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Joker93
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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.
Adam Landos said:i do not know hot to express the potential in momentum space
Hello and thanks for the reply. What about the potential? How will we convert that to momentum space.jtbell said:Hint: find the position operator in momentum space, analogous to the momentum operator in position space.
Adam Landos said: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.Adam Landos said:Hello and thanks for the reply. What about the potential? How will we convert that to momentum space.
Adam Landos said:What about the potential? How will we convert that to momentum space.
Adam Landos said: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.
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 [tex]\Phi(\bar{p},t) [/tex]samalkhaiat said:1) If you know Dirac’s notation
Start with [tex]i\partial_{t}|\Psi (t) \rangle = H (X , P) | \Psi (t) \rangle = \frac{P^{2}}{2m} |\Psi (t)\rangle + V(X) |\Psi (t) \rangle ,[/tex] where [itex]X[/itex] and [itex]P[/itex] are operators but [itex]x[/itex] and [itex]p[/itex] will be used to mean ordinary numbers. In fact, we will work in 3D, so the operators [itex]X = (X_{1},X_{2},X_{3})[/itex], [itex]P = (P_{1},P_{2},P_{3})[/itex], while [itex]x = (x_{1},x_{2},x_{3})[/itex] and [itex]p=(p_{1},p_{2},p_{3})[/itex]. Multiply from the left by the bra [itex]\langle p |[/itex] and use [tex]\langle p | \frac{P^{2}}{2m} = \frac{p^{2}}{2m}\langle p | .[/tex]
Introduce the wave function notation [itex]\langle p | \Psi (t) \rangle = \Psi (p ,t)[/itex]
[tex]i\partial_{t}\Psi (p,t) = \frac{p^{2}}{2m}\Psi (p,t) + \langle p | V(X) | \Psi (t)\rangle .[/tex] Now, let us work on the potential term. Insert the completeness relation [itex]\int d^{3}x \ |x\rangle \langle x | = 1[/itex] between [itex]\langle p|[/itex] and [itex]V(X)[/itex], then use
[tex]\langle x | V(X) | \Psi (t)\rangle = V(x) \langle x | \Psi (t)\rangle [/tex]
and
[tex]\langle p | x \rangle = e^{i x \cdot p } .[/tex]
So the potential term becomes
[tex]\langle p | V(X) | \Psi (t)\rangle = \int d^{3}x \ V(x) \ e^{i x \cdot p } \langle x | \Psi (t) \rangle .[/tex]
Now, insert the completeness relation [itex]\int d^{3}\bar{p} \ | \bar{p}\rangle \langle \bar{p} | = 1[/itex] between [itex]\langle x |[/itex] and [itex]| \Psi (t)\rangle[/itex] and use
[tex]\langle x | \bar{p}\rangle = e^{- i x \cdot \bar{p}}, \ \ \langle \bar{p}| \Psi (t)\rangle = \Psi ( \bar{p},t)[/tex]
So, the potential term becomes
[tex]\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) .[/tex]
The integral in the bracket is just the Fourier transform of [itex]V(x)[/itex]
[tex]V(p - \bar{p}) = \int d^{3}x \ V(x) \ e^{i x \cdot (p - \bar{p})} .[/tex]
So, we rewrite the potential term as
[tex]\langle p | V(X) | \Psi (t)\rangle = \int d^{3}\bar{p} \ V(p - \bar{p}) \Psi (p,t) ,[/tex]
and the whole Schrodinger’s equation becomes just an ordinary integral equation
[tex]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) .[/tex]
2) If you do not know the Dirac notation
Start with
[tex]i\frac{\partial}{\partial t}\Psi (x,t) = - \frac{1}{2m} \nabla^{2} \Psi (x,t) + V(x) \Psi (x,t) .[/tex]
Multiply by [itex]\exp (i x \cdot p )[/itex], integrate over [itex]x[/itex] and use
[tex] \Phi (p,t) = \int d^{3}x \ e^{i x \cdot p} \ \Psi (x,t) .[/tex]
In the differential term on the right, integrate by parts twice and neglect surface term:
[tex]\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) .[/tex]
Okay, now for the potential term, use
[tex]\Psi(x,t) = \int d^{3}y \ \Psi(y,t) \delta ( x - y) ,[/tex] and for the delta function, use the integral representation
[tex]\delta (x-y) = \int d^{3}\bar{p} \ e^{- i (x-y) \cdot \bar{p}} .[/tex]
So, after changing the order of integrations, the potential term can be written as
[tex]\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) .[/tex]
Well, the integrals in the brackets on right hand of this are just the Fourier transforms of [itex]V(x)[/itex] and [itex]\Psi (x,t)[/itex]. So we can rewrite the potential term as
[tex]\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) .[/tex]
And, the whole equation becomes
[tex]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) .[/tex]
Yes, clearly those were typos. Thanks, at least I know that you have read it all.Adam Landos said: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 [tex]\Phi(\bar{p},t) [/tex]
Oh, and thanks again for taking the time to derive it for me!
Schrodinger's equation in momentum space is a mathematical formula that describes the behavior of a quantum system in terms of its momentum. It is an extension of the original Schrodinger's equation, which describes the behavior of a quantum system in position space.
Schrodinger's equation in momentum space can be derived by applying a Fourier transform to the original Schrodinger's equation. This transforms the equation from position space to momentum space, allowing us to describe the system's behavior in terms of its momentum.
Schrodinger's equation in momentum space allows us to analyze quantum systems from a different perspective, which may provide insights and solutions that are not readily apparent in position space. It also simplifies certain calculations and makes it easier to describe certain physical phenomena, such as wave-particle duality.
Yes, Schrodinger's equation in momentum space is a fundamental equation in quantum mechanics and can be used to describe the behavior of any quantum system, regardless of its complexity.
Schrodinger's equation in momentum space is closely related to the Heisenberg uncertainty principle, which states that it is impossible to know both the position and momentum of a particle with absolute certainty. This is because the equation describes the probability of finding a particle in a certain momentum state, rather than a specific value.