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seen Apr 8 at 23:13

I write code in C++, Python, Java and now Haskell.


Apr
8
comment Why is the electric field operator normalized by a volume?
Also, unless there is a way to eliminate $V$, what value do I choose for $V$ when I don't obviously have a "box" to play with? That degree of freedom lets me have the theory say anything I want, doesn't it?
Apr
8
comment Why is the electric field operator normalized by a volume?
But here, if I let $V$ go to infinity, I get 0! Is that not a problem?
Apr
8
revised Why is the electric field operator normalized by a volume?
added 8 characters in body
Apr
8
asked Why is the electric field operator normalized by a volume?
Apr
3
asked Meaning of “vacuum state”?
Mar
25
accepted Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
Mar
24
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
Can you justify this? It may very well be, but it is rather surprising. It requires some explanation.
Mar
24
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
QMechanic - don't you think the first derivation I showed at the top, in the position basis, is bogus? The step one before last remove the integration on $dx'$ ignoring $x'$ in the integrand, for example. That can't be right, can it?
Mar
24
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
Oops - sorry! Let me take that back! I had missed that you put $i$ under in the expression for $\langle p|x\rangle$ :-) My excuse is that the characters are really small.
Mar
24
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
AhA! But then - just to be absolutely clear, what do you mean by "act on the bras rather than the kets"? Or rather, it doesn't seem to bring in anything decisive: you can remove $|x\rangle$ on both sides of the $\langle p|\hat{x}|x\rangle$ equation, but nothing is really gained that I can tell of, since we have a good definition for $\langle p|x\rangle$ that is easy to differentiate anyway.
Mar
24
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
QMechanic - I can see what you are saying, and you are probably right, but I can't seem to be able to use it to fix my derivations above. In particular, there is still a minus sign coming down from the exponential when applying $\hat{x}$ in differential form in the $p$-basis - unless I'm missing something?
Mar
23
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
The one before last step in the first derivation probably cannot be correct, because it is integrating over $p'$, but ignores the $p'$ before $\langle x\ |\ p'\rangle$.
Mar
23
revised Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
edited title
Mar
23
comment Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
That is the question! It's wrong in at least one place, and probably several, and I can't seem to figure the wrong step(s) :-)
Mar
23
revised Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
added 100 characters in body
Mar
23
asked Calculating $\langle x | \hat{x} | p \rangle$ in $p$ basis
Mar
23
comment CSCO and Hamiltonian
Is it true that the number of observables I need in a CSCO is equal to the number of symmetries in the Hamiltonian?
Mar
23
comment CSCO and Hamiltonian
Without knowing it, I had "smelled" there was something going on between degenerate energy levels and observables required to describe a system, but I got it "wrong" (i.e. it's not as obvious as in the case of the 2D SHO :-)
Mar
23
comment CSCO and Hamiltonian
:-) I meant "subtle" because I got the wrong impression by working on the 2D SHO in class: I thought the degeneracies resulted from having $n_x$ and $n_y$ straight into the Hamiltonian.
Mar
23
comment CSCO and Hamiltonian
Yes - I'm finding some info on the Wikipedia, that shows there is a relationship between the Hamiltonian and degeneracies via symmetries of the Hamiltonian - more subtle than I thought...