View Full Version : Parallel Transportation and Aharanov-Bohm Effect
Yi-Zen Chu; Yiren Qu
Oct15-04, 01:55 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>\nHello everyone\n\nI am just beginning to learn about differential geometry and general\nrelativity. I am wondering if there is a relationship between parallel\ntransportation of a tangent vector around a closed loop on some manifold\nand the fact that they do not return as the same vector, to the\nAharanov-Bohm effect in quantum mechanics where a charged fermion gains\na phase shift in its wavefunction by going round a solenoid?\n\nIn quantum mechanics am I correct that we replace the derivative del_k\nwith del_k - q A_k, where A is the vector potential? Can we view this as\nsome sort of covariant derivative? By promoting A to a vector operator\ndoes that mean, just as the Riemann curvature tensor is in some sense\nthe commutator of covariant derivatives, the commutator of the\ncomponents of the vector potential, say [A_q,A_r] also represents some\nsort of curvature "tensor"?\n\nDoes the Aharanov-Bohm effect give us some sort of geometrical\ninterpretation of either quantum mechanics itself, electromagnetism (esp\nthe meaning of the vector potential), or both? If so what is it?\n\nOn a similar note we say the momentum operator (or the derivative in\nquantum mechanics) is the generator of translation. When we replace del\nwith del - q A are we still able to view this as some sort of\ntranslation generator? How does A play a role here?\n\nThanks,\nYi-Zen\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 everyone
I am just beginning to learn about differential geometry and general
relativity. I am wondering if there is a relationship between parallel
transportation of a tangent vector around a closed loop on some manifold
and the fact that they do not return as the same vector, to the
Aharanov-Bohm effect in quantum mechanics where a charged fermion gains
a phase shift in its wavefunction by going round a solenoid?
In quantum mechanics am I correct that we replace the derivative del_k
with del_k - q A_k, where A is the vector potential? Can we view this as
some sort of covariant derivative? By promoting A to a vector operator
does that mean, just as the Riemann curvature tensor is in some sense
the commutator of covariant derivatives, the commutator of the
components of the vector potential, say [A_q,A_r] also represents some
sort of curvature "tensor"?
Does the Aharanov-Bohm effect give us some sort of geometrical
interpretation of either quantum mechanics itself, electromagnetism (esp
the meaning of the vector potential), or both? If so what is it?
On a similar note we say the momentum operator (or the derivative in
quantum mechanics) is the generator of translation. When we replace del
with del - q A are we still able to view this as some sort of
translation generator? How does A play a role here?
Thanks,
Yi-Zen
<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>\n\nYi-Zen Chu; Yiren Qu wrote:\n> I am just beginning to learn about differential geometry and general\n> relativity. I am wondering if there is a relationship between\nparallel\n> transportation of a tangent vector around a closed loop on some\nmanifold\n> and the fact that they do not return as the same vector, to the\n> Aharanov-Bohm effect in quantum mechanics where a charged fermion\ngains\n> a phase shift in its wavefunction by going round a solenoid?\n>\n> In quantum mechanics am I correct that we replace the derivative\ndel_k\n> with del_k - q A_k, where A is the vector potential? Can we view this\nas\n> some sort of covariant derivative? By promoting A to a vector\noperator\n> does that mean, just as the Riemann curvature tensor is in some sense\n\n> the commutator of covariant derivatives, the commutator of the\n> components of the vector potential, say [A_q,A_r] also represents\nsome\n> sort of curvature "tensor"?\n\nThe addition of local gauge invariance to a field, is the same as\nsaying that the original field has a curvature, AFAIU. The gauge-field\ntells you what happens to a particle when you go around in small loops.\nIn simple non-relativistic electromagnetics, you could consider spatial\ngauge invariance of the wave phase as the possibility of having a\nmagnetic field, in that a wavefront can curve/bend (and electric fields\nare manifestations of temporal gauge invariance). When you move up to\nquantum electrodynamics, perhaps the curvature analogue is normally\nabandoned not because it is the wrong math, but because of the probably\ncounter-intuitive concept of quantized curvature (the photon-field,\ni.e. the gauge-field, would then describe curvature which is created\nand annihilated all over the field).\n\nThe gut-feeling (no pun intended :) is that nature operates similar\nprinciples in both the standard model and general relativity as you\nnote, and you can probably draw a lot of analogues (I didn\'t look too\nclosely at the ones you wrote). This way of seeing it is hardly\nentertained in the litterature though, being that most books are about\nthe standard model _or_ general relativity.\n\n/bjorn\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>Yi-Zen Chu; Yiren Qu wrote:
> I am just beginning to learn about differential geometry and general
> relativity. I am wondering if there is a relationship between
parallel
> transportation of a tangent vector around a closed loop on some
manifold
> and the fact that they do not return as the same vector, to the
> Aharanov-Bohm effect in quantum mechanics where a charged fermion
gains
> a phase shift in its wavefunction by going round a solenoid?
