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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?
@NickLarsen now that will involve quantitative calculations. You must know the tensile strength of the material that makes the shell of ball. If the pressure inside is approximately atmospheric so it doesn't collapse or pop in air, then you would need to know the pressure of the partial vacuum (0 Torr in case of full vacuum, if that's possible), and find if the material can withstand such a pressure difference. In general, you can safely say that normal materials such as plastic wouldn't withstand the pressure gradients between say, a sputter ion vacuum on one side and air.
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 $\rho = \psi_1^* \psi_2$?
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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)