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seen Oct 6 '14 at 12:28

Learning from scratch


Apr
18
comment If I put a ping pong ball in a vacuum, would it pop?
if it has air inside, yes.
Apr
18
revised How do I correctly interpret $\rho = \psi_1^* \psi_2$?
added 4 characters in body; added 36 characters in body
Apr
18
awarded  Promoter
Apr
18
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
@David actually i corrected that factor after seeing your comment. I was copy pasting my intitial equation's latex to be quicker, and I had left the $i\hbar$ factor sticking to the $\partial_t(\cdots)$ so thanks :-P
Apr
18
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
@David yup.. done!
Apr
18
revised How do I correctly interpret $\rho = \psi_1^* \psi_2$?
deleted 21 characters in body
Apr
18
revised How do I correctly interpret $\rho = \psi_1^* \psi_2$?
added 888 characters in body
Apr
18
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
@David Ok. I will work out the steps for full scrutiny in another edit.
Apr
18
revised How do I correctly interpret $\rho = \psi_1^* \psi_2$?
added 116 characters in body; added 7 characters in body; deleted 1 characters in body
Apr
18
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
@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
18
revised How do I correctly interpret $\rho = \psi_1^* \psi_2$?
added 731 characters in body
Apr
16
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
I get it. Thanks. I was confused because I tried to technically interpret the word correlation.
Apr
16
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
@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
comment How do I correctly interpret $\rho = \psi_1^* \psi_2$?
@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
accepted How do I correctly interpret $\rho = \psi_1^* \psi_2$?
Apr
14
comment What is the difference between $|0\rangle $ and $0$?
wow! I see. thanks!
Apr
14
asked How do I correctly interpret $\rho = \psi_1^* \psi_2$?
Apr
14
comment What is the difference between $|0\rangle $ and $0$?
@Tedd Bunn one question: can't we have a state $|0\rangle$ where the ket represents a column vector in a particular basis where all components are zero? for an analogy in 3-space.. take a point with finite coordinates and shift the origin to that point, and in this new basis the point is represented as a 0-component vector.
Apr
12
comment Who works professionally on reformulation of QFT?
I think the upvotes to this answer serve a very bad example. (even though they're probably motivated by a notorious history of the OP)
Apr
9
comment Correspondence principle
@Noldorin I think you mean $-\frac{i}{\hbar}$ times the commutator