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$\def\pd#1#2{\frac{\partial#1}{\partial#2}}\def\braket#1#2{\langle#1|#2\rangle}\def\bra#1{\langle#1}\def\ket#1{#1\rangle}$I am given that $$\bra{\theta,\phi}|L_+|\ket{l,m} = \hbar e^{i\phi} \left(\pd{}{\theta}+i\cot{\theta}\pd{}{\phi}\right) \braket{\theta,\phi}{l,m}$$

Where $L_+ = \left(\pd{}{\theta}+i\cot{\theta}\pd{}{\phi}\right)$

Why can one simply pull the $L_+$ operator straight out of the braket?

Thanks.

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I don't know how to edit your post because I'm not certain what you were trying to write, but hopefully you can fix it yourself: To write $\langle \psi | \phi \rangle$, the code is " \langle \psi | \phi \rangle ". –  Ragib Zaman Aug 29 '11 at 13:49
    
\pd is $\partial$? –  J. M. Aug 29 '11 at 13:54
    
Turns out I don't have enough reputation points to edit your post anyway. Anyway, you may also want to use $\partial$, the code is " \partial ". Edit: @J.M. I was guessing that's what he meant as well. –  Ragib Zaman Aug 29 '11 at 13:55
    
@Ragib: you can "propose" an edit for later review by a user with sufficient rep. :) –  J. M. Aug 29 '11 at 14:02
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Your meaning seems to depend on the notation for $|l,m\rangle$ and $\langle \theta,\phi|$, which I assume is some sort of physics convention. If you have a definition for them, please include it. If your question is about them, perhaps asking on the physics stackexchange site would be better? –  Willie Wong Aug 29 '11 at 14:25

1 Answer 1

up vote 3 down vote accepted

I think you're missing a factor $\hbar\mathrm e^{\mathrm i\phi}$ in your equation for $L_+$.

That factor notwithstanding, that second equation doesn't really make sense in the context of the first equation. In the first equation, you're using fancy ket notation to denote the angular representation of the state.

On the left-hand side you're applying $L_+$ to an abstract state $|l,m\rangle$ and then multiplying by an angular "eigenstate" $\langle\theta,\phi|$ to get its angular representation, i.e. the wave function as a function of the angular variables.

On the right-hand side, you're first forming the angular representation by multiplying by $\langle\theta,\phi|$, and then applying a differential operator. That differential operator is what the $L_+$ operator looks like in the angular representation. But if you then write an equation that equates the abstract operator $L_+$ with its angular representation, you've defeated the entire purpose of the fancy notation in the first line that kept them distinct. It's quite usual to write things like your second equation, but only when you're implicitly assuming a certain standard representation, in this case the angular one. Once you apply the level of rigour to make the distinction you made in the first line, the second line is no longer correct. And asking why we can "simply pull the $L_+$ operator straight out of the braket" sort of misses the point -- if you rigorously distinguish it from its representation, then that's not what's being done, and if you aren't, then the whole $\langle\theta,\phi|$ ket thing makes little sense.

[Edit in response to the comments]

Mathematically speaking, a Hilbert space is just a complete inner product space. So "some space called the Hilbert space" isn't quite right -- a quantum-mechanical system has an associated state space, and that state space is a Hilbert space. Now you can think of that state space as the space of all possible wave functions in coordinate space if you want. The reason for abstracting from that viewpoint is that the space can equally well be described in many other ways. In particular, a state can be described as a linear combination of eigenstates of a suitable operator, for instance the Hamiltonian operator associated with the energy of the system. Once the state is conceptuatlized as a vector in a vector space, its representation as a wavefunction of the coordinates becomes in some sense an arbitrary one among many possibilities, since it corresponds to a particular choice of basis in that space. (Note, however, that in dealing with the measurement problem, some philosophers of physics have suggested that the spatial representation of the wave function should be accorded special status.)

Concerning the dimension of the Hilbert space, I'm not quite sure how to interpret your expression "arbitrarily many dimensions". There are two different notions of the dimension of a Hilbert space. In this context, what is usually called a "basis" of a vector space is called a "Hamel basis", and the cardinality of a Hamel basis (which is the same for all Hamel bases) is called the "Hamel dimension" of the space, whereas an orthonormal basis is sometimes called a "Hilbert basis", and the cardinality of a Hilbert basis (which is the same for all Hilbert bases) is called the "Hilbert dimension" of the space. For finite-dimensional Hilbert spaces, these two notions of dimension coincide (which makes it meaningful to talk about "finite-dimensional Hilbert spaces" without specifying which of the notions of dimension this refers to) but they need not in general; the Hamel dimension may be greater than the Hilbert dimension. For instance the Hamel dimension may be uncountable even if the Hilbert dimension is countable; this is the case for the Hilbert space of all square-summable sequences of complex numbers.

As I wrote, I'm not sure I understand exactly what sort of explanation you're looking for, so feel free to ask again, but please ask as specifically as possible.

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Thanks, I have indeed missed out the factor. Hmm... I copied this from some notes... :-( Do you mean that an operator is not associated with a representation? –  O. P. Aug 29 '11 at 15:19
    
@O. P.: Yes, an operator can be considered as an abstract operator just like a state can be considered as an abstract state. The transition from the concrete form of $L_+$ as a differential operator in the angular representation to $L_+$ as an abstract operator is precisely analogous to the transition from the concrete form of the wavefunction as a function of position coordinates to an abstract state. –  joriki Aug 29 '11 at 15:21
    
Thanks again :-) Would you mind explaining what bras and kets mean? I (kind of) know the "algebra" of these things and that they are supposed to be vectors in some space called the Hilbert space. But what are they? I have read the Wiki page on Bra-ket notation but I am not sure I can fully appreciate it. Thank you very much :-) –  O. P. Aug 29 '11 at 15:29
    
@O. P.: Perhaps you could say more specifically what part of the Wikipedia article is unclear to you or what aspect you want to know more about? The article looks fine to me; most of what I would have written as an explanation is already there. Also, I don't quite understand "they are supposed to be vectors in some space called the Hilbert space. But what are they?". The obvious answer is: "They are vectors in some space called the Hilbert space" :-) –  joriki Aug 29 '11 at 15:35
    
Hehe. Ok, that was a silly question. I am just rather confused. So the Hilbert space has arbitrarily many dimensions and its vectors are wavefunctions-made abstract? –  O. P. Aug 29 '11 at 15:41

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