View Full Version : Electron transport in metals
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Hello:\n\nI would appreciate if somebody knowledgeable in electron transport in\nmetals would help me answer the following question:\n\nConsider a bulk metal at 0K with no defects or impurities. The only\nsources of scattering of the electrons in metals are phonons, defects,\nand impurities. (I believe electron-electron scattering is not very\nsignificant.) So, our bulk metal is free of all three scatterers.\n\nHow do I find out what happens to the conduction electrons in such a\nmetal subjected to an external electric field?\n\nMany thanks,\nPeter\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Hello:
I would appreciate if somebody knowledgeable in electron transport in
metals would help me answer the following question:
Consider a bulk metal at 0K with no defects or impurities. The only
sources of scattering of the electrons in metals are phonons, defects,
and impurities. (I believe electron-electron scattering is not very
significant.) So, our bulk metal is free of all three scatterers.
How do I find out what happens to the conduction electrons in such a
metal subjected to an external electric field?
Many thanks,
Peter
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>radical environmentalism)\nseem to include both personalities of the leftist type and\npersonalities of thoroughly un-leftist types who ought to know better\nthan to collaborate with leftists. Varieties of leftists fade out\ngradually into varieties of non-leftists and we ourselves would often\nbe hard-pressed to decide whether a given individual is or is not a\nleftist. To the extent that it is defined at all, our conception of\nleftism is defined by the discussion of it that we have given in this\narticle, and we can only advise the reader to use his own judgment in\ndeciding who is a leftist.\n\n228. But it will be helpful to list some criteria for diagnosing\nleftism. These criteria cannot be applied in a cut and dried manner.\nSome individuals may meet some of the criteria without being leftists,\nsome leftists may not meet any of the criteria. Again, you just have\nto use your judgment.\n\n229. The leftist is oriented toward largescale collectivism. He\nemphasizes the duty of the individual to serve society and the duty of\nsociety to take care of the individual. He has a negative attitude\ntoward individualism. He often takes a moralistic tone. He tends to be\nfor gun control, for sex education and other psychologically\n"enlightened" educational methods, for planning, for affirmative\naction, for multiculturalism. He tends to identify with victims. He\ntends\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>radical environmentalism)
seem to include both personalities of the leftist type and
personalities of thoroughly un-leftist types who ought to know better
than to collaborate with leftists. Varieties of leftists fade out
gradually into varieties of non-leftists and we ourselves would often
be hard-pressed to decide whether a given individual is or is not a
leftist. To the extent that it is defined at all, our conception of
leftism is defined by the discussion of it that we have given in this
article, and we can only advise the reader to use his own judgment in
deciding who is a leftist.
228. But it will be helpful to list some criteria for diagnosing
leftism. These criteria cannot be applied in a cut and dried manner.
Some individuals may meet some of the criteria without being leftists,
some leftists may not meet any of the criteria. Again, you just have
to use your judgment.
229. The leftist is oriented toward largescale collectivism. He
emphasizes the duty of the individual to serve society and the duty of
society to take care of the individual. He has a negative attitude
toward individualism. He often takes a moralistic tone. He tends to be
for gun control, for sex education and other psychologically
"enlightened" educational methods, for planning, for affirmative
action, for multiculturalism. He tends to identify with victims. He
tends
Igor Khavkine
Dec14-04, 10:18 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\nOn Mon, 13 Dec 2004 15:12:34 +0000, Peter wrote:\n\n> I would appreciate if somebody knowledgeable in electron transport in\n> metals would help me answer the following question:\n>\n> Consider a bulk metal at 0K with no defects or impurities. The only\n> sources of scattering of the electrons in metals are phonons, defects, and\n> impurities. (I believe electron-electron scattering is not very\n> significant.) So, our bulk metal is free of all three scatterers.\n\nYou would be wrong in this assumption.\n\n> How do I find out what happens to the conduction electrons in such a\n> metal subjected to an external electric field?\n\nIf your assumptions are indeed justified, then the conduction electrons\nwould be accelerated by the electric field until something else scatters\nthem. And there will always be something to scatter them. For example, the\nboundary of the sample is always there. Also, it is impossible to\neliminate phonons completely unless the sample is at equilibrium (and\ncarrying a current is a non-equilibrium state). Because as soon as there\nis an inhomogeneity in the electron density, the electrostatic interaction\nwith the positive ion cores will come in to play, and this interaction is\njust the phonon interaction under another guise.\n\nHowever, if you do reach 0K (which is impossible, but lets humor this\nassumption for a bit) you will no longer have a normal metal, you\'ll have\na superconductor. A large fraction of all pure metals becomes\nsuperconducting at a low enough temperature. And those that have not been\nmade superconducting are believed to simply have a transition temperature\nthat is not achievable at the moment. Here\'s a table:\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/tables/supcon.html\n\nThis is the reason you can\'t neglect electron-electron interaction\n(although it is an effective interaction mediated by phonons).\n\nHope this helps.\n\nIgor\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Mon, 13 Dec 2004 15:12:34 +0000, Peter wrote:
> I would appreciate if somebody knowledgeable in electron transport in
> metals would help me answer the following question:
>
> Consider a bulk metal at 0K with no defects or impurities. The only
> sources of scattering of the electrons in metals are phonons, defects, and
> impurities. (I believe electron-electron scattering is not very
> significant.) So, our bulk metal is free of all three scatterers.
