noblegas
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gabbagabbahey said:no, h is an operator, e is a scalar, they cannot possibly be equal!
e=e^2
gabbagabbahey said:no, h is an operator, e is a scalar, they cannot possibly be equal!
noblegas said:e=e^2
gabbagabbahey said:No, \alpha E\varphi=E^2\varphi, so (since \alpha and E are both scalars!) \alpha E=E^2...so E=___?
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?
noblegas said:yes. E=0 or 1
Dick said:You are getting closer. E=0 solves it. Why do you think E=1 solves E*alpha-E*E=0?
noblegas said:E=1 ==> alpha =1 right?
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?
noblegas said:I've should have caught that ; Its been a long long ... long night.
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.