1(x)*ψ2(x) dx = ∫φ1(k)*φ2(k) dk

(Where the integrations are going from -∞ to ∞)

2. Relevant equations
1. Plancherel Theorem: ψ(x) = 1/√2π∫φ(k)eikx dk ⇔ φ(k) = 1/√2π∫ψ(x)e-ikx dx

3. The attempt at a solution
It is clear that Plancherel's theorem must be used to solve this.

We also know by the properties of a wavefunction, that the inner product is orthogonal therefore equal to 0.

Since I know the inner product on the right hand will be 0, I am not quite sure how to show that the left hand side is also equal to the right hand side. (I assume the 1/√2π factors will be removed by the inner product being 0?)

2. Oct 29, 2015

### blue_leaf77

Just plug in the above relation for $\psi_1$ and $\psi_2$ in the left hand side of the relation to be proved.

3. Oct 29, 2015

### MrPhoton

ψ1(x)* = 1/√2π∫φ*(k)e-ikx dk
ψ2(x) = 1/√2π∫φ(k)eikx dk

Therefore applying LHS of Initial equation, we get:

∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k)e-ikx φ(k)eikx dk
∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k) φ(k) dk
∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k) φ(k) dk = 0 (By Orthogonality)
∫ ψ1(x)*ψ2(x) dx = ∫φ*(k) φ(k) dk = 0 = RHS of initial equation

Something like this?

4. Oct 29, 2015

### nrqed

Watch out: in the expressions for psi_1 and for psi_2 you must use two different variables for the corresponding Fourier transforms. For example, use k in the FT of psi_1 and use k' in the FT of psi_2. If you have learned about the Dirac delta function, you will get the answer very easily.
By the way, you do not have (and should not) assume that the integral is zero. The proof is general, it does not require psi_1 and psi_2 to be orthogonal.

5. Oct 29, 2015

### blue_leaf77

You already made a couple of mistakes there. First, the wavefunctions in momentum space corresponding to $\psi_1(x)$ and $\psi_2(x)$ cannot be the same as Fourier transform is a linear operation. Second, you should just first plug in $\psi_1^*(x) = \int \phi_1^*(k)e^{-ikx} dk$ and $\psi_2(x) = \int \phi_2(k)e^{ikx} dk$ into the inner product expression, and the integration over $x$ should remain at this step.