Timeline for How do I correctly interpret $\rho = \psi_1^* \psi_2$?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2011 at 21:11 | comment | added | Ron Maimon | @Marek: sorry, you are absolutely right. The quantity is the probability amplitude for being in state 2 when you know you are in state 1. States in quantum mechanics are no mutually exclusive, as you know. | |
| Sep 3, 2011 at 22:20 | comment | added | Marek | @Ron: Kalitvianski's answer doesn't really explain anything and I have no idea what you have in mind either. The quantity in question can certainly be interpreted as overlap of wavefunctions but that's simply not a probability. It can be complex for goodness's sake... | |
| Aug 31, 2011 at 3:24 | comment | added | Ron Maimon | @Marek: -1 (sorry--- your answer is wrong and upvoted) this is explained properly below in Kalitvianski's answer--- the quantity in question is the probability, knowing that the particle is in state $\psi_1$ that it will be in state $\psi_2$. This is called the "overlap" of the two states. | |
| Apr 18, 2011 at 5:39 | comment | added | Marek | @yayu: continuity equation in quantum mechanics is always trivial in some sense. Say you have one particle. Then it's obvious that the probability that it is somewhere will always have to sum to unity. Continuity equation just restates this locally. If you have more particles, or more quantum states then obviously the local law will change. You can then interpret $\rho$ as probability density on some space (which will generally be higher dimensional) but its relation to the original position space is not apparent. | |
| Apr 18, 2011 at 2:31 | comment | added | yayu | @Marek sorry for dragging you back to this concerning a retrospective doubt. But i noticed that the quantity $\rho$ appeared to satisfy the continuity equation, and the interpretation says that it must be the probability density, albeit some strange kind of probability density. | |
| Apr 16, 2011 at 20:21 | comment | added | yayu | I get it. Thanks. I was confused because I tried to technically interpret the word correlation. | |
| Apr 16, 2011 at 7:44 | history | edited | Marek | CC BY-SA 3.0 |
Added some clarifications
|
| Apr 16, 2011 at 3:59 | comment | added | yayu | @marek then what would be an integral representation of $P_{1 \to 2} = {\left \Vert \left < \psi_1 | \psi_2 \right > \right \Vert^2 \over \left \Vert \psi_1 \right \Vert^2 \left \Vert \psi_2 \right \Vert^2}$ | |
| Apr 15, 2011 at 23:24 | comment | added | Marek | @yayu: no, it doesn't make any sense. Look, states in quantum theory can be extended in $x$ coordinate. When one state transitions into another the contribution to the probability of this process is from the correlation between all parts of the wavefunctions (i.e. correlations between $\psi_1(x)$ and $\psi_2(y)$ with $x \neq y$). You obtain only probability amplitude by a single integral. To get a genuine probability, you need a double interal (i.e. a square, as usual). | |
| Apr 15, 2011 at 17:12 | comment | added | yayu | @Marek thanks. one further question, does it make sense to say that $dP_{1\rightarrow 2} = \int (\psi_1^* \psi_2)^* \psi_1^* \psi_2 d^3r$... what does a probability of transition mean in a small volume element. Because states usually means a specifications of the values of $\psi$ in all points in space.. then how can we have a probability of state transition in a small volume?? | |
| Apr 15, 2011 at 17:07 | vote | accept | yayu | ||
| Apr 15, 2011 at 3:17 | history | edited | Marek | CC BY-SA 3.0 |
Removed the part about units
|
| Apr 15, 2011 at 3:16 | comment | added | Marek | @Ted: oh right, thanks. Probably just temporary loss of senses :) | |
| Apr 14, 2011 at 14:15 | comment | added | Ted Bunn | I don't know what you mean about not having the right units: the quantity in the question is indeed dimensionless, as a probability should be. You're right about the rest, especially the fact that this quantity is complex and so can't be a probability, which is the important thing. | |
| Apr 14, 2011 at 10:44 | history | answered | Marek | CC BY-SA 3.0 |