Proving Orthogonality of $\hat{A}|\Psi>$ with Anti-Unitary Operator $\hat{A}$

[peaco99]

Homework Statement

$\hat{A}$ is an anti-unitary operator, and it is known that $\hat{A^2}$= -$\hat{I}$, show that |$\Psi$> is orthogonal to $\hat{A}$|$\Psi$>

Homework Equations

I know that $\hat{A}$ can be represented by a unitary operator, $\hat{U}$, and the complex conjugation operator, $\hat{K}$, in the following way $\hat{A}$ = $\hat{U}$$\hat{K}$.

Also from the definition of anti-unitary operator <$\hat{A}$$\Psi$|$\hat{A}$$\Phi$> = <$\Psi$|$\Phi$>* = <$\Phi$|$\Phi$>

The Attempt at a Solution

I've tried starting with the inner product of psi and A*psi, and inserting "identities" of A and A's inverse, but then I realized in both my class and in sakurai, we have not defined the inverse of an anti-unitary operator, and sakurai warns about acting anti-unitary operators on bra's, I think he even says we/he will never do such an operation. So that's when I went to the definition of A being composed of a unitary operator and the complex conjugation operator, but then I get stuck with the inverse of the complex conjugation operator, and I'm not sure if that is even defined?

My thought is to somehow get A^2 into the inner product of psi and Apsi, then I would get a -1 after it acts on the ket, then compare the original inner product, with the final inner product having the minus sign, and declaring that that can't be true, so therefore the inner product is zero...Something along those lines...but...

All these inverses have got me stuck!

Please help.

Thanks.

JP

 

[gabbagabbahey]

$\hat{A}$ is an anti-unitary operator, and it is known that $\hat{A^2}$= -$\hat{I}$, show that |$\Psi$> is orthogonal to $\hat{A}$|$\Psi$>

I really don't think this is true in general. As a counter-example, consider the complex-conjugation operator itself, which is anti-unitary in the complex plane. What is its effect on the state $|\Psi\rangle=\begin{pmatrix}1 \\ 0 \end{pmatrix}$ (i.e what is $\hat{K}z$ for $z=1+0i$)?

Are you given any restrictions on $|\Psi\rangle$?

Edit: nvm, $\hat{K}^2=\hat{I}\neq -\hat{I}$ on the complex plane (although it is still anti-unitary). If $\hat{A}$ is anti-unitary and $\hat{K}^2=-\hat{I}$, the the desired result is true.

Hint: try computing $$\left(\langle\Psi|A\Psi\rangle\right)^{\dagger}$$

 

[peaco99]

No restrictions on psi, in fact it says to show this for all psi in the hilbert space.

All the problem states is that A is anti-unitary, A^2= -I, and show that A|psi> is orthogonal to |psi>.

 

[gabbagabbahey]

See my above edit ^^^

 

[peaco99]

Are you saying K^2=A^2= -I? Because the problem only states that A^2=-I, and the problem never mention what A is, other than it being antiunitary, I brought in the K operator as an attempt at solving the problem.

In regards to your hint, everytime I try and work with the inner product, or in your case the adjoint of the inner product, I get and inverse A operator, and in my class and in sakurai it says never to work with the inverse of antiunitary operators...so what I'm getting at is that I don't know how to handle/work with the inverse/adjoint of antiunitary operators.

Thanks again!

 

[gabbagabbahey]

Are you saying K^2=A^2= -I? Because the problem only states that A^2=-I, and the problem never mention what A is, other than it being antiunitary, I brought in the K operator as an attempt at solving the problem.

Sorry for any confusion, I meant that my earlier counterexample of $\hat{A}=\hat{K}$ in the complex plane was rubbish since $\hat{K}^2\neq-\hat{I}$.

In regards to your hint, everytime I try and work with the inner product, or in your case the adjoint of the inner product, I get and inverse A operator

You shouldn't need to use any inverses, just transposeconjugates for this. Can you show me where the inverse is popping up in your approach?

 

[peaco99]

I'm sorry I was equating the inverse with the adjoint(transpose conjugate). So my problem still becomes that when I get the transpose conjugate in the problem, I don't know how to handle it. Like I mentioned before, Sakurai and my class claims we will never defIne the way an antiunitary operator acts on bras. So, if/when I get the transpose conjugate of A, I can't pull it into a bra and then just say <psi|A^(*T)= <Apsi| ...sorry I'm on my phone so I can't use the tex notation.

Hope this makes sense, thanks.

 

[gabbagabbahey]

I'm sorry I was equating the inverse with the adjoint(transpose conjugate). So my problem still becomes that when I get the transpose conjugate in the problem, I don't know how to handle it. Like I mentioned before, Sakurai and my class claims we will never defIne the way an antiunitary operator acts on bras. So, if/when I get the transpose conjugate of A, I can't pull it into a bra and then just say <psi|A^(*T)= <Apsi| ...sorry I'm on my phone so I can't use the tex notation.

Hope this makes sense, thanks.

I've never given Sakurai much more than a cursory glance, so I'm not sure how much of what I'm about to write will make sense to you notationally:

Anyways, there is no need to act explicitly on the Bra with $\hat{A}^{\dagger}$, just assume that the operators act only on the Ket, in order

$$\left(\langle\Psi|\hat{A}|\Psi\rangle\right) ^{\dagger}=\langle\Psi|\hat{A}^{\dagger}|\Psi \rangle$$

What happens if you insert the identity operator $\hat{I}=-\hat{A}^{\dagger}\hat{A}$ before taking the adjoint?