You would be wrong in this assumption.
> How do I find out what happens to the conduction electrons in such a
> metal subjected to an external electric field?
If your assumptions are indeed justified, then the conduction electrons
would be accelerated by the electric field until something else scatters
them. And there will always be something to scatter them. For example, the
boundary of the sample is always there. Also, it is impossible to
eliminate phonons completely unless the sample is at equilibrium (and
carrying a current is a non-equilibrium state). Because as soon as there
is an inhomogeneity in the electron density, the electrostatic interaction
with the positive ion cores will come in to play, and this interaction is
just the phonon interaction under another guise.
However, if you do reach 0K (which is impossible, but lets humor this
assumption for a bit) you will no longer have a normal metal, you'll have
a superconductor. A large fraction of all pure metals becomes
superconducting at a low enough temperature. And those that have not been
made superconducting are believed to simply have a transition temperature
that is not achievable at the moment. Here's a table:
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/supcon.html
This is the reason you can't neglect electron-electron interaction
(although it is an effective interaction mediated by phonons).
Hope this helps.
Igor
Peter Wilson
Dec15-04, 12:36 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Thank you very much for the reply. I realize, I overlooked the fact\nthat the non-equilibrium nature of carrying a current can create an\ninhomogeneity in the electron density which in turn can trigger\nelectron-phonon interactions.\n\nCould you please explain (or point me to a good reference) how the\nelectron density changes when a metal carries a current? Also, how can\none calculate the "steady state" electron density? Would one have to\nsolve the Poisson equation? If so, what would be the correct boundary\nconditions? (I have posted related questions under the subject\n"Questions on electron transport" on Dec 14.)\n\nThe thing that bothered me initially about having no scatterers was the\nfollowing (although, now I understand that it is impossible to\neliminate scattering): If the electrons in the periodic potential are\naccelerated without any scattering, they would eventually reach the\nBrillouin zone edge after which their velocity is reversed. This would\nmean that the electrons would just be oscillating and not carry any\ncurrent. This is in contradiction to the fact that in the absence of\nany scattering, the metal should exhibit zero resistance. Assuming we\nhave some how eliminated scattering, could you please tell me what is\nincorrect with the above arguments?\n\nSincerely,\nPeter\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Thank you very much for the reply. I realize, I overlooked the fact
that the non-equilibrium nature of carrying a current can create an
inhomogeneity in the electron density which in turn can trigger
electron-phonon interactions.
Could you please explain (or point me to a good reference) how the
electron density changes when a metal carries a current? Also, how can
one calculate the "steady state" electron density? Would one have to
solve the Poisson equation? If so, what would be the correct boundary
conditions? (I have posted related questions under the subject
"Questions on electron transport" on Dec 14.)
The thing that bothered me initially about having no scatterers was the
following (although, now I understand that it is impossible to
eliminate scattering): If the electrons in the periodic potential are
accelerated without any scattering, they would eventually reach the
Brillouin zone edge after which their velocity is reversed. This would
mean that the electrons would just be oscillating and not carry any
current. This is in contradiction to the fact that in the absence of
any scattering, the metal should exhibit zero resistance. Assuming we
have some how eliminated scattering, could you please tell me what is
incorrect with the above arguments?
Sincerely,
Peter
Igor Khavkine
Dec16-04, 08:09 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Wed, 15 Dec 2004 18:36:49 +0000, Peter Wilson <peter21wilson@yahoo.com>\nwrote:\n\n> Thank you very much for the reply. I realize, I overlooked the fact that\n> the non-equilibrium nature of carrying a current can create an\n> inhomogeneity in the electron density which in turn can trigger\n> electron-phonon interactions.\n>\n> Could you please explain (or point me to a good reference) how the\n> electron density changes when a metal carries a current? Also, how can one\n> calculate the "steady state" electron density? Would one have to solve the\n> Poisson equation? If so, what would be the correct boundary conditions? (I\n> have posted related questions under the subject "Questions on electron\n> transport" on Dec 14.)\n\nIn the steady state, the macroscopic parameters describing the system\n(such as the electron density) are constant. And the lower, the\ntemperature, the more uniform they are at small scales. In essence, the\nelectron density can be taken as constant. The non-equilibrium nature of\nthe current carrying state is manifested by energy dissipation of\nmagnitude E.j, where E is the electric field and j is the current. When\nclose to equilibrium (small currents, or small electric fields), the\ndensity fluctuations can be calculated from the density-density\ncorrelation function. Basically, you have to evaluated a 4-point fermion\ncorrelation function, usually up to one loop order (if this tells you\nanything). "Close to equilibrium" covers most conditions we encounter in\nevery-day life. If you really want to know what happens far from\nequilibrium, I wouldn\'t be able to tell you anything about that.