Gamebm

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Conductivity and Probability flow

Apologies for my ignorance. I understand in classical electrodynamics conductivity means flow of electrons, now since here conductivity is understood in term of solution of Schrodinger's equation, what is conductivity then, which differs conductor from insulator. Now if one looks at the Bloch wave, as one approximation mentioned in this article, I will intuitively assume conductivity, the flow of electrons, is connected with the probability flow of electrons, that is (from the conservation of probability implied by the Schrodinger's equation)

${\vec {J}}={\frac {\hbar }{2im}}(\Psi ^{*}\nabla \Psi -\Psi \nabla \Psi ^{*})$

substituting the expression of Bloch wave, I get ${\vec {J}}={\frac {\hbar {\vec {k}}}{m}}$. So no matter where the electron is, it is moving like as a flow. Now I kinda guess, the summation of all the reciprocal vector ${\vec {k}}$ inside the Brillouin zone cancels out when the valence band is filled up. Is it true? Electrons are actually move collectively in the valance band but do not present macroscopic flow due to cancellation?

Then another question is, how to understand this in the case superconductivity? Gamebm (talk) 17:13, 27 November 2012 (UTC)

I don't think your calculation is right. With $\Psi =ue^{ikr}$, I get

$\Psi ^{*}\nabla \Psi -\Psi \nabla \Psi ^{*}=2ik|u|^{2}+u^{*}\nabla u-u\nabla u^{*}$

Can you please double-check? --Steve (talk) 23:29, 27 November 2012 (UTC)

Also as a side note, this won't help you understand superconductivity at all, since it doesn't arise from the band structure...a13ean (talk) 23:47, 27 November 2012 (UTC)

Thanks for pointing that out. I was study the topic and for some reason incorrectly assumed that u must be real. Now it seems the first term ($|u|^{2}$) will not change the conclusion since if one sums up all the k at a given spatial point, it is just a constant. However if the second term does remain, it will give a non-zero contribution to the flow (locally). Then I may understand that a filled band does present some flow locally since the charge flow calculated from QM does not vanish(? - is it measurable?). Now globally, since the second term is surface like (or due to its periodic behavior), it still may not lead to macroscopic flow, (since the length scale in our question is of lattice size.) and for the very same argument, the first term reduces to a irrelevant normalization constant.

I have no idea what superconductivity is. I was just trying to think of it following this line of thought, if all those electron pairs are in the same state (no idea what the wave function shall look like, will study it), and one is again legitimate to calculate its probability flow, but if everything is in the same state, and the resulting flow does not vanish (since it conducts not by excited to another state, but simply staying in this state?), then does this imply that there is some preferred direction in space? Gamebm (talk) 18:04, 28 November 2012 (UTC)

Electron motion is properly calculated using the formula for the group velocity of the electron wavepacket ... not the formula for probability current. They must be related ... I would not be surprised if they are identical ... but I'm not sure of the details off the top of my head.

As pointed out by A13ean, this line of thought will not lead you towards understanding superconductivity. Band structures are part of the "single-particle picture", where you by-and-large ignore the fact that electron quasiparticles interact with each other. But interactions between electron quasiparticles are the entire basis for superconductivity. --Steve (talk) 19:14, 28 November 2012 (UTC)

If you're interested in that bit, the last chapter of Ashcroft and Mermin gives an overview of how superconductivity arises from electron-electron interactions. a13ean (talk) 19:25, 28 November 2012 (UTC)

Few notes before and after some reading of Ziman.

(1) "group velocity" and "probability flow". Probability flow is just a kind of wave which flows, itself may have phase or group velocity. Group velocity is just something comes out when one summing up ${\vec {k}}$ when considering a dispersion relation.

(2) In the book of Ziman for instance, $u({\vec {r}})$ (in Bloch wave) is ignored when discussing electron flow, the arguments may come from when one discusses group velocity of wave package in QM. So somehow it turns out to be a somewhat (reasonable) approximation not to think of the contribution of $u({\vec {r}})$ in the first place, at least, when one considers group velocity.

