# How to Prove ∫ ψ1(x)*ψ2(x) dx = ∫φ1(k)*φ2(k) dk?

• MrPhoton
In summary: 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
MrPhoton

## Homework Statement

Show that ∫ ψ1(x)*ψ2(x) dx = ∫φ1(k)*φ2(k) dk

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

## Homework Equations

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

## 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?)

MrPhoton said:
ψ(x) = 1/√2π∫φ(k)eikx dk
Just plug in the above relation for ##\psi_1## and ##\psi_2## in the left hand side of the relation to be proved.

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?

MrPhoton said:
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?
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.

MrPhoton said:
∫ ψ1(x)*ψ2(x) dx = 1/2π∫φ*(k)e-ikx φ(k)eikx dk
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.

## 1. What is an inner product in the context of wavefunctions?

An inner product is a mathematical operation that takes two wavefunctions, represented by complex-valued functions, and produces a scalar quantity. In the context of wavefunctions, the inner product is used to determine the overlap or similarity between two wavefunctions.

## 2. Why is the inner product important in quantum mechanics?

The inner product is important in quantum mechanics because it allows us to calculate probabilities and make predictions about the behavior of quantum systems. By taking the inner product of a wavefunction with itself, we can determine the magnitude (or probability) of finding a particle in a specific state.

## 3. How is the inner product related to the concept of orthogonality?

In quantum mechanics, two wavefunctions are said to be orthogonal if their inner product is equal to zero. This means that the two wavefunctions do not overlap or share any common characteristics. Orthogonality is an important concept in quantum mechanics because it allows us to distinguish between different quantum states.

## 4. Can the inner product be extended to higher-dimensional systems?

Yes, the inner product can be extended to higher-dimensional systems. In this case, the inner product is known as a multi-dimensional inner product and takes into account the complex-valued functions in each dimension. This is important in quantum mechanics when dealing with systems with multiple particles or degrees of freedom.

## 5. How is the inner product used to prove the completeness of wavefunctions?

The inner product is used in the proof of wavefunction completeness by showing that any arbitrary function can be expressed as a linear combination of the eigenfunctions of a given operator. This means that the set of eigenfunctions is a complete set, and any function can be written as a sum of these eigenfunctions, with coefficients determined by taking the inner product with the original function.

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