As far as I understand, the momentum operator is:

$\hat{p} = -i \hbar \frac{\partial}{\partial \hat{q}}$

Where I'm not sure at this point if it's mathematically correct to talk about a derivative wrt. the position operator - but the point is, as far as I understand, that this equality is true in general, without taking the Schrodinger representation. It is not a derivative wrt the position eigenvalues.

Since $\hat{p}$ is a Hermitian operator, $\frac{\partial}{\partial \hat{q}}$ must be pure imaginary. So I think it's true that:

$\left( \frac{\partial}{\partial \hat{q}} |ψ> \right)^{\dagger} = <ψ|\left( \frac{\partial}{\partial \hat{q}} \right)^{\dagger} = -<ψ| \frac{\partial}{\partial \hat{q}}$

Similarly:

$\hat{H} |ψ> = i \hbar \frac{\partial |ψ>}{\partial t}$

Hence:

$\left( \frac{\partial}{\partial t} |ψ> \right)^{\dagger} = <ψ|\left( \frac{\partial}{\partial t} \right)^{\dagger} = - \frac{\partial}{\partial t} <ψ|$

But that is wrong. The minus shouldn't be there. What mistake am I making?
Also, can we say: $\hat{H} = i \hbar \frac{\partial }{\partial t}$ ?

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 Since $\hat{p}$ is a Hermitian operator, $\frac{\partial}{\partial \hat{q}}$ must be pure imaginary.
... in order to give real eigenvalues you are thinking? But surely you only need any resulting, additional, imaginary components to cancel out?

 $\hat{H} = -i\hbar\frac{\partial}{\partial t}$
... kinda, though we'd usually write:
$$\hat{H} = \frac{\hat{p}^2}{2m}$$

So for an energy eigenstate like $\Psi(x,t)=\psi_n(x) e^{iE_nt/\hbar}$ you'll see the two expressions give the same result.

notice that $\frac{\partial}{\partial t}$, in this example, is not purely imaginary yet the Hamiltonian operator is still giving real eigenvalues.

Recognitions:
 Quote by Loro Since $\hat{p}$ is a Hermitian operator, $\frac{\partial}{\partial \hat{q}}$ must be pure imaginary. So I think it's true that: $$\def\<{\langle} \def\>{\rangle} \left( \frac{\partial}{\partial \hat{q}} |ψ\> \right)^{\dagger} = \<ψ|\left( \frac{\partial}{\partial \hat{q}} \right)^{\dagger} = -\<ψ| \frac{\partial}{\partial \hat{q}}$$
Taking adjoints of such operators must be done more carefully. Consider an inner product ##C := \<\phi|P \psi\>##, where ##P## is the momentum operator. For ##P## to be selfadjoint we must have ##C = \<P \phi|\psi\>## also.

Writing ##C## in terms of Schrodinger wave functions, etc, (so that ##P## is represented as the derivative you described), this becomes
$$C := \<\phi|P \psi\> ~\equiv~ -i\hbar \int\! dq \; \phi^*(q) \partial_q \psi(q)$$
To make the derivative act on ##\phi## instead, we must use integration by parts, together with an assumption that the wave functions vanish sufficiently fast as ##q\to\infty##. This gives:
$$C = + i\hbar \int\! dq \; \Big( \partial_q \phi^*(q) \Big) \psi(q) ~=~ \int\! dq \; \Big(-i \hbar \partial_q \phi(q) \Big)^* \psi(q) ~\equiv~ \<P \phi|\psi\> ~,$$ as claimed.

HTH.

[Edit: last equation changed after typo pointed out in post#4 below.]

Recognitions:

 Quote by strangerep $$C = + i\hbar \int\! dq \; \Big( \partial_q \phi^*(q) \Big) \psi(q) ~=~ \int\! dq \; \Big(-i \hbar \partial_q \phi^*(q) \Big) \psi(q) ~\equiv~ \langle P \phi|\psi\rangle ~,$$ as claimed. HTH.
$$C = + i\hbar \int\! dq \; \Big( \partial_q \phi^*(q) \Big) \psi(q) ~=~ \int\! dq \; \Big(-i \hbar \partial_q \phi(q) \Big)^* \psi(q) ~\equiv~ \langle P \phi|\psi\rangle ~,$$

Recognitions:
 Quote by vanhees71 This final equation must read $$C = + i\hbar \int\! dq \; \Big( \partial_q \phi^*(q) \Big) \psi(q) ~=~ \int\! dq \; \Big(-i \hbar \partial_q \phi(q) \Big)^* \psi(q) ~\equiv~ \langle P \phi|\psi\rangle ~,$$
OOPS, yes! I changed my post #3 accordingly. Thanks for the correction.

Thanks for the replies.

 $+ i\hbar \int\! dq \; \Big( \partial_q \phi^*(q) \Big) \psi(q) ~=~ \int\! dq \; \Big(-i \hbar \partial_q \phi(q) \Big)^* \psi(q)$

But when you say that, aren't you assuming in this step, that:

$\partial_q ^{ \dagger} = \partial_q$

Which, as I understand, is the result we're trying to justify?

