# I How to prove Schrodinger's equation in momentum space?

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1. Mar 5, 2016

### Joker93

Specifically, i do not know hot to express the potential in momentum space. If someone would provide me with a link of source that has the proof in it, it would be appreciated.

2. Mar 5, 2016

### Staff: Mentor

Hint: find the position operator in momentum space, analogous to the momentum operator in position space.

3. Mar 5, 2016

### Joker93

Hello and thanks for the reply. What about the potential? How will we convert that to momentum space.

4. Mar 5, 2016

5. Mar 5, 2016

### blue_leaf77

In momentum space, you cannot consider the potential in a separate mathematical expression with the state it acts on. In position space, you have $V(x)\psi(x)$, then performing Fourier transform of this term will give you a convolution, $\int_{-\infty}^\infty \tilde{V}(p-p')\tilde{\psi}(p')dp'$ where $\tilde{V}(p)$ and $\tilde{\psi}(p)$ are the Fourier transforms of the potential and the state, respectively.

6. Mar 5, 2016

### Staff: Mentor

The potential is a function of position, right? Substitute the position operator for x. Consider how you do something similar for functions of momentum, e.g. p2, in position space.

7. Mar 5, 2016

### samalkhaiat

1) If you know Dirac’s notation
Start with $$i\partial_{t}|\Psi (t) \rangle = H (X , P) | \Psi (t) \rangle = \frac{P^{2}}{2m} |\Psi (t)\rangle + V(X) |\Psi (t) \rangle ,$$ where $X$ and $P$ are operators but $x$ and $p$ will be used to mean ordinary numbers. In fact, we will work in 3D, so the operators $X = (X_{1},X_{2},X_{3})$, $P = (P_{1},P_{2},P_{3})$, while $x = (x_{1},x_{2},x_{3})$ and $p=(p_{1},p_{2},p_{3})$. Multiply from the left by the bra $\langle p |$ and use $$\langle p | \frac{P^{2}}{2m} = \frac{p^{2}}{2m}\langle p | .$$
Introduce the wave function notation $\langle p | \Psi (t) \rangle = \Psi (p ,t)$
$$i\partial_{t}\Psi (p,t) = \frac{p^{2}}{2m}\Psi (p,t) + \langle p | V(X) | \Psi (t)\rangle .$$ Now, let us work on the potential term. Insert the completeness relation $\int d^{3}x \ |x\rangle \langle x | = 1$ between $\langle p|$ and $V(X)$, then use
$$\langle x | V(X) | \Psi (t)\rangle = V(x) \langle x | \Psi (t)\rangle$$
and
$$\langle p | x \rangle = e^{i x \cdot p } .$$
So the potential term becomes
$$\langle p | V(X) | \Psi (t)\rangle = \int d^{3}x \ V(x) \ e^{i x \cdot p } \langle x | \Psi (t) \rangle .$$
Now, insert the completeness relation $\int d^{3}\bar{p} \ | \bar{p}\rangle \langle \bar{p} | = 1$ between $\langle x |$ and $| \Psi (t)\rangle$ and use
$$\langle x | \bar{p}\rangle = e^{- i x \cdot \bar{p}}, \ \ \langle \bar{p}| \Psi (t)\rangle = \Psi ( \bar{p},t)$$
So, the potential term becomes
$$\langle p | V(X) | \Psi (t)\rangle = \int d^{3}\bar{p} \left( \int d^{3}x \ V(x) \ e^{i x \cdot (p - \bar{p})} \right) \ \Psi (\bar{p},t) .$$
The integral in the bracket is just the Fourier transform of $V(x)$
$$V(p - \bar{p}) = \int d^{3}x \ V(x) \ e^{i x \cdot (p - \bar{p})} .$$
So, we rewrite the potential term as
$$\langle p | V(X) | \Psi (t)\rangle = \int d^{3}\bar{p} \ V(p - \bar{p}) \Psi (p,t) ,$$
and the whole Schrodinger’s equation becomes just an ordinary integral equation
$$i\partial_{t}\Psi (p,t) = \frac{p^{2}}{2m} \Psi(p,t) + \int d^{3}\bar{p} \ V(p - \bar{p}) \Psi (p,t) .$$
2) If you do not know the Dirac notation
$$i\frac{\partial}{\partial t}\Psi (x,t) = - \frac{1}{2m} \nabla^{2} \Psi (x,t) + V(x) \Psi (x,t) .$$
Multiply by $\exp (i x \cdot p )$, integrate over $x$ and use
$$\Phi (p,t) = \int d^{3}x \ e^{i x \cdot p} \ \Psi (x,t) .$$
In the differential term on the right, integrate by parts twice and neglect surface term:
$$\int d^{3}x \ e^{i x \cdot p} \ \nabla^{2}\Psi (x,t) = \int d^{3}x \ \Psi (x,t) \nabla^{2}\left( e^{i x \cdot p} \right) = - p^{2} \Phi (p,t) .$$
Okay, now for the potential term, use
$$\Psi(x,t) = \int d^{3}y \ \Psi(y,t) \delta ( x - y) ,$$ and for the delta function, use the integral representation
$$\delta (x-y) = \int d^{3}\bar{p} \ e^{- i (x-y) \cdot \bar{p}} .$$
So, after changing the order of integrations, the potential term can be written as
$$\int d^{3}x \ V(x) \Psi(x,t) e^{i x \cdot p} = \int d^{3}\bar{p} \left( \int d^{3}x \ V(x) e^{i x \cdot (p - \bar{p}) } \right) \left( \int d^{3}y \ e^{ i y \cdot \bar{p}} \Psi(x,t) \right) .$$
Well, the integrals in the brackets on right hand of this are just the Fourier transforms of $V(x)$ and $\Psi (x,t)$. So we can rewrite the potential term as
$$\int d^{3}x \ V(x) \Psi(x,t) e^{i x \cdot p} = \int d^{3}\bar{p} \ V(p-\bar{p}) \Phi (p,t) .$$
And, the whole equation becomes
$$i \partial_{t} \Phi (p,t) = \frac{p^{2}}{2m} \Phi (p,t) + \int d^{3}\bar{p} \ V(p-\bar{p}) \Phi (p,t) .$$

Last edited: Mar 5, 2016
8. Mar 10, 2016

### Joker93

Thanks, but i should point a mistake. In the third line from the end, in the third bracket, it must be Ψ(y,t) rather than Ψ(x,t). Also, in the last line, it must be $$\Phi(\bar{p},t)$$
Oh, and thanks again for taking the time to derive it for me!

Last edited: Mar 10, 2016
9. Mar 10, 2016

### samalkhaiat

Yes, clearly those were typos. Thanks, at least I know that you have read it all.