Solve Evolution Operator: <J|U|E>=Exp[-iEt/h]*<J|E>??

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Homework Help Overview

The discussion revolves around the evolution operator in quantum mechanics, specifically the expression =Exp[-iEt/h]*, where U is the evolution operator and H is the Hamiltonian. Participants explore the implications of this equation and its derivation.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the meaning of the symbol "*" in the equation, questioning whether it represents multiplication or complex conjugation. There is also mention of using Taylor expansion to derive the equation.

Discussion Status

Some participants express gratitude for clarifications regarding the multiplication aspect of the equation. Others share insights about applying functions of operators to eigenstates, indicating a productive exchange of ideas. A new question about the definition of "two-level systems" in quantum mechanics is introduced, suggesting ongoing exploration of related concepts.

Contextual Notes

One participant notes a desire to avoid creating a new thread for a related question, indicating a preference for maintaining the discussion within the current context.

JK423
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[SOLVED] Evolution Operator

If U=Exp[-iHt/h] is the evolution operator, and H the hamiltonian one,
i can't figure out how to derive the following equation.
Let`s call |E> an eigenstate of H, E it`s eigenvalue and |J> a random basis.
Then:
<J|U|E>=Exp[-iEt/h]*<J|E>.

??
 
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JK423 said:
If U=Exp[-iHt/h] is the evolution operator, and H the hamiltonian one,
i can't figure out how to derive the following equation.
Let`s call |E> an eigenstate of H, E it`s eigenvalue and |J> a random basis.
Then:
<J|U|E>=Exp[-iEt/h]*<J|E>.

??

what is that "*" mean? complex conjugate (cc)?? or just multiplication?

Use

[tex]e^{x}=\sum_n x^n/n![/tex] and [tex]H|E\rangle = E|E\rangle[/tex]

then you can get that equation. I think that "*" is just a multiplication instead of a cc. :smile:
 
Ohh it`s just a multiplication! Sorry for that.. :rolleyes:
Thanks a lot... I would never have thought of expanding [tex]e^{x}[/tex]!
 
JK423 said:
Ohh it`s just a multiplication! Sorry for that.. :rolleyes:
Thanks a lot... I would never have thought of expanding [tex]e^{x}[/tex]!

It's a very useful trick~try to make good use of it~
 
JK423 said:
Ohh it`s just a multiplication! Sorry for that.. :rolleyes:
Thanks a lot... I would never have thought of expanding [tex]e^{x}[/tex]!

If you have an operator A and an eigenstate of a with eigenvalue "a" (so A|a> = a |a> ) then any function f(A) which can be Taylor expanded will give

f(A) |a> = f(a) |a>

in other words, applying the function of the operator on an eigenstate gives the function of the eigenvalue.
 
nrqed said:
If you have an operator A and an eigenstate of a with eigenvalue "a" (so A|a> = a |a> ) then any function f(A) which can be Taylor expanded will give

f(A) |a> = f(a) |a>

in other words, applying the function of the operator on an eigenstate gives the function of the eigenvalue.

Yeah, i just proved it. Very usefull!
Thanks for ur help!
 
Hello again!
A minor question came up, and i really don't want to make a new thread for it so i post it here.
In quantum mechanics, what`s the definition of the "two-level systems"? I understand that the state vector is in the form |Ψ>=a|1>+b|2>, where |1>,|2> is a basis of the state space.
Then i think of the particle in a box. The energy is quantized (lets say that the possible values are E1 and E2) while position x is continuous. So, in the first case we would have: |Ψ>=a|E1>+b|E2> and in the second one: |Ψ>=Integral(Ψ(x) |x> dx).
So if we use as a basis the eigenstates of the energy, our system would be a "two-level system". However, in {x} representation, we would have an "infinite-level system".

So what`s the definition of a "two-level system" since the number of levels depend on the basis we use?


*EDIT*: If it`s not permitted to ask irrelevant to the "Evolution Operator" questions, pls let me know so that i`ll make a new thread.
 
Last edited:

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