Wave mechanics: the adjoint of a hamiltonian

noblegas
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Homework Statement



The operator Q satisfies the two equations

Q^{\dagger}Q^{\dagger}=0 , QQ^{\dagger}+Q^{\dagger}Q=1

The hamiltonian for a system is

H= \alpha*QQ^{\dagger},

Show that H is self-adjoint

b) find an expression for H^2 , the square of H , in terms of H.

c)Find the eigenvalues of H allowed by the result from part(b) .

where \alpha is a real constant

Homework Equations


The Attempt at a Solution



QQ^{\dagger}=1-Q^{\dagger}Q

H=\alpha*QQ^{\dagger}=\alpha*(1-Q^{\dagger}Q)

self adjoint of an operator

(Q^{\dagger}\varphi,\phi)=(\varphi,Q\phi) Should I take the conjugate of the operator H?

b)H^2=(\alpha)*(1-Q^{\dagger}Q)(\alpha)*(1-Q^{\dagger}Q)=(\alpha)^2*(1-Q^{\dagger}Q-Q^{\dagger}Q+Q^{\dagger}Q^{\dagger}QQ.) since Q^{\dagger}Q^{\dagger}=0 then the expression for H^2 is : H^2=(\alpha)^2*(1-Q^{\dagger}Q-Q^{\dagger}Q=(\alpha)^2(1-2*Q^{\dagger}Q). Now what?
 
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a) any operator of the form cQQ^\dagger, c \in R, is self-adjoint. You don't need to use any other equation to prove this.

b) What does finding an expression for H^2 in terms of H mean? H^2 is an expression for H^2 in terms of H.
 
Preno said:
a) any operator of the form cQQ^\dagger, c \in R, is self-adjoint. You don't need to use any other equation to prove this.

b) What does finding an expression for H^2 in terms of H mean? H^2 is an expression for H^2 in terms of H.

a)show H=H^{\dagger}? if soH^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)
 
What is the adjoint of a product?
 
Preno said:
What is the adjoint of a product?

i don't know (PQ)^{\dagger}=Q^{\dagger}P^{\dagger}?
 
Well, yeah, so you just apply this rule to the definition of H and use the fact that (Q^\dagger)^\dagger=Q.
 
Preno said:
Well, yeah, so you just apply this rule to the definition of H and use the fact that (Q^\dagger)^\dagger=Q.

I did that: <br /> H^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)<br />
 
So did I apply the rule correctly? Hope there is no hard feelings
 
er... bump! I am having trouble with part b) H^2=HH=(\alpha)QQ^{\dagger}*(\alpha)QQ^{\dagger}=(\alpha)^2*Q^{\dagger}Q(1-QQ^{\dagger})=(\alpha)^2*(Q^{\dagger}Q-QQ^{\dagger}Q^{\dagger}Q. Using the fact that Q^{\dagger}Q^{\dagger}=0 H^2 the equals: H^2=(\alpha)^2*(Q^{\dagger}Q); Shouldn't (Q^{\dagger}Q) be squared?

c) H*\varphi=E*\varphi ; Not sure how to finish this problem.
 
  • #10
noblegas said:
er... bump! I am having trouble with part b) H^2=HH=(\alpha)QQ^{\dagger}*(\alpha)QQ^{\dagger}=(\alpha)^2*Q^{\dagger}Q(1-QQ^{\dagger})=

Shouldn't this be H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q)?
 
  • #11
noblegas: you need to pay more careful attention to non-commutativity. You seem to be randomly swapping Q Q^\dagger with Q^\dagger Q in your expressions... and in general that is simply not allowed when working with operators.


You should not be surprised to see Q behaving degenerately -- that's one of the big things that the identity Q^\dagger Q^\dagger = 0 tells you!

I don't know if it will help your intuition or not, but that means Q^\dagger will, heuristically speaking, behave something like an "infinitessimal" number whose square is negligible.


Now, rather than doing things haphazardly, you can try and systematically rewrite equations. The original identities say that Q^\dagger Q^\dagger is trivial, and that \{ Q Q^\dagger, Q^\dagger Q \} is a linearly dependent set. (What about QQ?) It might help if you can decide upon a standard form for any products of Q and Q^\dagger, and use these relations to convert any product into the standard form.
 
  • #12
gabbagabbahey said:
Shouldn't this be H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q)?

yes. Thats what I meant to write. Nevertheless the second term in the impression goes away.
. How should I approach the eigenvalue problem? I know that H*\varphi =E*\varphi ==&gt; \alpha*QQ^{\dagger}*\varphi=E*\varphi
 
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  • #13
noblegas said:
yes. Thats what I meant to write. Nevertheless the second term in the impression goes away.
.

Right, the second term goes away (since Q^{\dagger}Q^{\dagger}=0)...and the first term is____?
 
  • #14
Hurkyl said:
noblegas: you need to pay more careful attention to non-commutativity. You seem to be randomly swapping Q Q^\dagger with Q^\dagger Q in your expressions... and in general that is simply not allowed when working with operators.



