# How to Perform Fourier Transform on a Quantum State of Spin-1/2 Particle?

I am currently studying quantum mechanics and need help understanding how to perform the Fourier transform of a particular state. I have a spin-1/2 particle whose momentum and spin state at time $$t=0$$ is given by:

$$|\psi(0)\rangle = \frac{1}{(\pi \sigma^2)^{1/4}} \int_{-\infty}^{\infty} dx \, \exp\left[-\frac{x^2}{4 \sigma^2}\right] |x\rangle \otimes (\alpha |+\rangle + \beta |-\rangle)$$

Where $$|x\rangle$$ represents the position basis, and $$|+\rangle$$ and $$|-\rangle$$ are the spin up and down states, respectively. $$\alpha$$ and $$\beta$$ are complex numbers representing the spin state coefficients.

I'm trying to understand how to transform this state into the momentum (P) basis. Specifically, I need to calculate the Fourier transform of this state to find its representation in momentum space, but I'm uncertain how to handle the spin components in this transformation.

Could someone guide me through the process or suggest some useful references? I'm particularly interested in the correct approach to integrate the spin states into this transformation. Any help would be greatly appreciated!

• Quick answer: you treat each component independently. You can just pull the tensor product out of the integral and make the change of basis for the continuous variable Commented May 11 at 1:51
• To clarify: when performing the Fourier transform on $|\psi(0)\rangle$, each component (position and spin) is treated independently, correct? Does this mean I apply the Fourier transform only to the position part and keep the spin part unchanged? Can you confirm if my calculation below is correct? $\langle p | \psi(0) \rangle = \frac{1}{(\pi\sigma^2)^{1/4}} (\alpha |+\rangle + \beta |-\rangle) \int_{-\infty}^{\infty} dx \exp \left[ \frac{-x^2}{4\sigma^2} \right] \frac{e^{-ipx/\hbar}}{\sqrt{2\pi\hbar}}$ Commented May 11 at 2:04
• yes, you apply the Fourier transform only to the position part. I'm just checking really fast, so I might be missing something, but your expression seems correct Commented May 11 at 12:45

The fermion ladder operators carry all the degrees of freedom. If you define a basis as the eigenstates of some operator composed of the ladder operators, these eigenstates will also carry the same degrees of freedom. So, if we imagine that we can define a position basis in terms of an associated position operator and this position operator distinguishes the spin degree of freedom, then the eigenstates would be of the form $$|x,s\rangle$$ where $$s$$ represents a spin index. The Fourier transform will then be define for each spin index separately. In terms of such a representation, we can express the position basis as $$|x,s\rangle = \int |k,s\rangle \exp(i x k) dk ,$$ where $$k$$ is a wavenumber related to the momentum via the de Broglie relation. So, your expression in terms of $$|+\rangle$$ and $$|-\rangle$$ would contain the superposition $$|x,+\rangle\alpha + |x,-\rangle\beta$$. Now you can expand each term in terms of your momentum basis using the above integral expression to obtain the Fourier relationship. Hope that helps.