Timeline for Mahan's derivation of energy current of a free-particle system
Current License: CC BY-SA 4.0
29 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 14 at 18:17 | vote | accept | Redcrazyguy | ||
| Jul 14 at 16:01 | comment | added | Redcrazyguy | @RogerV. I've just submitted my answer, could you take a look and see if it's correct? | |
| Jul 14 at 16:00 | answer | added | Redcrazyguy | timeline score: 0 | |
| Jul 11 at 11:19 | history | reopened |
John Rennie Vincent Thacker Roger V. |
||
| Jul 11 at 10:33 | comment | added | Redcrazyguy | @RogerV. can you reopen my question so I can type up an answer | |
| Jul 10 at 15:56 | history | left closed in review |
Jon Custer Miyase Michael Seifert |
Original close reason(s) were not resolved | |
| Jul 10 at 9:40 | comment | added | Roger V. | I recommend starting with $\partial_t |\psi|^2 = \partial_t\psi^*\psi + \psi^*\partial_t\psi=...$, using time-dependent Schrödinger equation and reducing it to the continuity equation $\partial_t\rho + \nabla\cdot\mathbf{j}=0$. I am not sure all QM textbooks derive current this way, but this is done in Landau&Livshitz QM, if I am not mistaken. You could then derive energy current following similar steps. | |
| Jul 9 at 19:21 | comment | added | Redcrazyguy | @RogerV. I think I've found a way analogous to the derivation of probability current, but it requires treating the Hamiltonian differently depending (i.e. using $\nabla\cdot(\psi^*\nabla\psi)=\psi^*\nabla^2\psi+(\nabla\psi^*)\cdot(\nabla\psi)$ for one term and $H\psi=E\psi$ for another). It has worked out, but I don't think it brings me any closer on Mahan's derivation. | |
| Jul 8 at 20:31 | history | edited | Redcrazyguy | CC BY-SA 4.0 |
added 9 characters in body
|
| Jul 8 at 19:46 | comment | added | Redcrazyguy | @RogerV. I'm also thinking through the energy current in one particle QM, but isn't the plane wave usually considered a steady state solution? If it's losing energy then surely it cannot be steady state? Unless I'm not plugging it into the free particle Hamiltonian? | |
| Jul 8 at 19:44 | comment | added | Redcrazyguy | @RogerV. also in one particle QM isn't the particle current density usually defined (as the real part of the momentum)? I think in this question I'm trying to figure out the form of the density current $\textbf{j}$, not starting from $\textbf{j}$ and solving for the current | |
| Jul 8 at 18:51 | comment | added | Redcrazyguy | @RogerV. just to clarify, when you say "continuation and equation" do you mean the continuity equation (in the form $\nabla\cdot\textbf{j}=\partial_tE$ for energy)? I've managed to do particle current using Mahan's method (here: physics.stackexchange.com/a/820476/259935) | |
| Jul 8 at 14:43 | comment | added | Roger V. | For a hexagonal lattice you will have different dispersion relation and specific integration limits (typically over the Brillouin zone). Can you derive particle and energy current in standard one-particle QM, using continuation and equation and SE? It does give $v_\mathbf{p}\epsilon_\mathbf{p}$ for a plane wave (provided that we have appropriate BC, so that we can do integration by parts and assume that the integral over the boundary vanishes.) | |
| Jul 8 at 14:06 | comment | added | Redcrazyguy | @RogerV. I've changed it, but the same issues still stand 1. Vanishing due to parity is only true if your volume is symmetrically centred at the origin (e.g. I don't think this would hold for a system with a hexagonal lattice, since either way you rotate it you would either have two output paths on one side or an "even" Y shape) 2. If the terms with $\textbf{r}$ vanish then that gets rid of the $\epsilon_k$ term which is in Mahan's solution. | |
| S Jul 8 at 14:02 | review | Reopen votes | |||
| Jul 10 at 15:56 | |||||
| S Jul 8 at 14:02 | history | edited | Redcrazyguy | CC BY-SA 4.0 |
added 54 characters in body
Added to review
|
| Jul 7 at 18:08 | history | closed |
Roger V. Matt Hanson Miyase |
Needs details or clarity | |
| Jul 7 at 13:11 | review | Close votes | |||
| Jul 7 at 18:08 | |||||
| Jul 7 at 12:54 | comment | added | Roger V. | It is not about notation, but about the errors (or perhaps inconsistent use of natotation) in your Q., where you without justification replace vector (in bold font) by its magnitude. Once you fix this, the integral the you claim as divergent will vanish due to parity, as per my first comment. Then a more serious discussion would be possible. | |
| Jul 6 at 18:09 | comment | added | Redcrazyguy | @RogerV. the question has to do with actually evaluating the integral, I realise Mahan's notation isn't ideal but I've managed to understand what it is. However, I can't use the same method I used on particle currents on energy currents, hence the question. | |
| Jul 5 at 16:32 | comment | added | Roger V. | Integral $\int \mathbf{r}f(r)d^3r$ is going to have the same value regardless of the system of coordinates, in which it is evaluated. what is important here is that this is an integral of a vector function (essentially, three different integrals - one for each component) - not identical with $\int rf(r)d^3r$, which is a single integral of a scalar function. I really think that the problem here is poor handling of vectors. | |
| Jul 5 at 15:24 | history | edited | Redcrazyguy | CC BY-SA 4.0 |
added 2 characters in body
|
| Jul 5 at 15:20 | comment | added | Redcrazyguy | @RogerV. you're right there's a factor of two missing that I've now added back in, but that still doesn't change the overall outcome of the argument. Wouldn't $\int xdx=0$ only work for 1. Cartesian coordinates 2. Integral limits being $a,-a$? This wouldn't be something that's generally true, especially when we change to spherical coords where $r$ is always positive. All the $r$s here are $\textbf{r}$, and it's by a quirk that $\nabla\textbf{r}$ acts like the identity, it's not $\nabla\cdot\textbf{r}$. | |
| S Jul 5 at 6:53 | history | suggested | Soumyajit Datta | CC BY-SA 4.0 |
Improved formatting of the equations.
|
| Jul 4 at 20:26 | review | Suggested edits | |||
| S Jul 5 at 6:53 | |||||
| Jul 4 at 14:28 | comment | added | Roger V. | I think you lost a factor of 2 in expanding $\nabla^2(r\psi)$. Also $\int xdx=0$, since it is an asymmetric function (under some regularization caveats, which are implied when using plane waves.) $\mathbf{r}$ is a vector, whereas $r=|\mathbf{r}|$ is its magnitude (there might be some confusion between gradient $\nabla r$ and divergence $\nabla \cdot\mathbf{r}$) | |
| Jul 4 at 14:15 | history | edited | Redcrazyguy | CC BY-SA 4.0 |
edited body
|
| Jul 4 at 13:01 | history | edited | Tobias Fünke |
edited tags
|
|
| Jul 4 at 12:52 | history | asked | Redcrazyguy | CC BY-SA 4.0 |