 

[peaco99]

I don't know how you got $\hat{I}$= -$\hat{A}^\dagger\hat{A}$ , this would imply that $\hat{A}^2$= $\hat{A}^\dagger$, but anyways if you use this all I get is

$(<\Psi|\hat{A}|\Psi>)^\dagger = (<\Psi|-\hat{A}^\dagger\hat{A}\hat{A}|\Psi>)^\dagger = (<\Psi|-\hat{A}^\dagger\hat{A}^2|\Psi>)^\dagger = (<\Psi|-\hat{A}^\dagger -(1)|\Psi>)^\dagger= (<\Psi|\hat{A}^\dagger|\Psi>)^\dagger = <\Psi|\hat{A}|\Psi>$ => $\hat{A}^\dagger=\hat{A}$

So, two things about the above, 1)Nothing here jumps out at me telling me there is a contradiction so that therefore the states are orthogonal, 2) this also means A is Hermitian, I'm not sure if antiunitary operators can or cannot be Hermitian.

If I use $\hat{I}=\hat{A}^\dagger\hat{A}$, then

$<\Psi|\hat{A}|\Psi> =<\Psi|\hat{A}^\dagger\hat{A}\hat{A}|\Psi> = -<\Psi|\hat{A}^\dagger|\Psi>$

Again, I still don't see anything that says there is a contradiction...I must be missing something.

 

[gabbagabbahey]

I don't know how you got $\hat{I}$= -$\hat{A}^\dagger\hat{A}$

Sorry, I had a brain fart. I meant $\hat{I}=-\hat{A}^2$.

$$\langle\Psi|\hat{A}|\Psi\rangle=\left(\langle\Psi|\hat{A}^{\dagger}|\Psi\rangle\right)^{\dagger}=-\left(\langle\Psi|A^2\hat{A}^{\dagger}|\Psi\rangle\right)^{\dagger}$$

What do you get when you calculate the adjoint of the RHS (remember $(BC)^{\dagger}=C^{\dagger}B^{\dagger}$ )

 

[peaco99]

Well $\hat{A}\hat{A}^\dagger$ either equals +1 or -1 in either case i'd get:

$<\Psi|\hat{A}^\dagger|\Psi> = <\Psi|-\hat{A}^2\hat{A}^\dagger|\Psi> = <\Psi|-\hat{A}\hat{A}\hat{A}^\dagger|\Psi> =$

1) for the -1 case:

$<\Psi|-\hat{A}\hat{A}\hat{A}^\dagger|\Psi> = <\Psi|\hat{A}|\Psi>$

which would just tell me A is Hermitian, maybe this can't be true for antiunitary operators? and here in lies the contradiction

2) for the +1 case:

$<\Psi|-\hat{A}\hat{A}\hat{A}^\dagger|\Psi> = <\Psi|-\hat{A}|\Psi> = -<\Psi|\hat{A}|\Psi> = -(<\Psi|\hat{A}^\dagger|\Psi>)^\dagger$

I still don't see anything, sorry,

 

[gabbagabbahey]

Sorry, I edited my previous post for clarification^^^

Also, why would $\hat{A}\hat{A}^{\dagger}=-1$ be possible? Does Sakurai say this?

 

[peaco99]

No, Sakurai, or in my notes, is $\hat{A}\hat{A}^\dagger$ ever defined. I just assumed because it is anti"unitary", it is either 1 or -1, the -1 maybe coming from "funny" properties of antiunitary operators. So I'm guessing -1 isn't an option then

$<\Psi|\hat{A}|\Psi>$ = $(<\Psi|\hat{A}^\dagger|\Psi>)^\dagger$=$-(<\Psi|\hat{A}^2\hat{A}^\dagger|\Psi>)^\dagger$ = $-<\Psi|\hat{A} \hat{A}^{2\dagger}|\Psi> = -(<\Psi|\hat{A}\hat{A}^\dagger\hat{A}^\dagger|\Psi>)^\dagger = -(<\Psi|\hat{A}^\dagger|\Psi>)^\dagger = -(<\Psi|\hat{A}|\Psi>)$

beautiful, but how do I know for sure that $\hat{A}\hat{A}^\dagger = 1$ ?

Thanks so much

 

[gabbagabbahey]

but how do I know for sure that $\hat{A}\hat{A}^\dagger = 1$

Does Sakurai show that $\hat{K}\hat{K}^{\dagger}=1$? If so, just use $\hat{A}=\hat{U}\hat{K}$ for some unitary $\hat{U}$ and calculate $\hat{A}\hat{A}^\dagger$

 

[peaco99]

No he doesn't. But if I accept that to be true, Then I can see it. BTW, may I ask your physics education level?

Thanks.

 

[gabbagabbahey]

Just glanced through Sakurai, and I see that his definition of antiunitary is not as nice as the one I am used to seeing: An antiunitary operator is a unitary, antilinear operator so that $\hat{A}\hat{A}^{\dagger}=\hat{I}$ and $\hat{A}(c_1|\alpha\rangle + c_2|\beta\rangle) = c_1^*\hat{A}|\alpha\rangle + c_2^*\hat{A}|\beta\rangle$. Moreover, he even shies away from defining $\hat{A}^{\dagger}$ at all.

I suppose that there must be a proof directly utilizing $\langle\tilde{\alpha}|\tilde{\beta}\rangle = \langle\beta|\alpha\rangle$ that allows you to avoid using $\hat{A}^{\dagger}$ at all, if you are expected to tackle this problem using only what has been covered in Sakurai.

Edit: Try using the above property with $|\alpha\rangle=\hat{A}|\Psi\rangle$ & $|\beta\rangle=|\Psi\rangle$; that is, apply $\hat{A}$ to both those states and take the inner product.

As for my physics education, just BSc, formally. Afterwhich I decided that Acedemia was not for me (I like money too much), and became an SAP application developer. However, I often self-study graduate level material when time permits.