Schrödinger equation in momentum space

Question

In literature on an introduction to quantum mechanics which I am working through, there is a section which explains that a vector has different representations based on the basis you choose. It then makes a statement that

The same is true for the state of a system in quantum mechanics. It is represented by a vector, $\lvert\mathcal{S(t)}\rangle$, that lives "out there in Hilbert space," but we can express it with respect to any number of different bases. The wave function $\Psi(x,t)$ is actually the coefficient in the expansion of $\mathcal{S(t)}$ in the basis of position eigenfunctions: $$\Psi(x,t) = \langle x | \mathcal{S(t)} \rangle$$ (with $|x\rangle$ standing for the eigenfunction of $\hat{x}$ with eigenvalue $x$), whereas the momentum space wavefunction $\Phi(p,t)$ is the expansion of $| \mathcal{S} \rangle$ in the basis of momentum eigenfunctions: $$\Phi(p,t) = \langle p | \mathcal{S}(t) \rangle$$ (with $|p \rangle$ standing for the eigenfunction of $\hat{p}$ with eigenvalue $p$).

It then states that $\Psi$ and $\Phi$ contain the same information and describe the same state.

Question:

If we decide to work in the momentum space (or space with any other basis), how does this affect the time-dependent Schrodinger equation $$i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 \Psi}{\partial x^2} + V \Psi?$$ Would it be stated differently?

Answer 1

As a complement to another answer, I'll demonstrate moving from the coordinate representation (wave function) to the momentum representation.

Recalling that

$$\Psi(x,t) = \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$

and putting this expression into the (coordinate representation of the) TDSE, we have

$$i\hbar\frac{\partial}{\partial t}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) + V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)$$

We can move the partials inside the integrals but we must be careful with the potential. Noting that

$$\Phi(p,t) = \int dx \frac{e^{-i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \Psi(x,t)$$

we have

$$V(x)\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t) = \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)$$

But,

$$ \int dx'\frac{e^{-i\frac{p}{\hbar}x'}}{\sqrt{2\pi\hbar}}V(x')\Psi(x',t)= V(p) * \Phi(p,t)$$

where $*$ denotes convolution. Thus, we can write

\begin{align} \int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left(i\hbar\frac{\partial}{\partial t}\Phi(p,t)\right) &= -\int dp \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\Phi(p,t)\right)\\&\qquad + \int dp\frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}}\left( V(p) * \Phi(p,t)\right) \end{align}

leading to

$$\int dp \frac{e^{i\frac{p}{\hbar}x}}{\sqrt{2\pi\hbar}} \left\{ i\hbar\frac{\partial}{\partial t}\Phi(p,t) = -\frac{\hbar^2}{2m}\left(-\frac{p^2}{\hbar^2}\Phi(p,t) \right) + V(p)*\Phi(p,t)\right\}$$

and so

$$i\hbar\frac{\partial}{\partial t}\Phi(p,t) = \frac{p^2}{2m}\Phi(p,t) + V(p)*\Phi(p,t)$$

Answer 2

The safest way to start with is the representation-free Schrodinger equation, $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \hat{H}|\Psi(t)⟩. $$ Referring to your case, we take the separable Hamiltonian: $H=\frac{p^2}{2m}+V$ so that $$ i\hbar\frac{\partial}{\partial t}|\Psi(t)⟩= \left(\frac{\hat{p}^2}{2m}+V\right)|\Psi(t)⟩. $$ Now is the time the representation thing has to come into play. To work in momentum space, you have to project the equation to momentum basis (i.e. multiply with $\langle p|$). It can be seen that $\langle p|\Psi(t)\rangle=\Psi(p,t)$ is an eigenfunction of operator $p$, so that $$\hat{p}\Psi(p,t)=p\Psi(p,t)$$ (note that $p$ in the RHS is a number, not an operator). You can see here we no longer use $\hat{p}$ as a derivative of $x$ which is making no sense while operating to a function of $p$.

For the potential part, just make sure you have converted $V(x)$ to $V(p)$ by using $$\hat{x}=i\hbar\frac{\partial}{\partial p}. $$ (Check out How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?)