\n\n> The thing that bothered me initially about having no scatterers was the\n> following (although, now I understand that it is impossible to eliminate\n> scattering): If the electrons in the periodic potential are accelerated\n> without any scattering, they would eventually reach the Brillouin zone\n> edge after which their velocity is reversed. This would mean that the\n> electrons would just be oscillating and not carry any current. This is in\n> contradiction to the fact that in the absence of any scattering, the metal\n> should exhibit zero resistance. Assuming we have some how eliminated\n> scattering, could you please tell me what is incorrect with the above\n> arguments?\n\nYou are right to expect zero resistance, *if* all possible scattering is\neliminated. You are mistaken about the significance of the Brillouin\nzone, though. Lets take a minute to recall how it comes up. Suppose you\nwant to solve the time-independent Schrodinger equation in a periodic\npotential. Then your wave function psi(x) has to change in a definite\nway when it is translated by one period of the potential (one lattice\nvector if you will). It can only change by a phase, psi(Tx) = exp(i\nphi_T) psi(x), where T is the translation operator and phi_T is the\nphase change corresponding to it. This phase can be written as phi_T =\nk.T, for some reciprocal vector k. However, k and k\' that differ by a\nreciprocal lattice vector K, k = k\' + K, give the same phase since T.K =\n2 pi n, for some integer n. Hence it is sufficient to restrict k to lie\nin the first Brillouin zone of the reciprocal lattice. Moreover, we can\nconclude that the wave function can be represented as psi(x) = exp(i\nk.x) u_k(x), where u_k(x) is now periodic with the same periods as the\npotential. In other words u_k(Tx) = u_k(x) for any lattice translation\nT. The whole Schrodinger equation can now be written in terms of u_k(x)\nand is guaranteed to have discrete energy eigenvalues E_(k,n), indexed\nby an integer n. This is how energy bands appear in periodic potentials.\nWhen changing k continuously, E_(k,n) will change continuously for fixed\nn, which corresponds to moving within a given band. However, when\nchanging n, the energy of a state E_(k,n) will usually change\ndiscontinuously, which corresponds to jumping between bands.\n\nNow, all the possible reciprocal vectors k form precisely the first\nBrillouin zone that you referred to. However, we can clearly see that\nthey do not represent the physical momenta of the electrons. An electron\nwith physical momentum p is represented by the wave function\n\npsi(x) = exp(i p.x) = exp(i k.x) u_(k,p)(x) ,\n\nwhere k is a reciprocal vector in the first Brillouin zone, while\nu_(k,p)(x) is some linear combination of the functions u_(k,n)(x)\nrepresenting different bands. (Note that in general a physical momentum\neigenstate as above will not be an eigenstate of the crystal potential\nand hence will be spread over several bands.) So as p is increased, for\na while k will also increase, until it reaches the Brillouin zone\nboundary. Then, k will wrap around and start receding from the boundary\non the opposite side of the Brillouin zone. However, all this time, the\nfunction u_(k,p)(x) will also be changing, which can be seen as a change\nin the u_(k,n)(x) coefficients of its expansion. To put the above\ndiscussion in more physical terms, we can say that an accelerated\nelectron\'s physical momentum behaves as expected, its crystal momentum\nwill not exceed the boundary of the first Brillouin zone, but the\nelectron may jump between the available bands to compensate for the\nrestrictions on its crystal momentum.\n\nHope this helps.\n\nIgor\n\nP.S.: A short nomenclature note. Given a periodic potential, the\ndecomposition of the solutions of the Schroedinger equation, psi(x) =\nexp(i k.x) u_k(x), into a phase exp(i k.x) and a periodic part u_k(x) is\na consequence of Bloch\'s theorem. At least that is how it is referred to\nin condensed matter physics circles. An analogous decomposition is\npossible for solutions of the time-dependent Schrodinger equation with a\nperiodic (in time) external perturbation. This result comes up in atomic\nphysics and quantum optics and, in these circles, is referred to as\nFloquet\'s theorem. Those in the know sometimes call it the Bloch-Floquet\ntheorem. Both of these formulations are special cases of a theorem on\nthe decomposition of the solutions of a linear equation according to\ntheir transformation properties under the symmetries of the equations.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Wed, 15 Dec 2004 18:36:49 +0000, Peter Wilson <peter21wilson@yahoo.com>
wrote:
> Thank you very much for the reply. I realize, I overlooked the fact that
> the non-equilibrium nature of carrying a current can create an
> inhomogeneity in the electron density which in turn can trigger
> electron-phonon interactions.
>
> Could you please explain (or point me to a good reference) how the
> electron density changes when a metal carries a current? Also, how can one
> calculate the "steady state" electron density? Would one have to solve the
> Poisson equation? If so, what would be the correct boundary conditions? (I
> have posted related questions under the subject "Questions on electron
> transport" on Dec 14.)
