Wave mechanics: the adjoint of a hamiltonian

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The discussion revolves around proving that the Hamiltonian H, defined as H = αQQ†, is self-adjoint, where Q satisfies specific operator equations. Participants clarify that any operator of the form cQQ† is self-adjoint, and they derive the expression for H², concluding H² = α²QQ†. They explore the implications of this result for the eigenvalues of H, leading to the quadratic equation αE = E², which yields solutions E = 0 or E = α. The conversation emphasizes the importance of operator non-commutativity and the implications of the operator identities provided.
  • #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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