(3) The fact that the system is non-perturbative has not really anything to say about whether one can write down the state in turn of wave function. See for example, Eq.(11.55) of Ziman when flow in superconductor was discussed, the notion of wave function is used. The wave function is not of one particle, but the important thing is its symmetry in reciprocal space, see below.

(4) In the book of Ziman, it explains what I was asking for in the first place. In fact, my argument was more based on spacial isotropy of the wave function in reciprocal space, so that when summing over all possible momenta, isotropy implies that there does not exist a preferred direction, which implies no flow in coordinate space. In the conductor/semiconductor case, electron has to be exited to a high-energy state to break this symmetry, which in turn gives macroscopic flow. In the case of superconductivity, the argument is based on the fact that the ground state and energy gap themselves are solutions in respect to Fermi surface in stead of some fixed coordinates in reciprocal space. In fact, an external field may distort Fermi surface and make it anisotropic, therefore even though all the electron pairs are in the ground state, one may still have macroscopic electron flow, simply because the ground state is not isotropic as in the case of conductor. Gamebm (talk) 20:10, 2 December 2012 (UTC)

(1b) In the book of Ashcroft, it was actually show in Ch.12 and Appendix E that the group velocity and expectation of velocity operator (the later is proportional to probability flow) are the same thing. It can be understood as following: the physics of velocity operator comes (mainly) from probability flow, since the physics of probability flow, by definition, is related to the measurement of detected matter flow, it is not surprising at all when it turns out to be the the same as the velocity of the propagation of the maximum of wave function -- group velocity of wave function.

(4b) This cancellation of flow is explicitly discussed in the book of Ashcroft, it seems only sparkles to who is actually thinking of this. Gamebm (talk) 14:08, 18 September 2014 (UTC)

Conservation of energy and momentum

I have a simple question concerning the conservation law of the energy-momentum tensor, and I cannot figure out where did I make the mistake in the following calculations. Many thanks in advance!

Let us discuss a simple example. Considering a single free particle with 4-momentum $p^{\mu }$ in Cartesian coordinates. In the following, we will try to verify explicitly for a closed surface, one has

$I=\int d\sigma _{C\mu }T_{C}^{\mu 0}=0$

where the closed 3-surface $\sigma _{\mu }$ is defined by $\tau =\tau _{1},\tau _{2}$, $\eta =\eta _{1},\eta _{2}$, $x=x_{1},x_{2}$ and $y=y_{1},y_{2}$ with the following definitions

$\tau ={\sqrt {t^{2}-z^{2}}}$

$\eta ={\frac {1}{2}}\ln {\frac {t+z}{t-z}}$

Since the energy momentum tensor is evaluated in the Cartesian coordinates, the expression is not covariant, the vanishing contraction corresponds to the energy conservation in Cartesian coordinates. We use the following notations

$I=I_{\tau }+I_{\eta }+I_{x}+I_{y}$

As shown below, the 3-surface element at $\tau =const$ reads

$d\sigma _{C\mu }=dxdyd\eta \left({\begin{array}{c}\tau \cosh \eta \\0\\0\\-\tau \sinh \eta \end{array}}\right)$

and the integral on the 3-surface $\tau =\tau _{1},\tau _{2}$ yields

${I}_{\tau }=\int _{\tau _{1}}\tau dxdyd\eta (\cosh \eta T_{C}^{00}-\sinh \eta T_{C}^{30})-\int _{\tau _{2}}\tau dxdyd\eta (\cosh \eta T_{C}^{00}-\sinh \eta T_{C}^{30})$

Similarly, the 3-surface element at $\eta =const$ reads

$d\sigma _{C\mu }=dxdyd\eta \left({\begin{array}{c}\sinh \eta \\0\\0\\-\cosh \eta \end{array}}\right)$

and the integral on the 3-surface $\eta =\eta _{1},\eta _{2}$ yields

${I}_{\eta }=\int _{\eta _{1}}dxdyd\tau (\sinh \eta T_{C}^{00}-\cosh \eta T_{C}^{30})-\int _{\eta _{2}}dxdyd\tau (\sinh \eta T_{C}^{00}-\cosh \eta T_{C}^{30})$

