Wave mechanics: the adjoint of a hamiltonian

[Dick]

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?

 

[noblegas]

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?

 

[Dick]

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?

 

[noblegas]

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.

 

[Dick]

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.

 

[noblegas]

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.