@DanielSank Note that I wrote $\psi(x)$ in that expression - you hadn't yet decided to go bra-ket
But I thought we were past that - the bra-ket analogon is $\vert u\rangle = \exp(\mathrm{i}(\theta/L)x)\lvert \psi\rangle$.
user54412
@DanielSank Oh. Well you certainly need that convexity in order for left-propagating diffusion to balance rightward advection. You could probably be more specific by noting how the u_x term has to balance the u_xx term.
@DanielSank It's not an eigenvalue. That's the definition of $\lvert u \rangle$, we're defining it to be the action of the operator $\exp(\mathrm{i}(\theta/L)x)$ on $\vert \psi \rangle$.
More physically, we are shifting the momenta of the original state by $\theta/L$ to get a state that has 1, and not some phase, as eigenvalue of the translation by $L$.
@DanielSank Yeah, we defined $\lvert u \rangle$ to get rid of the $\theta$ -. we want a periodic wavefunction, so the state we want must have eigenvalue 1 under $T_n$.
@0celo7 I want to do something like have $F_\infty$ be the derivative of momentum with respect to asymptotic time $t$ rather than the mass's proper time $\tau$. The redshift factor should be $\mathrm{d}\tau/\mathrm{d}t$.
If you did solve it, you didn't explain it very well.
@user507974 I would guess by the name that physical chemistry means chemistry where you try to use first principles (i.e. physics) to understand the chemistry, i.e. reaction rates, bond lengths, etc.
user54412
^ that's my interpretation too
user54412
20:24
pchem is full of wavefunctions and such, and only really looks at small parts of a system
user54412
in some sense it's the opposite of stat mech -- much of non-pchem chemistry looks at aggregate behavior of many particles
@DanielSank No, the thing is that $k$ and $k+l$ are physically equivalent for $l$ a reciprocal lattice vector (since the associated $\theta$ then differs by $2\pi$).
Ok, sorry, I'm really trying to understand why fluctuations in $Q_\text{offset}$ lead to dephasing of our system.
Fluctuations in $Q_\text{offset}$ should be similar to fluctuations in $k$.
Different values of $k$ have different energy differences between the bands in the system, so I can sort of see how fluctuations in $Q_\text{offset}$ sort of lead to fluctuations in the quantum phase of a superposition state, but not really.
@DanielSank Oh, it does! Changing the $k$ in the Bloch state by a reciprocal doesn't do anything, but changing the $k$ of a momentum eigenstate by a reciprocal does change it (the momentum eigenvalue is different!)
Why does Susskind say that up and down are orthogonal to each other? I would say that they are opposite. He says it himself when says that measuring up=YES means that down = NO. In the classical world, if I measured x component, I would get information about y only if x and y are not perpendicular. Otherwise they should not correlate. Absolute correlation means that we are on the same axis. How the same axis can be orthogonal?
@ValentinTihomirov They are orthogonal in the sense that the inner product of an up and a down state is zero. It has nothing to do with the geometric meaning of "up" and "down".
@ValentinTihomirov A state being orthogonal to another means you have zero chance to detect one state as being the other. Is it not obvious that for a state that by definition has definite spin up, you have no chance to detect its spin being down?
@ValentinTihomirov Uh, no, the axes area not put anywhere. It's that the (Hilbert) space in which the states lie is not our actual space, it is an abstract vector space. Just because in our actual space the vectors that describe up and down are linearly dependent does not mean they have to be in the Hilbert space.
What was on the same axis, eg. up and down, various positions on X-axis, became orthogonal because they are exclusive. Meantime orthogonal things, like x and y, position and momentum, are defined through formerly other axis, which were orthogonal to each other, now can be expressed in terms of up-down axis or X axis.
How is it that multiplying a vector by a constant 1 is the same as acting by identity matrix? Is matrix a special case of a real number, is it vice versa or it is just a fluke?
@ValentinTihomirov It's just a consequence of both the identity matrix and the real number 1 being essentially defined by them doing nothing when multiplying them with something. I don't understand why this would make you think that a matrix is a special case of a real number or vice versa (but indeed, you can see every real number as a 1x1 matrix).
@ValentinTihomirov Oh, no, that's not what is happening here. Just because the identity matrix and the scalar multiplication with 1 do the same thing that doesn't mean they are equal as objects.
@user507974 You can (that is indeed one possible inclusion $\mathbb{R}\to\mathbb{R}^{n\times n}$), but that's not what is going on here (note that this inclusion does not send 1 to the identity)
However, what one can say is that you can think of scalar multiplication by a real number as multiplying by the matrix which has this real number as every diagonal entry.
@BernardMeurer Are you saying they rejoice because he's still single available for marriage or because he hasn't made the life of one of them miserable yet?
@0celo7 1. All I did was ask @BernardMeurer a clarifying question :P 2. The point is less that you asked me about algebra and more that you kept bothering me after I explicitly told you I wouldn't say more.