Expressions for the other two 3-surfaces are more intuitive, one has

${I}_{x}=\int _{x_{1}}dtdydzT_{C}^{10}-\int _{x_{2}}dtdydzT_{C}^{10}$

and

${I}_{y}=\int _{y_{1}}dtdxdzT_{C}^{20}-\int _{y_{2}}dtdxdzT_{C}^{20}$

The energy momentum tensor of a point-like particle $q$ in Cartesian coordinates possesses the following form

$T_{C}^{\mu \nu }={\frac {p^{\mu }p^{\nu }}{p^{0}}}\delta ^{3}({\vec {x}}-{\vec {x}}_{q})={\frac {p^{\mu }p^{\nu }}{p^{0}}}\delta (x-x_{q})\delta (y-y_{q})\delta (z-z_{q})$

where

$\delta (z-z_{q})={\frac {\delta (\eta -\eta _{q})}{\tau \sinh \eta }}={\frac {\delta (\tau -\tau _{q})}{\cosh \eta }}$

depending on whether one deals with 3-surface parameterized in z, $\eta$ or $\tau$. Substituting into the expressions of ${I}$, one gets

${I}_{x}=\int _{x_{1}}dtdydzT_{C}^{10}-\int _{x_{2}}dtdydzT_{C}^{10}=\int dt(p^{1}-p^{1})=0$

${I}_{y}=\int _{y_{1}}dtdxdzT_{C}^{20}-\int _{y_{2}}dtdxdzT_{C}^{20}=\int dt(p^{2}-p^{2})=0$

${I}_{\tau }=\int _{\tau _{1}}d\eta (p^{0}{\coth \eta }-p^{3})\delta (\eta -\eta _{q})-\int _{\tau _{2}}d\eta (p^{0}{\coth \eta }-p^{3})\delta (\eta -\eta _{q})=\left\{{\begin{matrix}p^{0}(\coth \eta _{a}-\coth \eta _{b})&{\textrm {theworld-linegoesacrossbothsurfaces}}\\0&{\textrm {theworld-linegoesacrossneithersurface}}\\{\textrm {othervalues}}&{\textrm {othercases}}\end{matrix}}\right.$

${I}_{\eta }=\int _{\eta _{1}}d\tau (p^{0}\tanh \eta -p^{3})\delta (\tau -\tau _{q})-\int _{\eta _{2}}d\tau (p^{0}\tanh \eta -p^{3})\delta (\tau -\tau _{q})=\left\{{\begin{matrix}p^{0}(\tanh \eta _{1}-\tanh \eta _{2})&{\textrm {theworld-linegoesacrossbothsurfaces}}\\0&{\textrm {theworld-linegoesacrossneithersurface}}\\{\textrm {othervalues}}&{\textrm {othercases}}\end{matrix}}\right.$

where $\eta _{a},\eta _{b}$ are the values of $\eta$ coordinates when the particle does go across the surfaces $\tau _{1},\tau _{2}$. Among the four possible surfaces discussed above, one notes that the particle always goes across two of them. In practice, one may always choose a big enough area of $\tau _{1},\tau _{2}$ surface, so that the particle goes across $\tau _{1},\tau _{2}$ surfaces and, one ignores the integral on the $\eta _{1},\eta _{2}$ surfaces. In a more general case, if the system consists of many point-like particles, one can usually find big enough area of $\tau _{1},\tau _{2}$ so that all the particles go through the $\tau _{1},\tau _{2}$ area. One sees that the above expression does not vanish as expected from the conservation of energy.

• corrections: The expression of :$\delta$ function above is not correct, it should consider contain the information of the world-line of the particle, but the above expression does not. The correct expression is

$\delta (z-z_{q})={\frac {\delta (\eta -\eta _{q})}{\tau \cosh \eta -v_{z}\tau \sinh \eta }}$

when parameterized in term of :$\eta$. The final result is, of course, :$I=0$.