In the steady state, the macroscopic parameters describing the system
(such as the electron density) are constant. And the lower, the
temperature, the more uniform they are at small scales. In essence, the
electron density can be taken as constant. The non-equilibrium nature of
the current carrying state is manifested by energy dissipation of
magnitude E.j, where E is the electric field and j is the current. When
close to equilibrium (small currents, or small electric fields), the
density fluctuations can be calculated from the density-density
correlation function. Basically, you have to evaluated a 4-point fermion
correlation function, usually up to one loop order (if this tells you
anything). "Close to equilibrium" covers most conditions we encounter in
every-day life. If you really want to know what happens far from
equilibrium, I wouldn't be able to tell you anything about that.
> The thing that bothered me initially about having no scatterers was the
> following (although, now I understand that it is impossible to eliminate
> scattering): If the electrons in the periodic potential are accelerated
> without any scattering, they would eventually reach the Brillouin zone
> edge after which their velocity is reversed. This would mean that the
> electrons would just be oscillating and not carry any current. This is in
> contradiction to the fact that in the absence of any scattering, the metal
> should exhibit zero resistance. Assuming we have some how eliminated
> scattering, could you please tell me what is incorrect with the above
> arguments?
You are right to expect zero resistance, *if* all possible scattering is
eliminated. You are mistaken about the significance of the Brillouin
zone, though. Lets take a minute to recall how it comes up. Suppose you
want to solve the time-independent Schrodinger equation in a periodic
potential. Then your wave function \psi(x) has to change in a definite
way when it is translated by one period of the potential (one lattice
vector if you will). It can only change by a phase, \psi(Tx) = \exp(i\phi_T) \psi(x), where T is the translation operator and \phi_T is the
phase change corresponding to it. This phase can be written as \phi_T =
k.T, for some reciprocal vector k. However, k and k' that differ by a
reciprocal lattice vector K, k = k' + K, give the same phase since T.K =
2 \pi n, for some integer n. Hence it is sufficient to restrict k to lie
in the first Brillouin zone of the reciprocal lattice. Moreover, we can
conclude that the wave function can be represented as \psi(x) = \exp(ik.x) u_k(x), where u_k(x) is now periodic with the same periods as the
potential. In other words u_k(Tx) = u_k(x) for any lattice translation
T. The whole Schrodinger equation can now be written in terms of u_k(x)
and is guaranteed to have discrete energy eigenvalues E_(k,n), indexed
by an integer n. This is how energy bands appear in periodic potentials.
When changing k continuously, E_(k,n) will change continuously for fixed
n, which corresponds to moving within a given band. However, when
changing n, the energy of a state E_(k,n) will usually change
discontinuously, which corresponds to jumping between bands.
Now, all the possible reciprocal vectors k form precisely the first
Brillouin zone that you referred to. However, we can clearly see that
they do not represent the physical momenta of the electrons. An electron
with physical momentum p is represented by the wave function
\psi(x) = \exp(i p.x) = \exp(i k.x) u_(k,p)(x) ,
where k is a reciprocal vector in the first Brillouin zone, while
u_(k,p)(x) is some linear combination of the functions u_(k,n)(x)
representing different bands. (Note that in general a physical momentum
eigenstate as above will not be an eigenstate of the crystal potential
and hence will be spread over several bands.) So as p is increased, for
a while k will also increase, until it reaches the Brillouin zone
boundary. Then, k will wrap around and start receding from the boundary
on the opposite side of the Brillouin zone. However, all this time, the
function u_(k,p)(x) will also be changing, which can be seen as a change
in the u_(k,n)(x) coefficients of its expansion. To put the above
discussion in more physical terms, we can say that an accelerated
electron's physical momentum behaves as expected, its crystal momentum
will not exceed the boundary of the first Brillouin zone, but the
electron may jump between the available bands to compensate for the
restrictions on its crystal momentum.
Hope this helps.
Igor
P.S.: A short nomenclature note. Given a periodic potential, the
decomposition of the solutions of the Schroedinger equation, \psi(x) =\exp(i k.x) u_k(x), into a phase \exp(i k.x) and a periodic part u_k(x) is
a consequence of Bloch's theorem. At least that is how it is referred to
in condensed matter physics circles. An analogous decomposition is
possible for solutions of the time-dependent Schrodinger equation with a
periodic (in time) external perturbation. This result comes up in atomic
physics and quantum optics and, in these circles, is referred to as
Floquet's theorem. Those in the know sometimes call it the Bloch-Floquet
theorem. Both of these formulations are special cases of a theorem on
the decomposition of the solutions of a linear equation according to
their transformation properties under the symmetries of the equations.