You should not be surprised to see Q behaving degenerately -- that's one of the big things that the identity Q^\dagger Q^\dagger = 0 tells you!

I don't know if it will help your intuition or not, but that means Q^\dagger will, heuristically speaking, behave something like an "infinitessimal" number whose square is negligible.


Now, rather than doing things haphazardly, you can try and systematically rewrite equations. The original identities say that Q^\dagger Q^\dagger is trivial, and that \{ Q Q^\dagger, Q^\dagger Q \} is a linearly dependent set. (What about QQ?) It might help if you can decide upon a standard form for any products of Q and Q^\dagger, and use these relations to convert any product into the standard form.
Sorry yahiko, I know matrices don't Commute. I mistakenly wrote down the wrong expression
 
  • #15
gabbagabbahey said:
Right, the second term goes away (since Q^{\dagger}Q^{\dagger}=0)...and the first term is____?

<br /> H^2=\alpha^2QQ^{\dagger}<br /> <br />

that can't be right because the QQ^{\dagger} is not squared
 
  • #16
Why would the QQ^{\dagger} term have to be squared?

Anyways, H^2=\alpha^2QQ^{\dagger}...and \alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha__?
 
  • #17
gabbagabbahey said:
Why would the QQ^{\dagger} term have to be squared?

Anyways, H^2=\alpha^2QQ^{\dagger}...and \alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha__?

Because HH=(\alpha)QQ^{\dagger}(\alpha)QQ^{\dagger}
 
  • #18
noblegas said:
Because HH=(\alpha)QQ^{\dagger}(\alpha)QQ^{\dagger}

Yes, but the conditions Q^{\dagger}Q^{\dagger}=0 and QQ^{\dagger}+Q^{\dagger}Q put restrictions on H, which allow you to simplify (QQ^{\dagger})^2 to \alpha QQ^{\dagger}...which is exactly what you've just shown.

It's no different from saying x^2=1 when x=1.

Anyways, H^2=\alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha__?
 
  • #19
gabbagabbahey said:
Yes, but the conditions Q^{\dagger}Q^{\dagger}=0 and QQ^{\dagger}+Q^{\dagger}Q put restrictions on H, which allow you to simplify (QQ^{\dagger})^2 to \alpha QQ^{\dagger}...which is exactly what you've just shown.

It's no different from saying x^2=1 when x=1.

Anyways, H^2=\alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha__?

ah! &lt;br /&gt; H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q=(\alpha*H)&lt;br /&gt;
 
  • #20
Right, H^2=\alpha H...Now use that for part (c)... operate on both sides of your eignvalue equation with H...what do you get?
 
  • #21
gabbagabbahey said:
Right, H^2=\alpha H...Now use that for part (c)... operate on both sides of your eignvalue equation with H...what do you get?

assumming that H*\varphi=E*\varphi

H^2=\alpha*H=\alpha*E
 
  • #22
noblegas said:
assumming that H*\varphi=E*\varphi

H^2=\alpha*H=\alpha*E

No, you can't just forget about \varphi like that.

H\varphi=E\varphi \implies H^2\varphi=H(E\varphi)=E(H\varphi)

But H^2=\alpha H[/tex], so...
 
  • #23
gabbagabbahey said:
No, you can't just forget about \varphi like that.

H\varphi=E\varphi \implies H^2\varphi=H(E\varphi)=E(H\varphi)

But H^2=\alpha H[/tex], so...
<br /> <br /> H\varphi=E\varphi \implies H^2\varphi=H(E\varphi)=E(H\varphi)=E^2\varphi
 
  • #24
noblegas said:
H\varphi=E\varphi \implies H^2\varphi=H(E\varphi)=E(H\varphi)=E^2\varphi

Yes, but you also know H^2=\alpha H, so...
 
  • #25
gabbagabbahey said:
Yes, but you also know H^2=\alpha H, so...

so \alpha H*\varphi=E^2*\varphi
 
  • #26
Right, and \alpha H\varphi= \alpha (H\varphi)=\alpha___?
 
  • #27
gabbagabbahey said:
Right, and \alpha H\varphi= \alpha (H\varphi)=\alpha___?

can't believe I left out alpha. \alpha*E^2 \varphi
 
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  • #28
No, H^2\varphi=\alpha H\varphi= \alpha E\varphi,, and you showed in post #23, H^2\varphi also must equal E^2\varphi, so \alpha E\varphi=E^2\varphi, right?

And therfor E=___?
 
  • #29
gabbagabbahey said:
no, h^2\varphi=\alpha h\varphi= \alpha e\varphi,, and you showed in post #23, h^2\varphi also must equal e^2\varphi, so \alpha e\varphi=e^2\varphi, right?

And therfor e=___?

e=h=e^2 sorry. don't know why latex will not capitalized E and H
 
  • #30
noblegas said:
e=h=e^2 sorry. don't know why latex will not capitalized E and H

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

e=e^2
 
  • #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?
 
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  • #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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