How Do Wavefunctions in Coordinate and Momentum Spaces Form an Orthonormal Set?

[einai]

What's the wave function in coordinate space Ψ_x0(x') of a particle (in 1-D) located at a certain position x₀? What about the wave function Φ_x0(p') in momentum space? Now, consider the totality of these wave functions for different values of x₀. Do they form an orthonormal set?

The only thing I know is that if I know Ψ_x0(x'), I can Fourier transform it to Φ_x0(p')? But what's Ψ_x0(x')? I'm really confused.

Thanks in advance!

 

[R0man]

The wavefunction of a particle is a mathematical representation of its quantum state. It is a complex-valued function that describes the probability amplitude of finding the particle at a particular position or momentum.

In coordinate space, the wavefunction is denoted as Ψx(x'), where x' represents the position of the particle and x represents the position variable. This means that the value of Ψx(x') at a specific x' gives the probability amplitude of finding the particle at that position.

Similarly, in momentum space, the wavefunction is denoted as Φp(p'), where p' represents the momentum of the particle and p represents the momentum variable. The value of Φp(p') at a specific p' gives the probability amplitude of finding the particle with that momentum.

Now, if we consider the totality of these wavefunctions for different values of x0, where x0 represents the position of the particle, then yes, they do form an orthonormal set. This means that they are perpendicular to each other and have a magnitude of 1, which is a requirement for a valid wavefunction.

To answer your question, if you know Ψx0(x'), then you can indeed Fourier transform it to obtain Φx0(p'). This is because the Fourier transform is a mathematical operation that converts a function in one domain (coordinate space in this case) to a function in another domain (momentum space). It allows us to switch between representations of the same wavefunction.

I hope this helps to clarify your confusion. If you have any further questions, please feel free to ask.