me@privacy.net
Dec25-04, 01:48 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>at 325° for 30 - 45 minutes.\n\n\n\nUmbilical Cordon Bleu\n\nNothing is so beautiful as the bond between mother and child,\nso why not consume it?\nChildren or chicken breasts will work wonderfully also.\n\n4 whole umbilical chords (or baby breasts, or chicken breasts)\n4 thin slices of smoked ham, and Gruyere cheese\nFlour\neggwash (milk and eggs)\nseasoned bread crumbs\n1 onion\nminced\nsalt\npepper\nbutter\nolive oil\n\nPound the breasts flat (parboil first if using umbilical\ncords so they won?t be tough).\nPlace a slice of ham and cheese on each, along with some minced onion\nthen fold in half, trimming neatly.\nDredge in flour, eggwash, then seasoned breadcrumbs;\nallow to sit for a few minutes.\nSauté in butter and olive oil until golden brown,\nabout 6 minutes on each side.\n\n\n\nShish Kababes\n\nAs old as the hills, this technique has employed seafood, beef, pork, lamb,\npoultry, and vegetables; just about anything can be grilled, and young humans\nare no exception!\n\nHigh quality marinade (Teriyaki and garlic perhaps)\n1 inch cubes of tender meat, preferably from the nursery\nOnions\nbell peppers\nWooden or metal skewers\n\nMarinate the meat overnight.\nGet the grill good and hot while placing meat, vegetables, and\nfruit such as pineapples or cherries on the skewers.\nDon?t be afraid to use a variety of meats.\nGrill to medium rare,\nserve with garlic cous-cous and sautéed asparagus.\nCoffee and sherbet for desert then walnuts, cheese, and port.\nCigars for the gentlemen (and ladies if they so desire)!\n\n\n\nCrock-Pot Crack Baby\n\nWhen the quivering, hopelessly addicted crack baby succumbs to death,\nget him immediately butchered and into the crock-pot, so that any\nremaining toxin\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>at 325° for 30 - 45 minutes.
Umbilical Cordon Bleu
Nothing is so beautiful as the bond between mother and child,
so why not consume it?
Children or chicken breasts will work wonderfully also.
4 whole umbilical chords (or baby breasts, or chicken breasts)
4 thin slices of smoked ham, and Gruyere cheese
Flour
eggwash (milk and eggs)
seasoned bread crumbs
1 onion
minced
salt
pepper
butter
olive oil
Pound the breasts flat (parboil first if using umbilical
cords so they won?t be tough).
Place a slice of ham and cheese on each, along with some minced onion
then fold in half, trimming neatly.
Dredge in flour, eggwash, then seasoned breadcrumbs;
allow to sit for a few minutes.
Sauté in butter and olive oil until golden brown,
about 6 minutes on each side.
Shish Kababes
As old as the hills, this technique has employed seafood, beef, pork, lamb,
poultry, and vegetables; just about anything can be grilled, and young humans
are no exception!
High quality marinade (Teriyaki and garlic perhaps)
1 inch cubes of tender meat, preferably from the nursery
Onions
bell peppers
Wooden or metal skewers
Marinate the meat overnight.
Get the grill good and hot while placing meat, vegetables, and
fruit such as pineapples or cherries on the skewers.
Don?t be afraid to use a variety of meats.
Grill to medium rare,
serve with garlic cous-cous and sautéed asparagus.
Coffee and sherbet for desert then walnuts, cheese, and port.
Cigars for the gentlemen (and ladies if they so desire)!
Crock-Pot Crack Baby
When the quivering, hopelessly addicted crack baby succumbs to death,
get him immediately butchered and into the crock-pot, so that any
remaining toxin
me@privacy.net
Dec25-04, 02:29 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>neonate stock (chicken, or turkey stock is fine)\ngarlic\nparsley\nfresh cracked black pepper\n\nSeason and sauté the cutlets in olive oil till golden brown, remove.\nAdd the garlic and onions and cook down a bit.\nAdd some lemon juice and some zest, then de-glaze with stock.\nAdd a little cornstarch (dissolved in cold water) to the sauce.\nYou are just about there, Pour the sauce over the cutlets,\ntop with parsley, lemon slices and cracked pepper.\nServe with spinach salad, macaroni and cheese (homemade) and iced tea...\n\n\n\nSpaghetti with Real Italian Meatballs\n\nIf you don?t have an expendable bambino on hand,\nyou can use a pound of ground pork instead.\nThe secret to great meatballs, is to use very lean meat.\n\n1 lb. ground flesh; human or pork\n3 lb. ground beef\n1 cup finely chopped onions\n7 - 12 cloves garlic\n1 cup seasoned bread crumbs\n½ cup milk, 2 eggs\nOregano\nbasil\nsalt\npepper\nItalian seasoning, etc.\nTomato gravy (see index)\nFresh or at least freshly cooked spaghetti or other pasta\n\nMix the ground meats together in a large bowl,\nthen mix each of the other ingredients.\nMake balls about the size of a baby?s fist\n(there should be one lying around for reference).\nBake at 400°for about 25 minutes -\nor you could fry them in olive oil.\nPlace the meatballs in the tomato gravy, and simmer for several hours.\nServe on spaghetti.\nAccompany with green salad, garlic bread and red wine.\n\n\n\nNewborn Parmesan\n\nThis classic Sicilian cuisine can easily be turned into Eggplant Parmesan\nIf you are planning a vegetarian meal. Or you could just as well use veal -\nafter all, you have to be careful - Sicilians are touchy about their young\nfamily members...\n\n6 newborn or veal cutlets\nTomato gravy (see index)\n4 cups mozzarella, 1cup parmesan, 1cup romano\nSeasoned br\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>neonate stock (chicken, or turkey stock is fine)
garlic
parsley
fresh cracked black pepper
Season and sauté the cutlets in olive oil till golden brown, remove.
Add the garlic and onions and cook down a bit.
