##[H,x^2]=[\frac{p^2}{2m},x^2]=\frac{1}{2m}[p^2,x^2]=p[p,x^2]+[p,x^2]p##
##[p,x^2]=-2i\hbar x##
from that
##[H,x^2]=-2i\hbar px-2i\hbar xp##
from that
##[H,[H,x^2]]=p[\frac{p^2}{2m},-2i\hbar x]+[\frac{p^2}{2m},-2i\hbar x]p+p[\frac{1}{2}m\omega^2x^2,-2i\hbar x]+[\frac{1}{2}m\omega^2x^2,-2i\hbar...
In Dirac definition ##\delta(x)## is ##\infty## when ##x=0##, and ##0## when ##x\neq 0##. My question is when I have some ##\alpha \delta(x)## could I interpretate this like function which have value ##\alpha## in point ##x=0##?
For example you have spin ##S=\frac{7}{2}## and for example ##J=10## quantum Heisenberg model. And you have Monte Carlo simulation code for classical Heisenberg ##S=\infty##. What should you use for ##J## in classical Heisenberg model Monte Carlo code?
I'm speaking in general. I just see that in some papers some authors uses for example four sublattices for body centered cubic lattice. Why not two? I don't understand this.
In ordinary definition antiferromagnet lattice has to sublattices, one with spins up, and one with of spin down in ##T=0##. Why in some cases people discuss situations with four or even more subblatices? Do you have explanation for this? Some references maybe?
In quantum Heisenberg model
\hat{H}=-J\sum_{\vec{n},\vec{m}}\hat{\vec{S}}_{\vec{n}}\cdot \hat{\vec{S}}_{\vec{m}}
##J## can be obtained from dispersion experiment. For large spin ##S## even classical Heisenberg model is good for description of Curie temperature for example. Is that with the same...
What is refered as one Monte Carlo step. In all books, papers people is written that was performed ##5 \cdot 10^{6}## MCS on all system sizes. Or the time is measured in MCS. But what is refered as one MCS? For example in MC simulation of Ising model what is a one MCS?
Thanks for you're answer. I suppose like
##df(x,y)=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy##
special case
##\frac{df}{dx}dx=df(x)##
I made a mistake. But I'm asking when you get that ##(\frac{d^n}{dx^n})_{x_0}##? Please answer my question if you know. In
e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...
you never have ##x_0##.
I have a problem with that. So
f(x+\alpha)=f(x)+\alpha f'(x)+...
My problem is that we have ##\frac{df}{dx}## and that isn't value in some fixed point ##x##. This is the value in some fixed point ##(\frac{df}{dx})_{x_0}##.
Is then in eq
##y'(x)=y(x)##, ##y(x)## value of function or function? :D Question for WannabeNewton.
micromass ok called whatever you want. Why isn't equal ##df=\frac{df}{dx}##?
e^{\alpha\frac{d}{dx}}=1+\alpha\frac{d}{dx}+\frac{\alpha^2}{2!}\frac{d^2}{dx^2}+...=\sum^{\infty}_{n=0}\frac{\alpha^n}{n!}\frac{d^n}{dx^n}
Why this is translational operator?
##e^{\alpha\frac{d}{dx}}f(x)=f(x+\alpha)##
Linear operator A is defined as
A(C_1f(x)+C_2g(x))=C_1Af(x)+C_2Ag(x)
Question. Is A=5 a linear operator? I know that this is just number but it satisfy relation
5(C_1f(x)+C_2g(x))=C_15f(x)+C_25g(x)
but it is also scalar.
Is function ##A=x## linear operator? It also satisfy...
One more question. By solving Sroedinger eq we get
##\varphi_n(x)=\sqrt{\frac{2}{a}}\sin \frac{n\pi x}{a}##
##E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}##
But for example solution of Sroedinger eq is also
##C_1sin\frac {\pi x}{a}+C_2\sin\frac{2\pi x}{a}##
in which state is particle?
What is the energy...
Well I agree with second reason. But you can normalized state in such way that
\int^{\infty}_{-\infty}|\psi(x,t)|dx=1
Right?
And the second one. Why current is defined like
\vec{j}=\frac{\hbar}{2im}(\psi^*\frac{d}{dx}\psi-\psi\frac{d}{dx}\psi^*)
I believed that theory is consistent. But why Bohm took ##|\psi(x,t)|^2##? Is there some bond with
\vec{j}=\frac{\hbar}{2im}(\psi^*\frac{d}{dx}\psi-\psi\frac{d}{dx}\psi^*)