>
> In quantum mechanics am I correct that we replace the derivative
del_k
> with del_k - q A_k, where A is the vector potential? Can we view this
as
> some sort of covariant derivative? By promoting A to a vector
operator
> does that mean, just as the Riemann curvature tensor is in some sense
> the commutator of covariant derivatives, the commutator of the
> components of the vector potential, say [A_q,A_r] also represents
some
> sort of curvature "tensor"?
The addition of local gauge invariance to a field, is the same as
saying that the original field has a curvature, AFAIU. The gauge-field
tells you what happens to a particle when you go around in small loops.
In simple non-relativistic electromagnetics, you could consider spatial
gauge invariance of the wave phase as the possibility of having a
magnetic field, in that a wavefront can curve/bend (and electric fields
are manifestations of temporal gauge invariance). When you move up to
quantum electrodynamics, perhaps the curvature analogue is normally
abandoned not because it is the wrong math, but because of the probably
counter-intuitive concept of quantized curvature (the photon-field,
i.e. the gauge-field, would then describe curvature which is created
and annihilated all over the field).
The gut-feeling (no pun intended :) is that nature operates similar
principles in both the standard model and general relativity as you
note, and you can probably draw a lot of analogues (I didn't look too
closely at the ones you wrote). This way of seeing it is hardly
entertained in the litterature though, being that most books are about
the standard model _or_ general relativity.
/bjorn
Yi-Zen Chu
Oct17-04, 01:46 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>\n\n"Yi-Zen Chu; Yiren Qu" <y_i_-_z_e_n_._c_h_u_@_y_a_l_e_._e_d_u> wrote in message >\n> In quantum mechanics am I correct that we replace the derivative del_k\n> with del_k - q A_k, where A is the vector potential? Can we view this as\n> some sort of covariant derivative? By promoting A to a vector operator\n> does that mean, just as the Riemann curvature tensor is in some sense\n> the commutator of covariant derivatives, the commutator of the\n> components of the vector potential, say [A_q,A_r] also represents some\n> sort of curvature "tensor"?\n\nI must correct myself here regarding what I said about the commutator\nof the "covariant derivative" defined above.\n\n[del_mu - q A_nu, del_nu - q A_mu]\n= -q(del_mu A_nu - del_nu A_mu) + q^2[A_mu, A_nu]\n\nSo we really have two terms. The first term is commonly defined as\nF_{mu,nu}, I believe, and it exists even in the classical case when A,\nthe vector potential is not an operator. The second term involving the\ncommutator is purely quantum mechanical since it vanishes if A is just\na (vector) function.\n\nI\'d really appreciate if someone could tell me more about the\ngeometric significance of both terms, in either (or both) the\nclassical and quantum case. Can we view these as the analogue of the\nRiemann curvature tensor, and if so in what sense?\n\nThanks!\n\nYi-Zen\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>"Yi-Zen Chu; Yiren Qu" <y_{i_}-_z_e_n_._c_h_u_@_y_a_l_e_._e_d_u> wrote in message >
> In quantum mechanics am I correct that we replace the derivative del_k
> with del_k - q A_k, where A is the vector potential? Can we view this as
> some sort of covariant derivative? By promoting A to a vector operator
> does that mean, just as the Riemann curvature tensor is in some sense
> the commutator of covariant derivatives, the commutator of the
> components of the vector potential, say [A_q,A_r] also represents some
> sort of curvature "tensor"?
I must correct myself here regarding what I said about the commutator
of the "covariant derivative" defined above.
[del_mu - q A_{nu}, del_nu - q A_{mu}]= -q(del_mu A_{nu} - del_nu A_{mu}) + q^2[A_{mu}, A_{nu}]
So we really have two terms. The first term is commonly defined as
F_{\mu,\nu}, I believe, and it exists even in the classical case when A,
the vector potential is not an operator. The second term involving the
commutator is purely quantum mechanical since it vanishes if A is just
a (vector) function.
I'd really appreciate if someone could tell me more about the
geometric significance of both terms, in either (or both) the
classical and quantum case. Can we view these as the analogue of the
Riemann curvature tensor, and if so in what sense?
Thanks!
Yi-Zen
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