Add some lemon juice and some zest, then de-glaze with stock.
Add a little cornstarch (dissolved in cold water) to the sauce.
You are just about there, Pour the sauce over the cutlets,
top with parsley, lemon slices and cracked pepper.
Serve with spinach salad, macaroni and cheese (homemade) and iced tea...
Spaghetti with Real Italian Meatballs
If you don?t have an expendable bambino on hand,
you can use a pound of ground pork instead.
The secret to great meatballs, is to use very lean meat.
1 lb. ground flesh; human or pork
3 lb. ground beef
1 cup finely chopped onions
7 - 12 cloves garlic
1 cup seasoned bread crumbs
½ cup milk, 2 eggs
Oregano
basil
salt
pepper
Italian seasoning, etc.
Tomato gravy (see index)
Fresh or at least freshly cooked spaghetti or other pasta
Mix the ground meats together in a large bowl,
then mix each of the other ingredients.
Make balls about the size of a baby?s fist
(there should be one lying around for reference).
Bake at 400°for about 25 minutes -
or you could fry them in olive oil.
Place the meatballs in the tomato gravy, and simmer for several hours.
Serve on spaghetti.
Accompany with green salad, garlic bread and red wine.
Newborn Parmesan
This classic Sicilian cuisine can easily be turned into Eggplant Parmesan
If you are planning a vegetarian meal. Or you could just as well use veal -
after all, you have to be careful - Sicilians are touchy about their young
family members...
6 newborn or veal cutlets
Tomato gravy (see index)
4 cups mozzarella, 1cup parmesan, 1cup romano
Seasoned br
Peter Wilson
Dec25-04, 03:28 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Thank you, once again, for your reply. It was definitely helpful. (I\nunderstood everything except the 4-point fermion correlation function.)\nI have some more questions that are related to the semiclassical\ntreatment of electrons in metals:\n\n1. What exactly do we mean by \'an electron\' having a position and a\nmomentum (within the uncertainty principle) under the semiclassical\ntreatment AND lying in a particular band? If I understood you\ncorrectly, it is impossible for the electron to have a definite\nmomentum as well as a definite band index. Does the momentum in the\nsemiclassical approximation refer to the crystal momentum?\n\n2. Is the following correct: Only the crystal momentum of the electron\nobeys Newton\'s laws corresponding to just the external field. The real\nmomentum\'s behavior is much more complex as it depends on the\ninteraction with the ions too.\n\n3. Is it true that every electron has a crystal momentum and a real\nmomentum? Or, is it that the semiclassical "particle" that has a\ncrystal momentum, is in fact an artificial way (or a different way) of\nlooking at all the electrons in a different basis set, if I may call it\nthat way? In other words, the semiclassical particle having the crystal\nmomentum, is not really an electron. Am I correct?\n\n4. This probably sounds silly, but, what exactly is an electron? I am\nso very confused. I am looking for a good reference that will help me\nunderstand the classical limit of an electron which is quantum\nmechanically described as a wave packet. I guess, I should first do\nthis for a free electron and then for an electron in a periodic\npotential.\nPlease help me.\n\nMerry Christmas to everybody!\n\nMany thanks,\nPeter\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Thank you, once again, for your reply. It was definitely helpful. (I
understood everything except the 4-point fermion correlation function.)
I have some more questions that are related to the semiclassical
treatment of electrons in metals:
1. What exactly do we mean by 'an electron' having a position and a
momentum (within the uncertainty principle) under the semiclassical
treatment AND lying in a particular band? If I understood you
correctly, it is impossible for the electron to have a definite
momentum as well as a definite band index. Does the momentum in the
semiclassical approximation refer to the crystal momentum?
2. Is the following correct: Only the crystal momentum of the electron
obeys Newton's laws corresponding to just the external field. The real
momentum's behavior is much more complex as it depends on the
interaction with the ions too.
3. Is it true that every electron has a crystal momentum and a real
momentum? Or, is it that the semiclassical "particle" that has a
crystal momentum, is in fact an artificial way (or a different way) of
looking at all the electrons in a different basis set, if I may call it
that way? In other words, the semiclassical particle having the crystal
momentum, is not really an electron. Am I correct?
4. This probably sounds silly, but, what exactly is an electron? I am
so very confused. I am looking for a good reference that will help me
understand the classical limit of an electron which is quantum
mechanically described as a wave packet. I guess, I should first do
this for a free electron and then for an electron in a periodic
potential.
Please help me.
Merry Christmas to everybody!
