partial traces , fermion antisymmetry and density matrix [Archive]

Charles J. Quarra

May17-04, 08:00 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>Hi,\n\n\nUsually when one is first brought the problem in Quantum Mechanics to\ndeal with systems of more than one particle, one appeals to the notion\nof tensor product spaces. So if i have a system A of wavefunction\n\n\n\\psi_{A}(x_{A}) with basis |e_{A}(E_{i})>\nwith equivalent formulation in matrix density \\rho(x_{A,0} ,\nx_{A,1})=\\conjugate \\psi_{A}(x_{A,0}) \\psi_{A}(x_{A,1})\n\nand a wavefunction of system B\n\n\\psi_{B}(x_{B}) with basis |e_{B}(E_{j})>\nwith equivalent formulation in matrix density \\rho(x_{B,0} ,\nx_{B,1})= \\conjugate \\psi_{B}(x_{B,0}) \\psi_{B}(x_{B,1})\n\nthen QM says that the A+B system space is the one spanned by the\nbasis\n\n|e_{A}(E_{i})> |e_{B}(E_{j})>\n\nthe wavefunction of the coupled system is (assuming fermion fields)\n\n\\psi_{A}(x_{A})\\psi_{B}(x_{B}) - \\psi_{A}(x_{B})\\psi_{B}(x_{A})\n\n\nwhich in the equivalent matrix formalism gives us\n\n\\rho_{A}(x_{A,0},x_{A,1})\\rho_{B}(x_{B,0}, x_{B,1}) -\n\\rho_{A}(x_{A,0},x_{B,1})\\rho_{B}(x_{B,0},x_{A ,1}) -\n\\rho_{A}(x_{B,0},x_{A,1})\\rho_{B}(x_{A,0},x_{B ,1}) +\n\\rho_{A}(x_{B,0},x_{B,1})\\rho_{B}(x_{A,0},x_{ A,1})\n\nwhich is also totally antisymmetric relative to permutations between\nsystem A and B\n\n\n\nHowever, when you want to retrieve the partial systems in when the\ndiscrepancies between the wavefunction and the matrix density\nformalisms start to arise, because now, if you do a partial trace of\nthe A+B matrix density, on for example, you get something like (feel\nfree to correct me if i screwd up)\n\n\\rho_{A}(x_{A,0},x_{A,1}) + \\rho_{B}(x_{A,0},x_{A,1}) - sum ( K ,\n\\rho_{A}(x_{A,0} , K)\\rho_{B}(K , x_{A,1}) + \\rho_{B}(x_{A,0} ,\nK)\\rho_{A}(K , x_{A,1}) )\n\nwhich obviously depends on the wavefunction of both spaces\n\nThe problem is that i started with a uncoupled system (one of which\nthe wavefunction is factorizable in the tensor space), but now my\npartial trace is not anymore uncoupled\n\nObviously the problem happened when i assumed the field is symmetric,\nbut its likely that happens the same in the boson case. Im confused\nwith the relationship between these two steps: (anti)symmetrization\nand partial tracing.\n\nI suppose that i was expecting that the partial trace got me the\n\\rho_{A} density alone (which is pure, hence it can be expressed as a\nwavefunction on A).\n\n\nA density/wavefunction that depends upon an antisimetrization step\nbefore partial trace looks very bad. This should be reversible, since\nwe are talking of an uncoupled system. Any ideas out there what is\nwrong here in this reasoning?\n\n\nCharles Quarra\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>Hi,

Usually when one is first brought the problem in Quantum Mechanics to
deal with systems of more than one particle, one appeals to the notion
of tensor product spaces. So if i have a system A of wavefunction

\psi_{A}(x_{A})[/itex] with basis |e_{A}(E_{i})>
with equivalent formulation in matrix density \rho(x_{A,0} ,x_{A,1})=\conjugate \psi_{A}(x_{A,0}) \psi_{A}(x_{A,1})

and a wavefunction of system B

\psi_{B}(x_{B}) with basis |e_{B}(E_{j})>
with equivalent formulation in matrix density \rho(x_{B,0} ,x_{B,1})= \conjugate \psi_{B}(x_{B,0}) \psi_{B}(x_{B,1})

then QM says that the A+B system space is the one spanned by the
basis

|e_{A}(E_{i})> |e_{B}(E_{j})>

the wavefunction of the coupled system is (assuming fermion fields)

\psi_{A}(x_{A})\psi_{B}(x_{B}) - \psi_{A}(x_{B})\psi_{B}(x_{A})

which in the equivalent matrix formalism gives us

\rho_{A}(x_{A,0},x_{A,1})\rho_{B}(x_{B,0},x_{B,1}) -\rho_{A}(x_{A,0},x_{B,1})\rho_{B}(x_{B,0},x_{A,1}) -\rho_{A}(x_{B,0},x_{A,1})\rho_{B}(x_{A,0},x_{B,1}) +\rho_{A}(x_{B,0},x_{B,1})\rho_{B}(x_{A,0},x_{A,1} )

which is also totally antisymmetric relative to permutations between
system A and B

However, when you want to retrieve the partial systems in when the
discrepancies between the wavefunction and the matrix density
formalisms start to arise, because now, if you do a partial trace of
the A+B matrix density, on for example, you get something like (feel
free to correct me if i screwd up)

\rho_{A}(x_{A,0},x_{A,1}) + \rho_{B}(x_{A,0},x_{A,1}) - sum ( K ,
[itex]\rho_{A}(x_{A,0} , K)\rho_{B}(K , x_{A,1}) + \rho_{B}(x_{A,0} ,K)\rho_{A}(K , x_{A,1}) )

which obviously depends on the wavefunction of both spaces

The problem is that i started with a uncoupled system (one of which
the wavefunction is factorizable in the tensor space), but now my
partial trace is not anymore uncoupled

Obviously the problem happened when i assumed the field is symmetric,
but its likely that happens the same in the boson case. Im confused
with the relationship between these two steps: (anti)symmetrization
and partial tracing.