Surface element and initial energy

To evalute the surface element, one may transform (the orientation of) a surface element in hyperbolic coordinates $\tau =const.$, which is a unitary vector in hyperbolic coordinates, into Cartesian coordinates (here we use :$B$ to indicate hyperbolic coordinates and :$C$ to indicate Cartesian coordinates.)

$e_{B}^{\mu }=\left({\begin{array}{c}1\\0\\0\\0\end{array}}\right)$

$e_{C\mu }=\eta _{\lambda \mu }e_{C}^{\lambda }=\eta _{\mu \lambda }{\frac {\partial x_{C}^{\lambda }}{\partial x_{B}^{\nu }}}e_{B}^{\nu }=\left({\begin{array}{cccc}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}}\right)\left({\begin{array}{cccc}\cosh \eta &0&0&\tau \sinh \eta \\0&1&0&0\\0&0&1&0\\\sinh \eta &0&0&\tau \cosh \eta \end{array}}\right)\left({\begin{array}{c}1\\0\\0\\0\end{array}}\right)=\left({\begin{array}{c}\cosh \eta \\0\\0\\-\sinh \eta \end{array}}\right)$

It is straightforward to verify at this point $|e_{C\mu }|=|e_{B\mu }|=1$. The area of the surface element can be calculated as following

$d{\vec {r}}_{1}\equiv {\frac {\partial x_{C}^{\mu }}{\partial \eta }}d\eta =\left({\begin{array}{c}\tau \sinh \eta d\eta \\0\\0\\\tau \cosh \eta d\eta \end{array}}\right)$

$d{\vec {r}}_{2}\equiv {\frac {\partial x_{C}^{\mu }}{\partial x}}dx=\left({\begin{array}{c}0\\dx\\0\\0\end{array}}\right)$

and

$d{\vec {r}}_{3}\equiv {\frac {\partial x_{C}^{\mu }}{\partial y}}dy=\left({\begin{array}{c}0\\0\\dy\\0\end{array}}\right)$

for completeness, we also write down here

$d{\vec {r}}_{0}\equiv {\frac {\partial x_{C}^{\mu }}{\partial \tau }}d\tau =\left({\begin{array}{c}\cosh \eta d\tau \\0\\0\\\sinh \eta d\tau \end{array}}\right)$

$|d\sigma _{C\mu }|_{(\tau =const.)}\equiv |d\sigma _{C\mu }|=|d\sigma _{C}^{\mu }|=|d{\vec {r}}_{1}||d{\vec {r}}_{2}||d{\vec {r}}_{3}|=\tau dxdyd\eta$

which had used the fact that the three vectors are mutually orthogonal. In fact, this also implies

$|d\sigma _{B\mu }|=\tau dxdyd\eta \rightarrow |d\sigma _{C\mu }|=\tau dxdyd\eta$

where $d{\vec {r}}$ are the 4-vectors in Cartesian coordinates. One sees that the size of the surface element defined by $dx,dy,d\eta$ does not change.

As a side note, in the above, the surface element was calculated by (1) assuming that it s a vector, therefore its components in Cartesian coordinates was obtained by coordinate transformation (2) the module of the surface element was obtained by using the fact the surface $\tau =const.$ is parameterized in terms of orthogonal coordinates, therefore all displacements corresponding to each variables ($\tau ,\eta ,x,y$) are orthogonal, the total area of the 3-surface element is simply a product of the module of three displacements.

