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.


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?

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    $\begingroup$ Of course it has to be stated differently, differentiating $\Phi$ with respect to $x$ doesn't make any sense! Just read on; whatever you're reading should explain how to express operators in different bases, too. $\endgroup$
    – ACuriousMind
    Commented Apr 13, 2016 at 19:08
  • $\begingroup$ @ACuriousMind Might be stupid question but further on there is an example. Example: Imagine a system in which there are just two linearly independent states: $$|1\rangle =\begin{pmatrix}1\\0\end{pmatrix}~~~\text{and}~~~|2\rangle=\begin{pmatrix}0\\1\end{pmatrix}$$ The most general state is a normalized linear combination: $$\ \mathcal{S} = a|1\rangle + b|2\rangle= \begin{pmatrix}a\\b\end{pmatrix}$$ with $|a|^2+|b|^2=1$. The (time-dependent) Schrodinger equation says $i\hbar \frac{d}{dt}| \mathcal{S} \rangle = H |\mathcal{S} \rangle$. Why is the Schrodinger equation unchanged in this example? $\endgroup$
    – Alex
    Commented Apr 13, 2016 at 19:22
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    $\begingroup$ It doesn't change because everything there is in matrix or vector form. If you expanded things out into components (like $H_{12}$ and so on) those components would depend on the basis. $\endgroup$
    – knzhou
    Commented Apr 13, 2016 at 19:22
  • $\begingroup$ @knzhou Thanks for your response. Why does writing in matrix or vector form allow you not to change the Shrodinger equation? And what exactly do you mean by expanding it out in component form? $\endgroup$
    – Alex
    Commented Apr 13, 2016 at 19:28
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    $\begingroup$ When you say e.g. there's a rotation $M$ which, being applied to a vector $\vec v$, gives you a new vector $\vec p$, you can write it in two ways: $$M\vec v=\vec p,\tag1$$ $$\sum_j M_{ij}v_j=p_i.\tag2$$ The first one is coordinate-independent (general), while the second one is written in coordinates and will change for a particular set of basis vectors you choose (e.g. $\vec x,\vec y,\vec z$ vs $\vec a,\vec b,\vec c$), when you write out the matrix and vector exactly. $\endgroup$
    – Ruslan
    Commented Apr 13, 2016 at 19:40

2 Answers 2


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

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


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

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

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


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?)

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    $\begingroup$ Thanks. Is the full working: $$ i \hbar \frac{\partial}{\partial t} | \Psi(t) \rangle = \frac{\hat{p}^2}{2m}| \Psi(t) \rangle + V| \Psi(t) \rangle \\ \therefore i \hbar \langle p| \frac{\partial}{\partial t} | \Psi(t) \rangle = ( \frac{1}{2m})\langle p| \hat{p}^2 | \Psi(t) \rangle + V(p)\langle p | \Psi(t) \rangle \\ \therefore i \hbar \frac{\partial}{\partial t} \langle p | \Psi(t) \rangle = \frac{p^2}{2m} \langle p| \Psi(t) \rangle + V(p)\langle p| \Psi(t) \rangle \\ \therefore i \hbar \frac{\partial}{\partial t}\Psi(p,t) = \frac{p^2}{2m}\Psi(p,t) + V(p)\Psi(p,t)$$ $\endgroup$
    – Alex
    Commented Apr 14, 2016 at 11:57
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    $\begingroup$ @Alex, momentum space potential $V(p)$ is convolved with the momentum space wave function. $\endgroup$ Commented Apr 14, 2016 at 15:56

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