# From position space to momentum space

Lets say I have a state vector $\left|\Psi(t)\right\rangle$ in a position space with an orthonormal position basis. If I now use an operator $\hat{p}$ on this basis I will get basis which corresponds to a momentum space and projections of a $\left|\Psi(t)\right\rangle$ on these base vectors will now be $\left|\Psi(p,t)\right\rangle$?

In other words. Do operators transform basis or a state vector or both?

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– Qmechanic May 26 '13 at 10:12
Having a basis is extremely useful, but a vector in a vector space exists, independent of the basis. So, $|\Psi(t)\rangle$ is an element of a Hilbert space and that's that. It's just a function of time. The existence of a basis of a vector space is given by the axiom of choice, and we can decompose a vector in terms of basis vectors, but I reiterate that that does not define a vector. Btw, $|x\rangle$ is not strictly a basis set of the Hilbert space, not being normalizable... – nervxxx May 26 '13 at 14:57

The wavefunction vector $|\Psi (t) \rangle$ is supposed to be a function of time only. When you write $| \Psi (t) \rangle$ you are not considering the projection of the wavefunction nor on the position neither on the momentum space, but just the state of the system at time $t$, which is nothing but a postulate of Quantum Mechanics. You will have the wavefunction in coordinate (or momentum or any other observable) once you project your state vector on a basis of the observable you have chosen. For instance, in coordinate space: $$\langle \mathbf{x} | \Psi (t) \rangle := \Psi (\mathbf{x},t)$$. which is the probability amplitude of finding my system (here we have just one coordinate, so we suppose we are dealing with a single particle system) at position $\mathbf{x}$ at time $t$. If you want to switch from coordinate space to momentum space, i.e. you want to have the following probability amplitude: $$\langle \mathbf{p} | \Psi (t) \rangle = \tilde{\Psi}(\mathbf{p} ,t)$$ (where we have used $\tilde{\Psi}$ to mean that is not the same function of $\mathbf{p}$ as $\Psi$ was for $\mathbf{x}$), we can write like this: $$\tilde{\Psi}(\mathbf{p},t)=\int\,d\mathbf{x} \langle\mathbf{p}|\mathbf{x}\rangle\langle\mathbf{x}|\Psi(t)\rangle$$ for each $t$, having inserted the completeness relation of the space coordinate observable. Now, knowing that $\langle \mathbf{p} | \mathbf{x} \rangle = \frac{1}{\sqrt{2\pi \hbar}}\exp(\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{x}),$ you find that that projection of wavefunction in momentum space is the fourier transform of the coordinate-space wave function.

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But how do i get base vectors for momentum if i know base vectors of position space? – 71GA May 26 '13 at 11:17
It's the same thing, you have to make a fourier transform of the coordinate-space eigenstates, i.e.: $$|\mathbf{p}\rangle=\int \,d\mathbf{x} |\mathbf{x} \rangle \langle \mathbf{x} | \mathbf{p} \rangle$$ – user24959 May 26 '13 at 12:01

When you express a state vector in terms of a basis, you're expressing it as a sum of other state vectors:

$$\sum_i c_i | \Psi_i \rangle$$

If you have some linear operator that acts on a state to transform it into some other state or into some other basis, acting on the individual basis vectors and then doing the sum is the same as applying the operator to the total state.

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