# 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?

• Commented May 26, 2013 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... Commented May 26, 2013 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.
• It's the same thing, you have to make a fourier transform of the coordinate-space eigenstates, i.e.: $$|\mathbf{p}\rangle=\int \,d\mathbf{x} |\mathbf{x} \rangle \langle \mathbf{x} | \mathbf{p} \rangle$$ Commented May 26, 2013 at 12:01
$$\sum_i c_i | \Psi_i \rangle$$