Many thanks,
Peter
Igor Khavkine
Dec26-04, 02:46 PM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Peter Wilson <peter21wilsonYahooCom> wrote:\n> Thank you, once again, for your reply. It was definitely helpful. (I\n> understood everything except the 4-point fermion correlation function.)\n\nYou\'ll understand what I meant once you learn what Feynman diagrams are\nand how they are used.\n\n> I have some more questions that are related to the semiclassical\n> treatment of electrons in metals:\n>\n> 1. What exactly do we mean by \'an electron\' having a position and a\n> momentum (within the uncertainty principle) under the semiclassical\n> treatment AND lying in a particular band? If I understood you\n> correctly, it is impossible for the electron to have a definite\n> momentum as well as a definite band index. Does the momentum in the\n> semiclassical approximation refer to the crystal momentum?\n\nAn electron cannot have definite values for any pair from position,\nmomentum, (crystal momentum, band index). Semiclassically, we refer to\nthe crystal momentum.\n\n> 2. Is the following correct: Only the crystal momentum of the electron\n> obeys Newton\'s laws corresponding to just the external field. The real\n> momentum\'s behavior is much more complex as it depends on the\n> interaction with the ions too.\n\nYes.\n\n> 3. Is it true that every electron has a crystal momentum and a real\n> momentum? Or, is it that the semiclassical "particle" that has a\n> crystal momentum, is in fact an artificial way (or a different way) of\n> looking at all the electrons in a different basis set, if I may call it\n> that way? In other words, the semiclassical particle having the crystal\n> momentum, is not really an electron. Am I correct?\n\nThere are states with definite crystal momentum and band index, there\nare states with definite momentum, there are states with definite\nposition. All of these sets of states are complete, orthogonal and form\nbases for the Hilbert space of states. A particular state can be\ndecomposed in any one basis. The tates that approximate classical\nparticles (the semiclassical states) generally have non-trivial\ndecompositions in any one of these bases. A semiclassical state in a\nlattice potential is as much an "electron" as a wavepacket in free\nspace. Either one is "fuzzy" compared to a classical particle.\n\n> 4. This probably sounds silly, but, what exactly is an electron? I am\n> so very confused. I am looking for a good reference that will help me\n> understand the classical limit of an electron which is quantum\n> mechanically described as a wave packet. I guess, I should first do\n> this for a free electron and then for an electron in a periodic\n> potential.\n\nThis kind of question is always hard to answer without delving into\nquestions about what something "really is". I find that, rather than\ntrying to refine a nebulous definition, it is more helpful to see how\nvarious precise descriptions of the same nebulous concept relate to\neach other.\n\nConsider the example in question. We have a classical description of an\nelectron as a point particle with definite position and momentum which\nobeys Newton\'s second law of motion. On the other hand, the quantum\ndescription is that of a wave function, or a state in a Hilbert space,\nwith the dynamics given by Schroedinger\'s equation.\n\nHere is how we make the link. Write down the wave function as\n\npsi(x,t) = rho(x,t) exp(i S(x,t)),\n\nwhere rho(x,t)^2 = |psi(x,t)|^2. Rewrite the Schroedinger equation in\nterms of these new variables and drop the terms that have too many\nhbar\'s. What you have now is a system of coupled PDE\'s for rho(x,t) and\nS(x,t). One equation tells you that rho(x,t) evolves as if it were a\nfluid density where the fluid element at (x,t) has momentum dS/dx.\nWhile S(x,t) satisfies the Hamilton-Jacobi equation. In other words,\ndS/dx is in fact the momentum of a classical particle moving in the\ngiven potential. Thus if rho(x,t) is strongly peaked, it\'s peak follows\nthe trajectory of a classical particle.\n\nThe above discussion gives you reason to interpret a narrow wave packet\nas a semiclassical state approximating a classical particle, up to hbar\ncorrections. The expectation values for position and momentum of this\nstate will also approximate those of the classical particle. Hopefully,\nthe link between the classical and quantum descriptions of an electron\nshould be clearer to you, without having to define what an electron\n"really is".\n\nAs a closing remark. I can tell you why the semiclassical approximation\nuses the crystal momentum instead of the physical momentum. Momentum is\na vector constant of motion for a free particle and whose rate of\nchange is given by the sum of forces acting on it. "Free" here means\nabsence of external forces or potentials. When an electron is inside a\nsolid, only the forces external to the solid are considered as\n"external" (such as an electromagnetic field). Since crystal momentum\nis conserved when ignoring such external froces, if we forget that the\nsolid is actually there, we can pretend that a semiclassical conduction\nelectron is approximated a classical particle with crystal momentum\nplaying the role of physical momentum. But then its dispersion relation\nis no loger p^2/2m, but the more complicated crystal dispersion\nrelation. And in its equations of motion only forces external to the\nsolid are taken into account.\n\nAll of this can be made much more precise if you sit down and simply\ncalculate the rate of change for the expectation value of the physical\nmomentum and the crystal momentum. You will recover Newton\'s second\nlaw, but with different potentials in each case.\n\nHope this helps.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Peter Wilson <peter21wilsonYahooCom> wrote:
> Thank you, once again, for your reply. It was definitely helpful. (I
> understood everything except the 4-point fermion correlation function.)
You'll understand what I meant once you learn what Feynman diagrams are
and how they are used.
> I have some more questions that are related to the semiclassical
> treatment of electrons in metals:
>
> 1. What exactly do we mean by 'an electron' having a position and a
> momentum (within the uncertainty principle) under the semiclassical
> treatment AND lying in a particular band? If I understood you
> correctly, it is impossible for the electron to have a definite
> momentum as well as a definite band index. Does the momentum in the
> semiclassical approximation refer to the crystal momentum?
