i've been trying to do some _simple_ (i guess) find-the-hermitian-conjugate problems and i'm already having some trouble =/

so, for Q = x in
< f | Q g >; f, g functions of x in C[-inf, inf]

the hermitian conjugate Q# such that

< Q# f | g > = < f | Q g >

is x right? so Q = x is a hermitian operator itself?

what about Q = d/dx?

so far i've got

< f | Q g> = g f* - < Q f | g> = < Q# f | g>

(using integration by parts)

but what is Q# ??

thanks for your time.
any help appreciated =)

Right, although your domain looks suspicious to me (it's very important that you consider the domain for these problems). See below.

This isn't even eligible to be Hermitian (it's missing a factor of i), so we expect it to be anti-Hermitian. In other words, we expect Q^&dagger; = -Q

very good. You haven't made a mistake, but the mapping you have defined here Q# is not a linear operator, and therefore can't possible be the Hermitian adjoint (notice that 0#&ne;0). There is a theorem that every linear operator has a Hermitian adjoint, so why doesn't this theorem work in this example? The problem is that the theorem only holds in Hilbert spaces, and the space C[-&infin;,&infin;] is not a Hilbert space. You have to make some kind of restriction to your space, like restrict yourself to periodic functions, or functions which vanish at infinity appropriately quickly. whatever it is you do, will make that term vanish above, and you will have Q^&dagger;=-Q

Can you see why your space is not a Hilbert space? Can you imagine why this is a necessary condition for the existence of the hermitian adjoint operator?

sure

thanks lethe.

so to be in Hilbert space the functions must be square-integrateble? (or complete inner product space...still a bit unsure what that means)

and C[-inf, inf] is definitely not the case.

mm okay i see that now.
but for square-integrateble functions, you're doing < f | Q f > and i see that why f must be in Hilbert space...but
i guess that means Hilbert space doesn't just include the square integratebles huh...okay mm

mmm....and isn't Q = d/dx a linear operator?
thanks.
sorry i'm slow.

yes, that makes it an inner product space.
complete means that every Cauchy sequence converges. We achieve this by making sure we're using Lebesgue integrals, and equivalance classes of functions that are equal almost everywhere.

bangarang. This space is probably (or can be made) complete, but inner product, ain't gonna happen.


The Hilbert space L²(R) is made up solely of (equivalence classes of) square integrable functions. Why do you think it includes something else?

yes, it's a linear operator. what isn't linear is the map that takes Q to Q# as you defined it above. Observe that under your definition, the adjoint of the zero operator is not zero, so this operation # can't be linear.
no sweat. Feel free to ask if you're still not getting it.

mmm.....
for < Q# f | g > = < f | Q g >
Q# is not Q's adjoint?

well, this isn't how you defined Q# above. recall, you said:

This defines a mapping Q# which depends on Q, but this mapping is not the Hermitian adjoint (which doesn't exist on this domain). Nor is Q# linearly (or even anti-linearly, which is more appropriate here). Consider the case Q=0. If Q# were linearly related to Q, then we would expect Q#=0 as well. But obviously all you have to do is choose as test functions f and g so that f(&infin;)g(&infin;)*&ne;f(-&infin;)g(-&infin;)* and then Q#&ne;0. The point is, if you don't take care of your boundary conditions, you don't have a Hilbert space, and you don't have an adjoint either (the two facts are of course related).

oh hmm....
i thought i was just working out the < f | Qg > and trying to set it equal to < Q# f| g >

mm =/
i guess what i was trying to ask was, what is the adjoint of d/dx....if it has one?

oh well....let me think about this tomorrow...
thanks for all your help!

think about it, good. the answer is of course already given somewhere in this thread, you know that right? now you just need to think about why it's true.

ah alright. just woke up.

seems like i've already worked it out....as it states
and now i get it....haha that took me a while =_=

thanks lethe!