Wave mechanics: the adjoint of a hamiltonian

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Homework Help Overview

The problem involves the operator Q and its adjoint, focusing on the Hamiltonian H defined as H = αQQ†. The task is to demonstrate that H is self-adjoint, find an expression for H² in terms of H, and determine the eigenvalues of H based on the previous results. The context is within wave mechanics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the self-adjoint nature of operators of the form cQQ† and question the implications of finding H² in terms of H. There are inquiries about the adjoint of a product and the non-commutativity of operators. Some participants express confusion about the simplification of expressions involving Q and Q†.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem. Some have provided insights into the properties of the operators involved, while others are working through the implications of the equations presented. There is no explicit consensus yet, but productive guidance has been offered regarding the handling of operator products and the implications of the identities involving Q and Q†.

Contextual Notes

Participants are navigating the complexities of operator algebra, particularly the implications of the identities Q†Q† = 0 and QQ† + Q†Q = 1. There is an ongoing examination of how these identities affect the calculations and interpretations of H and H².

  • #31
gabbagabbahey said:
no, h is an operator, e is a scalar, they cannot possibly be equal!

e=e^2
 
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  • #32
noblegas said:
e=e^2

No, \alpha E\varphi=E^2\varphi, so (since \alpha and E are both scalars!) \alpha E=E^2...so E=___?
 
  • #33
gabbagabbahey said:
No, \alpha E\varphi=E^2\varphi, so (since \alpha and E are both scalars!) \alpha E=E^2...so E=___?

E=E^2/(/alpha) alpha isn't a matrix so I can divide alpha to the other side?
 
  • #34
E and alpha are numbers, yes. You want to solve alpha*E=E*E. That's the same as alpha*E-E*E=0. It's a quadratic equation. It has two solutions. Can you factor it?
 
Last edited:
  • #35
Dick said:
E and alpha are numbers, yes. You want to solve alpha*E=E*E. That's the same as alpha*E-E*E=0. It's a quadratic equation. It has two solutions. Can you factor it?

yes. E=0 or 1
 
  • #36
noblegas said:
yes. E=0 or 1

You are getting closer. E=0 solves it. Why do you think E=1 solves E*alpha-E*E=0?
 
  • #37
Dick said:
You are getting closer. E=0 solves it. Why do you think E=1 solves E*alpha-E*E=0?

E=1 ==> alpha =1 right?
 
  • #38
noblegas said:
E=1 ==> alpha =1 right?

I had really hoped you would figure out that E*alpha-E*E=0 means E*(E-alpha)=0. So E=0 or E-alpha=0. So E=0 or E=alpha. But that doesn't seem to be happening. There is nothing in the problem that requires E=1 any more than there is that E=56, is there?
 
  • #39
Dick said:
I had really hoped you would figure out that E*alpha-E*E=0 means E*(E-alpha)=0. So E=0 or E-alpha=0. So E=0 or E=alpha. But that doesn't seem to be happening. There is nothing in the problem that requires E=1 any more than there is that E=56, is there?

I've should have caught that ; Its been a long long ... long night.
 
  • #40
noblegas said:
I've should have caught that ; Its been a long long ... long night.

Granted, a long night. So alpha is sort of the fixed constant in the problem, right? You want to solve for E given the value of alpha. Are we agreed that either E=0 or E=alpha? If so then gabbagabbahey and everybody else can take a nap.
 
  • #41
Dick said:
Granted, a long night. So alpha is sort of the fixed constant in the problem, right? You want to solve for E given the value of alpha. Are we agreed that either E=0 or E=alpha? If so then gabbagabbahey and everybody else can take a nap.

yes.
 

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