 Proving this more carefully: $<\phi | \left( -i\hbar \hat{\partial_q} \right) |\psi > = \int dq' <\phi | q' >< q' | \left( -i\hbar \hat{\partial_q} \right) |\psi > = -i\hbar \int dq' \phi^{*} \partial_{q'} \psi$ $<\phi | \left( -i\hbar \hat{\partial_q} \right)^{\dagger} |\psi > = \int dq' <\phi | q' >< q' | \left( -i\hbar \hat{\partial_q} \right)^{\dagger} |\psi > = +i\hbar \int dq' \phi^{*} \partial_{q'}^{\dagger} \psi$ And demanding that $\hat{p}$ is self-adjoint, both expressions have to be equal. Hence: $\int dq' \phi^{*} \partial_{q'}^{\dagger} \psi = -\int dq' \phi^{*} \partial_{q'} \psi$ $\int dq' <\phi |q'> = -\int dq' <\phi | q'>$ $<\phi | \hat{\partial_{q}}^{\dagger} |\psi> = - <\phi | \hat{\partial_{q}} |\psi>$ And since this works for any $\phi$ and ψ: $\hat{\partial_{q}}^{\dagger} = -\hat{\partial_{q}}$
 Recognitions: Science Advisor Sorry for interrupting but please note that 1) a wave function ψ(q) in q-space or ψ(p) p-space is always a projection of the abstract state |ψ> on the position-eigenstate but only on wave functions ψ(q) = . An expression like ∂q|ψ> is ill-defined!!
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Also, the time derivative 'operator' mentioned in the OP is not an operator in the usual sense, it merely depicts the infinitesimal change suffered by a state of the (rigged) Hilbert space as time passes. It's the "q" or the "p" in the argument of the wavefunction with respect to which we apply the mathematical theory of linear operators in (rigged) Hilbert spaces, not "t". The time is merely a parameter, q and p are the true variables. Also, in functional analysis there's no derivative of an operator with respect to another operator, it's with respect to a parameter/variable which is a scalar, not a vector of a (rigged) Hilbert space, neither an operator on it.

Recognitions:
exactly;

and of course

 Quote by Loro ... can we say: $\hat{H} = i \hbar \frac{\partial }{\partial t}$ ?
is wrong; we can't

the time-dep. SE containing the wave functions says that acting with i∂t on a certain state |ψ> has the same effect as acting with H on that state; but this is not true for arbitrary states, only for solutions of the time-dep. SE.

simple example from algebra: suppose you have a 2-vector v and a 2*2 matrix M; now you can write down an eigenvalue equation (M-λ)v = 0; of course that does not imply that M = λ always holds b/c you can always construct vectors u for which (M-λ)u ≠ 0

 Quote by Loro Also, can we say: $\hat{H} = i \hbar \frac{\partial }{\partial t}$ ?
I really don't think you can. The TDSE states that:

$$H |\psi \rangle = i \hbar \frac{\partial}{\partial t} |\psi \rangle$$

I might be wrong, but I don't think it necessarily implies what you said. In fact, way I see it, what you are suggesting is that the TDSE is:

$$H |\psi \rangle = H |\psi \rangle$$

A completely trivial statement, which carries no interesting physical information. The TDSE is only physically interesting exactly because what you said does not hold. I think you can read the TDSE as "the energy of a state does not change (LHS) when an infinitesimal amount of time goes by" (mayne someone can formulate this better).

Thanks.

So one thing I think I've understood:

$<q'| \hat{p} |\psi> = - i \hbar \frac{\partial \psi}{\partial q'}$ holds for any ψ

Whereas:

$<q'| \hat{H} |\psi> = i \hbar \frac{\partial \psi}{\partial t}$ holds only for particular ψ's

And that's why we can't write this expression for the Hamiltonian, but we can for momentum.

 2) the derivative operator ∂q never acts on bras |ψ> but only on wave functions ψ(q) = . An expression like ∂q|ψ> is ill-defined!!
But Dirac does it all the time in "Principles of Quantum Mechanics". It would be great if someone more experienced could confirm that, but as far as I understood it - at the beginning of chapter 22 - he defines $\hat{\partial_q}$ such that:

$\hat{\partial_q} |\psi> = | \hat{\partial_q} \psi >$

Where:

$<q' | \hat{\partial_q} \psi > \equiv \partial_q' \psi$

And then he shows, that $\hat{p} = -i \hbar \hat{\partial_q}$

 Recognitions: Science Advisor let's make a simple example: think about |ψ> as an ordinary 3-vector in R³; think about { is the n-th component of |ψ> w.r.t. does not depend on n, but does! so changing q slightly and checking how |ψ> changes does not make any sense; let's look at n=1, then at n=2, and let's see how |ψ> changes; of course it doesn't; only ψn = depends on n b/c it is the projection on the n-th direction; so in QM the vector |ψ> itself does not have an n-dependence but you are right, |ψ>(t)> carries t-dependence

 Quote by Loro Thanks. So one thing I think I've understood: $= - i \hbar \frac{\partial \psi}{\partial q'}$ holds for any ψ Whereas: $= i \hbar \frac{\partial \psi}{\partial t}$ holds only for particular ψ's
Yes, the first one is the definition of the momentum operator (in q representation) and the second one is the SE (it is not the definition of the hamiltonian).

 Ok thanks everyone for explaining to me the bit about the Hamiltonian. I understand the difference now. Now coming back to the discussion of the $\hat{\partial_q}$ Ok I think we're actually getting closer to my problem. I understand the argument. Dirac actually says in the book that: $\hat{\partial_q} |\psi> = 0$ Where $|\psi>$ is the standard ket. I didn't understand it - but perhaps it is so, for the reasons that you've explained. But what about this - if in particular: $|\psi> = |q'>$ (a position eigenket) Then can we still say: $\hat{\partial_q} |q'> = 0$ ? And I can always expand any $|\psi>$ as a linear combination of $|q'>$'s
 Recognitions: Science Advisor I know it's bold but I don't agree with Dirac's explanation; it's confusing

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Homework Help
 I know it's bold but I don't agree with Dirac's explanation; it's confusing
Nothing wrong with that - you can agree or disagree with someone for any reason you like: personal confusion is fine. As long as you don't say that the explanation is wrong on the grounds that you find it too confusing ;)

Nah, that just means it's not the sort of explanation that would be helpful for you. BFD: there are different learning styles. Yours is not a good fit to Dirac's teaching style.