How to prove Schrodinger's equation in momentum space?

[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.

 

[jtbell]

i do not know hot to express the potential in momentum space

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

 

[Joker93]

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

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

 

[drvrm]

What about the potential? How will we convert that to momentum space.

well i personally have not done calculations in momentum space but i can suggest you a site where such transformations have been done -i think its related to Fourier Transforms from coordinate to momentum space or vice versa.. see

<http://www.phys.kyushu-u.ac.jp/fb20/PresentationFiles/Parallel-VIIIb/FB20_Yamaguchi.pdf >

.

 

[blue_leaf77]

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

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.

 

[jtbell]

What about the potential? How will we convert that to momentum space.

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.

 

[samalkhaiat]

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.

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 Schrödinger’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
Start with
$$i\frac{\partial}{\partial t}\Psi (x,t) = - \frac{1}{2m} \nabla^{2} \Psi (x,t) + V(x) \Psi (x,t) .$$
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
$$i \partial_{t} \Phi (p,t) = \frac{p^{2}}{2m} \Phi (p,t) + \int d^{3}\bar{p} \ V(p-\bar{p}) \Phi (p,t) .$$

 

[Joker93]

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 Schrödinger’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
Start with
$$i\frac{\partial}{\partial t}\Psi (x,t) = - \frac{1}{2m} \nabla^{2} \Psi (x,t) + V(x) \Psi (x,t) .$$
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
$$i \partial_{t} \Phi (p,t) = \frac{p^{2}}{2m} \Phi (p,t) + \int d^{3}\bar{p} \ V(p-\bar{p}) \Phi (p,t) .$$

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!

 

[samalkhaiat]

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!

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