Can Someone Please Explain Hermitian Conjugates To Me?

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Can Someone Please Explain Hermitian Conjugates To Me!?

I'm working on some problems about the Hermitian of a Harmonic Oscillator - I keep coming across the Hermitian written in a form with A[dagger]A and similar things - when I've looked in textbooks and online I find it explained using Bra-ket notation, which I also struggle to understand.

Can someone please help me - I've got a lot of problems to do and I don't want to be sat here all night trying to figure this one thing out.

TIA

Xx
 

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  • #2
Matterwave
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I think you should be more specific on what you are getting confused with. Are you confused with the notation or the concept of a Hermitian conjugate? Otherwise, I would suggest you read some intro QM texts like Griffiths.
 
  • #3
Fredrik
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What you need to know to be able to use the concept is that the adjoint of A, denoted by ##A^*## or ##A^\dagger##, is the unique linear operator such that
$$\langle x,Ay\rangle=\langle A^*x,y\rangle$$ for all x,y in the Hilbert space. A more complete answer would explain how we can be sure that there's exactly one such operator, but that part is kind of hard. It requires you to understand the Riesz representation theorem for Hilbert spaces. It's not super hard, but it's hard enough that you won't be able to fully understand it tonight.
 
  • #4
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The Hermitian conjugate of a matrix is very easy. You just need to take the complex conjugate of each entry and then transpose the matrix. The Hermitian conjugate of an operator is somewhat more difficult.

Since you mention [itex]A^\dagger A[/itex], it makes me think of positive(-definite) operators. Every positive operator can indeed be written in that form, but that's far from trivial.
 
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Its more expressing the hamiltonian of a quantum harmonic oscillator in terms of a variable - i.e

showing that:

H = p^2/2m + 1/2mw^2x^2

can be expressed as

(hbar)(w) [adagger*a + 1/2]

Where a = sqrt(mw/2hbar)x + i*sqrt(1/2mwhbar)p

where x and p are the position and momentum operators
 
  • #6
Fredrik
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Doesn't the solution to that appear in all the introductory books? If you're trying to follow a proof and is stuck on a detail, you should ask about that detail (preferably with a link to the page at google books).

Edit: One thing you will need to know is that the adjoint operation ##A\mapsto A^*## is conjugate linear in the sense that for all complex numbers c and all linear operators A, we have ##(cA)^*=c^*A^*##. In particular, for any two self-adjoint A and B, we have $$(A+iB)^*=A^*+(iB)^*=A^*-iB^*=A-iB.$$
 
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  • #7
dextercioby
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Fredrik;3768278[... said:
. In particular, for any two self-adjoint A and B, we have $$(A+iB)^*=A^*+(iB)^*=A^*-iB^*=A-iB.$$
This of course is valid only for bounded operators.
 
  • #8
Fredrik
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Is it? I didn't even know that. I just checked the notes I've made on the adjoint operation, and they end right as I've started to explain the definition of the adjoint operation for unbounded operators, with a comment to myself that I need to check if it satisfies identities like that.

Anyway, physics books completely ignore this issue. I would guess that many of the authors don't know this. The solution I have seen definitely requires the readers to treat x-ip as the adjoint of x+ip.
 
  • #9
dextercioby
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See it for yourself, the x and p are self-adjoint on individual non-equal domains in L^2 (R) and are only essentially self-adjoint on the intersection of their domains. Hence one should use operatorial inclusion instead of operatorial equality when using a and a^dagger. This finesse is indeed almost everywhere neglected and one uses the = sign in a quite abusive way.
 
  • #10
Fredrik
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OK, I think I see what you mean. We have
$$\langle (A+iB)^*x,y\rangle=\langle (A^*-iB^*)x,y\rangle$$ for all x,y such that this equality and the calculation that proves it make sense. But we can't conclude that ##(A+iB)^*=A^*-iB^*##, because these operators may have different domains. When A and B are self-adjoint, this is not a problem.

In a calculation like
$$x^2+a^2p^2=(x+iap)(x-iap)+ia[x,p],$$ there's some abuse of notation going on, because the domains of ##x^2## and ##p^2## are smaller than the domains of x and p.
 

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