Is a Local Potential Always Diagonal in Quantum Mechanics?

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SUMMARY

If V is a local potential in quantum mechanics, then the matrix element is indeed diagonal. This conclusion arises from the definition of a local potential, which depends solely on the position of the particle and not on the position of the interacting particle. The Lippmann-Schwinger equation requires the potential to be local and diagonal to ensure consistency in solving for the wave function. Non-diagonal potentials, such as those involving external electromagnetic fields, do not qualify as local potentials.

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jonas_nilsson
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Hello,

in my lecture notes I have made a note that if V is a local potential, then <r|V|r'> is diagonal. Is this true, and how come?

It popped up with the Lippmann-Schwinger equation, and I know it's important that V is local, but does really locality implicate diagonality?


Jonas
 
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jonas_nilsson said:
Hello,

in my lecture notes I have made a note that if V is a local potential, then <r|V|r'> is diagonal. Is this true, and how come?

It popped up with the Lippmann-Schwinger equation, and I know it's important that V is local, but does really locality implicate diagonality?


Jonas

I think no. Take the classical schroedinger equation with an external electromagnetic field (A,V). The interaction 1/2m(A.P+P.A+A^2) (e=c=1) is not diagonal in |r>.
However, you may say that the interaction is not a potential and a potential is of the form V(r) (grad V(r)= f(r) for a local local interaction) hence the diagonality of a potential.

Therefore it is a matter of words.

Seratend.
 


Hello Jonas,

Yes, it is true that if V is a local potential, then <r|V|r'> is diagonal. This is because a local potential is one that only depends on the position of the particle, and not on the position of the particle it is interacting with. This means that for any given position r, the potential V will have the same value regardless of the position r' of the interacting particle. This results in the matrix element <r|V|r'> being non-zero only when r = r', making it diagonal.

As for the connection to the Lippmann-Schwinger equation, the equation is used to solve for the wave function of a particle interacting with a potential. In order for the equation to be valid, the potential must be local and therefore diagonal. This is because the Lippmann-Schwinger equation relies on the potential being well-defined at every point in space, and a non-diagonal potential would lead to inconsistencies in the equation.

I hope this helps clarify the concept of locality and diagonality in potentials. Let me know if you have any further questions.


 

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