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The eigenfunctions (eigenstates) for the momentum of a particle are given by the plane waves $$\phi(x,t) = \sin(kx - \omega t)$$ If we sum a large number of these waves in a range from $0$ to $k_m$, we get something like this: $$\psi(x, t) = \int_0^{k_m} \sin(kx -\omega t)dk = \frac{-\cos(k_m~x - \omega t) + \cos(\omega t)}{x} .$$

This represents a wave-packet concentrated at $x=0$. The bigger $k_m$ is, the narrower and higher the packet gets.

If we integrate over ALL $k$-values, from $-\infty$ to $+\infty$ we get an extremely narrow and infinitly high pattern when we plot it, and we refer to it by the delta-function $\delta(x)$. This detla-function is a position Eigenstate.

But why when we integrate $\delta(x)$ for all $x$ values, we don't get the momentum Eigenstates back? Insted we get $$\int_{-\infty}^{+\infty}\delta(x)dx = 1$$ and not a sine wave. What should I do to get back the momentum eingenstate?

Qmechanic
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Anky Physics
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First of all $\phi(x,t) = \sin(kx - \omega t)$ is not a momentum eigenstate. Its a super position of of $\exp(i(kx - \omega t))$ and $\exp(-i(kx - \omega t))$, which are momentum eigenstates for $+k$ and $-k$. However depending on how you define $\omega$ one of them does not fullfill the free particle Schrödinger equation, because their time development should be identical as they are both kinetic energy eigenstates with the same eigenvalue. Now since for your question time development is irrelevant I suggest we look at the state $\exp(i(kx))$ instead.

Your equation now takes the form $$\psi(x, t) = \int_{-k_m}^{k_m} \exp(ikx)dk = \frac{\exp(ikx) - \exp(-ikx)}{ix} = \frac{\sin(kx)}{x}$$

and in the limit of $k \rightarrow \infty$ you get $2 \pi \delta(x)$.

So it behaves similar similar to what you described.

What we found is that integrating over all momentum eigenstates, is the same as taking the Fourier transform of a constant in momentum space and yields a delta function in position space, which is a position eigenstate. Now I am not exactly sure what you expect happening when you integrate over the delta function, except that it should be normalized, if one would have used normalized momentum states, which I didn’t (hence the factor $2 \pi $). But to recover the wavefunction in momentum space you need to take the Fourier transform of the delta function, which yields a constant in momentum space and is exactly what you expect since you started out by integrating over all momentum eigenstates.

Zaph
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