Wave mechanics: the adjoint of a hamiltonian

[noblegas]

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?

 

[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 $H$.

 

[noblegas]

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 so$$H^{\dagger}=((\alpha)QQ^{\dagger})^{\dagger}=QQ^{\dagger}(\alpha)$$

 

[Preno]

What is the adjoint of a product?

 

[noblegas]

What is the adjoint of a product?

i don't know $$(PQ)^{\dagger}=Q^{\dagger}P^{\dagger}$$?

 

[Preno]

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

 

[noblegas]

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

 

[noblegas]

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

 

[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.

 

[gabbagabbahey]

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 $H^2=\alpha^2QQ^{\dagger}(1-Q^{\dagger}Q)$?

 

[Hurkyl]

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.

 

[noblegas]

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

 

[gabbagabbahey]

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____?

 

[noblegas]

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

 

[noblegas]

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

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

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

 

[gabbagabbahey]

Why would the $QQ^{\dagger}$ term have to be squared?

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

 

[noblegas]

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

 

[gabbagabbahey]

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, $H^2=\alpha^2QQ^{\dagger}=\alpha(\alpha QQ^{\dagger})=\alpha$__?

 

[noblegas]

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

 

[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?

 

[noblegas]

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 $$H*\varphi=E*\varphi$$

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

 

[gabbagabbahey]

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 [itex]H^2=\alpha H[/tex], so...

 

[noblegas]

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 [itex]H^2=\alpha H[/tex], so...

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

 

[gabbagabbahey]

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

Yes, but you also know $H^2=\alpha H$, so...

 

[noblegas]

Yes, but you also know $H^2=\alpha H$, so...

so $$\alpha H*\varphi=E^2*\varphi$$

 

[gabbagabbahey]

Right, and $\alpha H\varphi= \alpha (H\varphi)=\alpha$___?

 

[noblegas]

Right, and $\alpha H\varphi= \alpha (H\varphi)=\alpha$___?

can't believe I left out alpha. $$\alpha*E^2 \varphi$$

 

[gabbagabbahey]

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 $E=$___?

 

[noblegas]

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

 

[gabbagabbahey]

$e=h=e^2$ 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!

 

[noblegas]

no, $h$ is an operator, $e$ is a scalar, they cannot possibly be equal!

$$e=e^2$$

 

[gabbagabbahey]

$$e=e^2$$

No, $\alpha E\varphi=E^2\varphi$, so (since $\alpha$ and $E$ are both scalars!) $\alpha E=E^2$...so $E=$___?

 

[noblegas]

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?

 

[Dick]

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]

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