# Why fourier transforms in QM

• pivoxa15
In summary, the use of Fourier transforms in quantum mechanics implies that the psi function operates with a continuous range of frequencies rather than discrete ones. This is because momentum is not quantized in quantum mechanics. Fourier transforms can only be applied to free particles because they have a continuous range of energies. For particles that are not free, Fourier transforms cannot be used as the frequencies are not continuous. In quantum mechanics, when a measurement is made, the possible measured values are the eigenvalues of the corresponding eigenvalue equation. For momentum, the eigenvalue equation has solutions that can take any real value, resulting in a continuous range of frequencies. This is true regardless of whether the particle is trapped or free. However, momentum is related to energy, which is
pivoxa15
Fourier transforms imply that the waves are added up with a continuous range of frequencies hence wavelengths instead of discrete numbers.

In QM Fourier transforms are used which imply the psi function employs a continuous rather than discrete range of frequencies. Why is this? Is it because psi wave is not a mechanical wave so it dosen't matter wheither the frequencies are continuous or discrete?

However, I have only seen Fourier transforms done on plane waves hence on free particles which can obtain a continuous range of energies via E=hf. So f takes on a continuous range hecne both wavenumer, k and period length lamda will all be continous. That is why the Fourier transform can be done on free particles. However, if the particle is not free than Fourier transforms cannot be used as the frequencies will not be continous, wavenumber and period will not be continous.

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The frequencies are continuous rather than discrete simply because momentum is not quantized.

Remember, in QM, when a measurement is made, the only possible measured values are the eigenvalues of the eigenvalue equation. In the case of momentum, the momentum operator is $\hat{p}=-i\hbar\partial_x$, and so the eigenvalue equation $$\hat{p}u_p(x)=pu_p(x)[/itex] has solutions $u_p(x)=(2\pi\hbar)^{(-1/2)}\exp(ipx/\hbar)$ and p can take any real value. And how do we know what value a measurement will yield? Well be first expand the wave function in terms of the eigenfunctions of the quantity we want to measure. In the case of the momentum, since there are an infinity of eigenfunctions, the expansion actually looks like a Fourier integral [tex]\psi_(x)=\int_{-\infty}^{+\infty}\phi(p)u_p(x)dp$$

and indeed it is one because $u_p(x)$ is actually a complex exponential in px.

And $|\phi(p)|^2$ is the probability distribution for measuring p.

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So it has nothing to do with whether a particle is trapped or free. Momentum is always continuous and never quantised for a particle in whatever situation? But momentum is related to energy and energy is quantised in for example the square well. So momentum in the square is also quantised? Hence momentum is not always continous?

## 1. Why do we use Fourier transforms in quantum mechanics?

The use of Fourier transforms in quantum mechanics allows us to express the state of a quantum system in terms of its constituent energy states. This is particularly useful in solving the time-independent Schrödinger equation, as it allows us to find the energy eigenstates and eigenvalues of a system.

## 2. How do Fourier transforms relate to the uncertainty principle?

The uncertainty principle states that we cannot simultaneously know the exact position and momentum of a particle. In quantum mechanics, the position and momentum of a particle are related through the wavefunction, which can be expressed as a Fourier transform. This means that the uncertainty in position and momentum can be mathematically described through Fourier transforms.

## 3. Can Fourier transforms be used in all quantum systems?

Yes, Fourier transforms can be used in all quantum systems, as they are a fundamental tool in understanding the behavior of quantum particles. However, the specific use and application of Fourier transforms may vary depending on the system and the problem at hand.

## 4. How do Fourier transforms help with the visualization of quantum systems?

Fourier transforms allow us to transform a wavefunction from its original position space to momentum space, which can often provide a clearer picture of the quantum system. This is because in momentum space, the wavefunction is represented as a series of discrete energy states, making it easier to visualize and analyze.

## 5. Are there any limitations to using Fourier transforms in quantum mechanics?

While Fourier transforms are a powerful tool in quantum mechanics, they do have some limitations. They are most useful in systems with discrete energy states, and may not be as effective in systems with continuous energy spectra. Additionally, the use of Fourier transforms in complex systems with multiple particles can become mathematically challenging.

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