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Regards,

Chris

- Thread starter kq6up
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- #26

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Regards,

Chris

- #27

strangerep

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The 1st step looks wrong. You've taken the derivative of ##\Psi^\dagger##, not ##\Psi##.$$\Psi = e^{ikx}$$ [...]

The right side equals:

$$\left< { -i\hbar \frac { \partial }{ \partial x } \Psi }|{ \Psi } \right> =\left< { -i\hbar (-i)\Psi }|{ \Psi } \right> = [\cdots] $$

But perhaps you've already seen this when you did it in terms of an integral?

(Oh, and I think you left out a "k", but that's not the problem.)

- #28

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But when it's in the bra isn't it supposed to be the conjugate?

Chris

Chris

- #29

kith

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No. The difference between the bra and the ket are denoted by the brackets, not by what's inside. Using your notation, [itex]| \Psi \rangle = |e^{ikx} \rangle[/itex] and [itex]\langle \Psi | = \langle e^{ikx} |[/itex]. The conjugate comes in when you write [itex]\langle \Phi | \Psi \rangle[/itex] as an integral.But when it's in the bra isn't it supposed to be the conjugate?

But your notation is strange. [itex]e^{ikx}[/itex] implies that you are working in the position or momentum basis while [itex]| \Psi \rangle [/itex] denotes a general ket vector. The standard notation for the connection between these concepts is [itex]e^{ikx} = \langle x | \Psi \rangle =: \Psi(x)[/itex] where [itex]\Psi(x)[/itex] is called the wavefunction.

- #30

Fredrik

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No. If A is an operator and f a square integrable function, then Af means the same thing no matter where you put it.But when it's in the bra isn't it supposed to be the conjugate?

Chris

The bra that corresponds to ##A|\alpha\rangle## is ##\langle\alpha|A^\dagger##, but I see no place to use that in your calculation.

What does a product of a bra and an operator even mean? It's defined by

$$\big(\langle\alpha|A\big)|\beta\rangle =\langle\alpha|\big(A|\beta\rangle\big).$$ The product ##\langle\alpha|A## is defined by saying that this equality holds for all ##|\beta>##.

The bra that corresponds to an arbitrary vector (i.e. function) f is the map ##g\mapsto \langle f,g\rangle##, which can be written as ##\langle f,\cdot\rangle##. So the bra that corresponds to Af is the map ##\langle Af,\cdot\rangle##. This map takes an arbitrary ##g## to ##\langle Af,g\rangle##, which is equal to ##\langle f,A^\dagger g\rangle##. In bra-ket notation, the map that takes ##g## to ##\langle f,A^\dagger g\rangle## is written as ##\langle f|A^\dagger##. In inner product notation, we'd have to use something super ugly like ##\langle f,A^\dagger(\cdot)\rangle## or just invent a new symbol for it.

- #31

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I think what I need here is a good textbook that hand holds me through this formalism for both the operators and the bra-ket notation. None of the books I have really gives a clear treatment from the bottom up -- just bits and pieces only, and I have looked a plenty of books the past few days. Mary Boas' text, Math Methods -- Riley & Hobson (which is a great book, but not fined grained enough in this area). I am using Griffiths to learn QM, and the text is very clear, but when you get to the problem set one gets the sense that a lot of what is expected in the problem set is not really developed in the text itself.

Suggestions?

Edit: Zettili looks like he develops these details in a much more thorough way. I think I will go through his chapter 2, and report back.

Thanks,

Chris

Suggestions?

Edit: Zettili looks like he develops these details in a much more thorough way. I think I will go through his chapter 2, and report back.

Thanks,

Chris

Last edited:

- #32

strangerep

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There's a reason I put a "permanent" suggestion in my (2nd) signature line below.Suggestions?

- #33

bhobba

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I second that.There's a reason I put a "permanent" suggestion in my (2nd) signature line below.

Ballentine had a strong effect on me clearing up all sorts of issues.

Thanks

Bill

- #34

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Fast forward to 2014. I decide to brush up and cover ground that I missed as an undergrad. Math was my first priority. Mary Boas' Methods was recommended, and I have been plowing through her text since Febuary doing most of the problems in the set. Doing quite well with though the going is slow. After finishing her 100 pages of linear algebra (which I have never taken before) I started to peek ahead at some QM, and found that I could actually begin to understand it. It definitely was time for a break from that math book. In the past two weeks I have done more QM problems than I have ever done as an undergrad -- with the added bonus of actually mostly understanding what I am doing. Sorry for all of the detail, but I thought it necessary to provide a basis for where I am at mathematically speaking.

I mostly understand the physics involved, but some of this QM math is really strange to me. However, I am resolved to learn it no matter how far back I have to go to build the correct foundation to get comfortable with the math that is involved. I will take a look at Ballentine today, and it does look less presumptuous in the opening math section, but maybe I will need a course that teaches me real/complex analysis and/or whatever folks learn things related to Hilbert space and other abstract generalized vector stuff. Also, most of the mathy/proofy statements here that are not a part of just practically doing lower division fairly straight forward physics stuff is *mostly* lost on me.

Thanks for your time,

Regards,

Chris

- #35

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There are times when self instruction can really be a bear, but it is nice to go at one's own pace.

Thank you and regards kind sirs,

Chris

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