Recent content by laser1

so it is only 0, 4, 8, ...?

hmm yes good point, the relations are valid throughout. So ##|\phi_7\rangle## exists, so ##|\phi_3\rangle## exists, which means ##|\phi_{-1}\rangle## exists, which can't be possible, so that whole ladder can't exist?

Let us consider the following Hamiltonian $$ \hat{H}=\hbar\omega\left(\hat{b}^\dagger \hat{b}+\frac{1}{2}\right). $$ The self-adjoint operator ##\hat b## and its adjoint ##\hat b^\dagger## fulfil the (unusual) commutation relation $$ [\hat b,\hat b^\dagger]=\hat b\hat b^\dagger-\hat...

for d), I am a bit confused. I have two trains of thoughts here any thoughts on which answer is correct, and why the other one is incorrect? Both seem like valid solutions to me. Or is the question ambiguous? thanks

ah ya it's 0 if it's outside isn't it so the 2nd derivative isn't continuous :(. Cheers

When I do this I keep getting a negative answer. Why? My workings: ##<O> = \int \psi* O \psi dx## in general. And ##\hat{p}=\frac{\hbar}{i} \frac{d}{dx}## so ##\hat{p^2} = -\hbar^2 \frac{d^2}{dx^2}##... And by plugging in ##\Psi##, I get ##<p^2>=-\frac{10\hbar^2}{3L^2}##. Any thoughts on why...

Thanks, I guess the justification is "a unit sphere by definition" then!

why unit sphere rather than an arbitrary sphere?

Where does the ##\sin \theta d\theta d\phi## come from? I had thought it came from the fact that the jacobian in spherical coordinates is ##r^2 \sin \theta## but maybe I was wrong.

In the solutions, I must integrate ##Y_1^0 Y_1^{\pm 1}##. But my concern is when it states that the differential element (I think area?) is ##d\theta d\phi##. I know that as then ##dA=r^2 \sin \theta d\theta d\phi##. But the solution only states ##\sin \theta d\theta d\phi##. Why is this? Also...

my lecturer uses the notation ##dF/dy## in the second last term. I am confused why it is not ##\partial F/\partial y## instead.

sorry, edited!

The above image is from my lecturer's notes. My concern is when it seems like my lecturer has split up the dF/dx term into dF/dy y' + dF/dy' y''. Why is it this as opposed to ##\frac{\partial F}{\partial y}## etc.? Or would this not matter, because y is an independent variable, and hence, the...

psi is given above. I have checked multiple times but can't find my mistake. Thank you!

Maybe instead of using the fact that integral of rho over all space is -e, I integrate it explicitly? I get the same answer, though.