On the other hand, the surface element can be calculated by using some more general formulae, which reads

$d\sigma _{\mu }=\epsilon _{\mu \alpha \beta \gamma }e_{1}^{\alpha }e_{2}^{\beta }e_{3}^{\gamma }d^{3}y$

where $\epsilon _{\mu \alpha \beta \gamma }={\sqrt {-g}}[\mu ,\alpha ,\beta ,\gamma ]$ is the Levi-Civita tensor, $y_{1},y_{2},y_{3}$ parameterize the 3-surface, and $e_{1}^{\alpha }=\partial x^{\alpha }/\partial y_{1},e_{2},e_{3}$ describe the connection between the coordinates and its parameters. For instance, see the discussion near (3.2.2) of . In our case of the Cartesian coordinates, ${\sqrt {-g}}=1$, therefore $d\sigma _{C\mu }=[\mu ,\alpha ,\beta ,\gamma ]e_{1}^{\alpha }e_{2}^{\beta }e_{3}^{\gamma }d^{3}y$. It is straightforward to verify that one obtains the same result, namely,

$d\sigma _{C\mu }=dxdyd\eta \left({\begin{array}{c}\tau \cosh \eta \\0\\0\\-\tau \sinh \eta \end{array}}\right)$

The energy momentum tensor

Up to this point, we have not discussed the explicit form of energy momentum tensor. According to ,

${\begin{aligned}T_{C}^{\mu \nu }=\sum _{i}{\frac {p_{i}^{\mu }p_{i}^{\nu }}{p_{i}^{0}}}\delta ^{3}({\vec {x}}-{\vec {x}}_{i})\end{aligned}}$

Integrating on the $t=t_{0}$ surface, one obtains

${\begin{aligned}E_{C}({\text{IC}})=\int _{\sigma _{\mu }(t=const.)}T_{C}^{\mu 0}=\int dxdydzT_{C}^{00}=\sum _{i}p_{i}^{0}\end{aligned}}$

which is as expected. --Gamebm (talk) 19:05, 28 April 2014 (UTC)

Your question is far from simple (at least for someone like me who does not do these calculations often). I have not checked everything you did above, so I cannot testify to its correctness. But I do see one possible source for your problem — your division by $p^{0}$ in the expression for the stress-energy tensor. The article says "In special relativity, ...". This implies that you are using the kind of Cartesian coordinates in which the Lorentz transformation is expressed. But you are not using such a system of coordinates. Accordingly, you should adjust the formula to make the factor of p by which you are dividing orthogonal to the surface in terms of which the delta-functions are expressed, for example, the $\tau =\tau _{1}$ surface. JRSpriggs (talk) 05:17, 2 May 2014 (UTC)

Hi JRSpringgs, I found my mistake and corrected it, it came from the $\delta$ function, as a result, the integral cancels out perfectly. I would not say the problem is not simple, it is the energy conservation problem of a free particle, the geodesic of a particle in flat space-time. I was careful about the energy momentum tensor, that part was all right. Again, thanks for the help! --Gamebm (talk) 12:09, 5 May 2014 (UTC)

WikiProject Investment

Hey there! Id like to invite you to the WikiProject Investment

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Cheers! WikiEditCrunch (talk) 19:01, 24 August 2017 (UTC)

Pseudospectrum and black hole quasi-normal mode (in)stability, arXiv:2004.06434v4, by José Luis Jaramillo et al

(69)

这里我们以RW势场主方程在hyperboloidal坐标下的形式,以及对应的乌龟坐标下平面波的形式推导的自洽性为例,给出一个hyperboloidal坐标变换的具体例子.

这里的坐标变换包括下述内容. 第一个是RW度规标准的乌龟坐标变换

$r_{*}=r+\ln(r-1)$

第二个是对乌龟坐标的标度变换,因为我们取视界为1,故标度因子$\lambda =4M=2$

${\bar {x}}={\bar {r}}={\frac {r_{*}}{2}}$

${\bar {t}}={\frac {t}{2}}$

第三个是把空间坐标转换到紧致的区间$[0,1]$的形式

$x=1-{\frac {1}{r}}$

注意到我们直接通过径向坐标$r$(而非${\bar {x}}$)来更直接的建立这个关系,乌龟坐标仍然是重要的,因为变换后的时间$\tau$无穷远处类光必须在乌龟坐标下建立. 与文中不同,我们这里让视界与空间无穷远分别对应"0"和"1",而非对调的形式.

