# Wave mechanics: the adjoint of a hamiltonian

1. Sep 15, 2009

### noblegas

1. The problem statement, all variables and given/known data

The operator Q satisfies the two equations

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

The hamiltonian for a system is

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

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

2. Relevant equations

3. The attempt at a solution

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

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

$$(Q^{\dagger}\varphi,\phi)=(\varphi,Q\phi)$$ 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?

Last edited: Sep 15, 2009
2. Sep 15, 2009

### Preno

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 $H$.

3. Sep 15, 2009

### noblegas

a)show $$H=H^{\dagger}$$? if so$$H^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)$$

4. Sep 15, 2009

### Preno

What is the adjoint of a product?

5. Sep 15, 2009

### noblegas

i dunno $$(PQ)^{\dagger}=Q^{\dagger}P^{\dagger}$$?

6. Sep 15, 2009

### Preno

Well, yeah, so you just apply this rule to the definition of H and use the fact that $(Q^\dagger)^\dagger=Q$.

7. Sep 15, 2009

### noblegas

I did that: $$H^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)$$

8. Sep 15, 2009

### noblegas

So did I apply the rule correctly? Hope there is no hard feelings

9. Sep 16, 2009

### noblegas

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. Sep 16, 2009

### gabbagabbahey

Shouldn't this be $H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q)$?

11. Sep 16, 2009

### Hurkyl

Staff Emeritus
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. Sep 16, 2009

### noblegas

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 $$H*\varphi =E*\varphi ==> \alpha*QQ^{\dagger}*\varphi=E*\varphi$$

Last edited: Sep 16, 2009
13. Sep 16, 2009

### gabbagabbahey

Right, the second term goes away (since $Q^{\dagger}Q^{\dagger}=0$)....and the first term is____?

14. Sep 16, 2009

### noblegas

Sorry yahiko, I know matrices don't Commute. I mistakenly wrote down the wrong expression

15. Sep 16, 2009

### noblegas

$H^2=\alpha^2QQ^{\dagger}$

that can't be right because the $$QQ^{\dagger}$$ is not squared

16. Sep 16, 2009

### gabbagabbahey

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. Sep 16, 2009

### noblegas

Because $$HH=(\alpha)QQ^{\dagger}(\alpha)QQ^{\dagger}$$

18. Sep 16, 2009

### gabbagabbahey

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. Sep 16, 2009

### noblegas

ah! [tex] $H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q=(\alpha*H)$

20. Sep 16, 2009

### gabbagabbahey

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?