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bio website anuars.wordpress.com
location Mexico
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visits member for 2 years, 10 months
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Graduate Physics Student at Department of Complex Systems.

Institute of Physics, National Autonomous University of Mexico (UNAM).

$$\frac{\partial P(\mathbf{r},t)}{\partial t}=\sum_{\mathbf{r}'}[p(\mathbf{r},\mathbf{r}')P(\mathbf{r}',t)-p(\mathbf{r}',\mathbf{r})P(\mathbf{r},t)]$$


3h
comment Wavefunctions in different Hilbert spaces
@Timaeus is right, there aren't lots of different Hilbert spaces, but there are lots of different basis, say $\{|x\rangle\}$ or $\{|p\rangle\}$. And BTW, $\psi(x,y):=\langle x, y |\psi \rangle$.
3h
comment Wavefunctions in different Hilbert spaces
@march Actually, the correct expression would be writing $| \psi \rangle$ at the left hand side (without the $x$ in $\psi(x)$, because $| \psi \rangle$ does not depend on x, but when you project at the position space you have: $\psi(x):=\langle x| \psi \rangle$).
5h
asked Proof of: if $\hat H | a_n\rangle=a_n|a_n\rangle$ then $f(\hat H)| a_n\rangle=f(a_n)|a_n\rangle$?
1d
comment Continuum analogue of $ \langle \psi | \psi \rangle = \sum _i a_i^* a_i$
@user12262 I think I understood your comment. You're just saying that in your notation, the derivation of the result would be more clear, right? By the way, the derivative $d/dx$ of the definite integral $\int_a^b f(x)dx$ (with $a$ and $b$ constants) is zero because when you make the integral, you get just a constant. I think you referred to the fundamental calculus theorem by this: $(d/dx)\int_a^xf(x')dx'=f(x)$, which is not the same as the first derivative. Nevertheless I understood your point.
1d
comment Continuum analogue of $ \langle \psi | \psi \rangle = \sum _i a_i^* a_i$
@ACuriousMind I'm not asking for you to check my work. I'm asking for experience in this problem, I have no clear idea of what it asks for. Therefore your comment is not constructive.
1d
comment Continuum analogue of $ \langle \psi | \psi \rangle = \sum _i a_i^* a_i$
@qfzklm When I said "maps to" I meant from the discrete case to the continuum case (talking about spectrum of eigenstates). I don't understand the reason you say $a_i$ are constants, actually they are constants but I think that has noting to do with my question. And of course $| \psi \rangle$ maps to $\psi(x)$ when you take $\langle x| \psi \rangle$.
1d
comment Continuum analogue of $ \langle \psi | \psi \rangle = \sum _i a_i^* a_i$
@user12262 Can you suggest me a book where I can find your replacement, please? I've never seen why is that possible or where it comes from. Thanks in advance!
2d
asked Continuum analogue of $ \langle \psi | \psi \rangle = \sum _i a_i^* a_i$
Jun
18
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May
12
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May
11
awarded  Popular Question
Mar
17
awarded  Popular Question
Dec
8
awarded  Popular Question
Dec
6
asked Work done by heating a balloon with $H_2$ ideal gas
Dec
1
comment Are there diffrent understandings of dimensions?
As you said we certainly live in 3 SPATIAL dimensions. And using only these 3 spatial dimensions and time we can describe phenomena involving a set of physical properties likes mass, charge, electric/magnetic moment, current, refraction index, etc.
Nov
25
asked Examples of systems with linear response behavior
Nov
4
awarded  Yearling
Oct
5
accepted Calculation of the partition function for a classical 2D gas lying on the surface of a sphere of constant radius $R$
Sep
28
answered Calculation of the partition function for a classical 2D gas lying on the surface of a sphere of constant radius $R$
Sep
24
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