An electron cannot have definite values for any pair from position,
momentum, (crystal momentum, band index). Semiclassically, we refer to
the crystal momentum.
> 2. Is the following correct: Only the crystal momentum of the electron
> obeys Newton's laws corresponding to just the external field. The real
> momentum's behavior is much more complex as it depends on the
> interaction with the ions too.
Yes.
> 3. Is it true that every electron has a crystal momentum and a real
> momentum? Or, is it that the semiclassical "particle" that has a
> crystal momentum, is in fact an artificial way (or a different way) of
> looking at all the electrons in a different basis set, if I may call it
> that way? In other words, the semiclassical particle having the crystal
> momentum, is not really an electron. Am I correct?
There are states with definite crystal momentum and band index, there
are states with definite momentum, there are states with definite
position. All of these sets of states are complete, orthogonal and form
bases for the Hilbert space of states. A particular state can be
decomposed in any one basis. The tates that approximate classical
particles (the semiclassical states) generally have non-trivial
decompositions in any one of these bases. A semiclassical state in a
lattice potential is as much an "electron" as a wavepacket in free
space. Either one is "fuzzy" compared to a classical particle.
> 4. This probably sounds silly, but, what exactly is an electron? I am
> so very confused. I am looking for a good reference that will help me
> understand the classical limit of an electron which is quantum
> mechanically described as a wave packet. I guess, I should first do
> this for a free electron and then for an electron in a periodic
> potential.
This kind of question is always hard to answer without delving into
questions about what something "really is". I find that, rather than
trying to refine a nebulous definition, it is more helpful to see how
various precise descriptions of the same nebulous concept relate to
each other.
Consider the example in question. We have a classical description of an
electron as a point particle with definite position and momentum which
obeys Newton's second law of motion. On the other hand, the quantum
description is that of a wave function, or a state in a Hilbert space,
with the dynamics given by Schroedinger's equation.
Here is how we make the link. Write down the wave function as
\psi(x,t) = \rho(x,t) \exp(i S(x,t)),
where \rho(x,t)^2 = |\psi(x,t)|^2. Rewrite the Schroedinger equation in
terms of these new variables and drop the terms that have too many
\hbar's. What you have now is a system of coupled PDE's for \rho(x,t) and
S(x,t). One equation tells you that \rho(x,t) evolves as if it were a
fluid density where the fluid element at (x,t) has momentum dS/dx.
While S(x,t) satisfies the Hamilton-Jacobi equation. In other words,
dS/dx is in fact the momentum of a classical particle moving in the
given potential. Thus if \rho(x,t) is strongly peaked, it's peak follows
the trajectory of a classical particle.
The above discussion gives you reason to interpret a narrow wave packet
as a semiclassical state approximating a classical particle, up to \hbar
corrections. The expectation values for position and momentum of this
state will also approximate those of the classical particle. Hopefully,
the link between the classical and quantum descriptions of an electron
should be clearer to you, without having to define what an electron
"really is".
As a closing remark. I can tell you why the semiclassical approximation
uses the crystal momentum instead of the physical momentum. Momentum is
a vector constant of motion for a free particle and whose rate of
change is given by the sum of forces acting on it. "Free" here means
absence of external forces or potentials. When an electron is inside a
solid, only the forces external to the solid are considered as
"external" (such as an electromagnetic field). Since crystal momentum
is conserved when ignoring such external froces, if we forget that the
solid is actually there, we can pretend that a semiclassical conduction
electron is approximated a classical particle with crystal momentum
playing the role of physical momentum. But then its dispersion relation
is no loger p^2/2m, but the more complicated crystal dispersion
relation. And in its equations of motion only forces external to the
solid are taken into account.
All of this can be made much more precise if you sit down and simply
calculate the rate of change for the expectation value of the physical
momentum and the crystal momentum. You will recover Newton's second
law, but with different potentials in each case.
Hope this helps.
Igor
Arnold Neumaier
Dec27-04, 07:32 AM
<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Peter Wilson wrote:\n\n> 4. This probably sounds silly, but, what exactly is an electron? I am\n> so very confused. I am looking for a good reference that will help me\n> understand the classical limit of an electron which is quantum\n> mechanically described as a wave packet. I guess, I should first do\n> this for a free electron and then for an electron in a periodic\n> potential.\n\nClassical electrons are not very meaningful.\nOne loses indistinguishability and spin, both of which are very\nimportantn to model electrons in practice.\nThis is why it is difficult to find a good reference.\n\n\nArnold Neumaier\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form"> View this Usenet post in original ASCII form </a></div><P></jabberwocky>Peter Wilson wrote:
> 4. This probably sounds silly, but, what exactly is an electron? I am
> so very confused. I am looking for a good reference that will help me
> understand the classical limit of an electron which is quantum
> mechanically described as a wave packet. I guess, I should first do
> this for a free electron and then for an electron in a periodic
> potential.
Classical electrons are not very meaningful.
One loses indistinguishability and spin, both of which are very
importantn to model electrons in practice.
This is why it is difficult to find a good reference.
Arnold Neumaier
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