由上面的定义,我们首先给出空间的坐标变换

${\bar {x}}={\frac {1}{2}}r_{*}={\frac {1}{2}}\left(r+\ln(r-1)\right)={\frac {1}{2}}\left({\frac {1}{1-x}}+\ln \left({\frac {1}{1-x}}-1\right)\right)={\frac {1}{2}}\left({\frac {1}{1-x}}+\ln x-\ln(1-x)\right)\equiv g(x)$

由此,我们建立了$x$和${\bar {x}}$之间的关系,这就是(6)式第二行中定义的$g$函数.

最后我们需要给出$h$函数的形式. 我们要求它在视界处渐进的与乌龟坐标一致,但在空间无穷远处与乌龟坐标或者径向坐标差一个负号. 具体的,由乌龟坐标的具体形式,我们通过在$r$项前增加一个负号来构造下面的函数

${\frac {1}{2}}\left(-r+\ln(r-1)\right)={\frac {1}{2}}\left(-{\frac {1}{1-x}}+\ln x-\ln(1-x)\right)\equiv {\bar {h}}(x)$

事实上,我们注意到虽然在空间无穷远时${\frac {1}{1-x}}\gg \ln(1-x)$,但是为了追求在空间无穷远时$h$函数与乌龟坐标完全一致,我们可以同时改变$\ln(1-x)$的符号,这样我们得到(69)的第一行中的定义.注意到视界对应$\sigma \to 1$,且$\sigma \equiv (1-x)$. 同时上述定义不改变在视界处$h$函数的渐进行为. 具体的,我们有

${\frac {1}{2}}\left(-{\frac {1}{1-x}}+\ln x+\ln(1-x)\right)\equiv h(x)$

${\bar {t}}=\tau -h(x)$

我们指出,这里定义的$h$函数与最初的定义arXiv:1102.2451v2的(1)差了一个负号!

现在我们来构造乌龟坐标视界和空间无穷远处的平面波解,并且证明这两个解同样是hyperboloidal坐标下的渐进解. 在空间无穷远处的右行波具有形式

$e^{i\omega (r_{*}-t)}$

按上述定义,我们有

$e^{i\omega (r_{*}-t)}=e^{2i\omega ({\bar {x}}-{\bar {t}})}=e^{i\omega \left[\left({\frac {1}{1-x}}+\ln x-\ln(1-x)\right)-2\left(\tau -{\frac {1}{2}}\left(-{\frac {1}{1-x}}+\ln x+\ln(1-x)\right)\right)\right]}=e^{-2i\omega \tau }e^{2i\omega \ln x}=e^{-2i\omega \tau }x^{2i\omega }$

类似的,在视界处

$e^{i\omega (-r_{*}-t)}=e^{2i\omega (-{\bar {x}}-{\bar {t}})}=e^{-2i\omega \tau }e^{-2i\omega \left({\frac {1}{1-x}}-\ln(1-x)\right)}=e^{-2i\omega \tau }e^{-{\frac {2i\omega }{1-x}}}(1-x)^{2i\omega }$

把上面的结果扣去时间因子$e^{-i\lambda \omega \tau }=e^{-2i\omega \tau }$后,剩余的空间部分应该满足hyperboloidal坐标中把RW有效势取为零后主方程. 后者可以通过在主方程中代入$\ell =0,s=1$实现,通过Mathematica我们得到方程

$(4\omega ^{2}+(1-x)(1+(i+2\omega )^{2}))\phi (x)-((1-x)(2+(1-x)(-3+4i\omega ))-2i\omega )\phi '(x)+(1-x)^{2}x\phi ''(x)=0$

它的通解正是$\phi (x)=C_{1}x^{2i\omega }+C_{2}e^{-{\frac {2i\omega }{1-x}}}(1-x)^{2i\omega }$.

本文档除了包括推导,疑惑外,做读书重点的记录 ${\displaystyle }$

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本文档除了包括推导,疑惑外,做读书重点的记录 ${\displaystyle }$