I suppose that i was expecting that the partial trace got me the
\rho_{A} density alone (which is pure, hence it can be expressed as a
wavefunction on A).

A density/wavefunction that depends upon an antisimetrization step
before partial trace looks very bad. This should be reversible, since
we are talking of an uncoupled system. Any ideas out there what is
wrong here in this reasoning?

Charles Quarra

Charles J. Quarra

May21-04, 10:40 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>\ndisposablemailaccountfornews@yahoo.com.ar (Charles J. Quarra) wrote in message news:<bc979c06.0405151808.583c62b6@posting.google. com>...\n> Hi,\n>\n>\n> Usually when one is first brought the problem in Quantum Mechanics to\n> deal with systems of more than one particle, one appeals to the notion\n> of tensor product spaces. So if i have a system A of wavefunction\n>\n>\n> \\psi_{A}(x_{A}) with basis |e_{A}(E_{i})>\n> with equivalent formulation in matrix density \\rho(x_{A,0} ,\n> x_{A,1})=\\conjugate \\psi_{A}(x_{A,0}) \\psi_{A}(x_{A,1})\n>\n> and a wavefunction of system B\n>\n> \\psi_{B}(x_{B}) with basis |e_{B}(E_{j})>\n> with equivalent formulation in matrix density \\rho(x_{B,0} ,\n> x_{B,1})= \\conjugate \\psi_{B}(x_{B,0}) \\psi_{B}(x_{B,1})\n>\n> then QM says that the A+B system space is the one spanned by the\n> basis\n>\n> |e_{A}(E_{i})> |e_{B}(E_{j})>\n>\n> the wavefunction of the coupled system is (assuming fermion fields)\n>\n> \\psi_{A}(x_{A})\\psi_{B}(x_{B}) - \\psi_{A}(x_{B})\\psi_{B}(x_{A})\n>\n>\n> which in the equivalent matrix formalism gives us\n>\n> \\rho_{A}(x_{A,0},x_{A,1})\\rho_{B}(x_{B,0},x_{B,1 }) -\n> \\rho_{A}(x_{A,0},x_{B,1})\\rho_{B}(x_{B,0},x_{A,1 }) -\n> \\rho_{A}(x_{B,0},x_{A,1})\\rho_{B}(x_{A,0},x_{B,1 }) +\n> \\rho_{A}(x_{B,0},x_{B,1})\\rho_{B}(x_{A,0},x_{A,1 })\n>\n> which is also totally antisymmetric relative to permutations between\n> system A and B\n>\n>\n>\n> However, when you want to retrieve the partial systems in when the\n> discrepancies between the wavefunction and the matrix density\n> formalisms start to arise, because now, if you do a partial trace of\n> the A+B matrix density, on for example, you get something like (feel\n> free to correct me if i screwd up)\n>\n> \\rho_{A}(x_{A,0},x_{A,1}) + \\rho_{B}(x_{A,0},x_{A,1}) - sum ( K ,\n> \\rho_{A}(x_{A,0} , K)\\rho_{B}(K , x_{A,1}) + \\rho_{B}(x_{A,0} ,\n> K)\\rho_{A}(K , x_{A,1}) )\n>\n> which obviously depends on the wavefunction of both spaces\n>\n> The problem is that i started with a uncoupled system (one of which\n> the wavefunction is factorizable in the tensor space), but now my\n> partial trace is not anymore uncoupled\n>\n> Obviously the problem happened when i assumed the field is symmetric,\n> but its likely that happens the same in the boson case. Im confused\n> with the relationship between these two steps: (anti)symmetrization\n> and partial tracing.\n>\n> I suppose that i was expecting that the partial trace got me the\n> \\rho_{A} density alone (which is pure, hence it can be expressed as a\n> wavefunction on A).\n>\n>\n> A density/wavefunction that depends upon an antisimetrization step\n> before partial trace looks very bad. This should be reversible, since\n> we are talking of an uncoupled system. Any ideas out there what is\n> wrong here in this reasoning?\n>\n\n\n\nI\'ve read this question again and i\'ve found with my question is\nill-formulated: of course that even if you begin with N 1-particle\nwavefunctions , once you form a slater determinant to get a physical\nN-fermion wavefunction, its impossible to reobtain the original\n1-particle wavefunctions (in the 2-particle case this can be easily\nviewed since you would need also the symmetric part to reconstruct the\nindividual factors, in N>=3 i guess you would need a list of\neven-permutation-positive products, but that is another history)\n\nHowever, Now that i realize i cannot obtain univocally a set of N\n1-particle wavefunctions of which slater determinant is equal to my\noriginal N-fermion wavefunction, i do the question: can i pick \'some\'\nof those sets of N 1-particle wavefunctions and constructs them\nalgorithmically? can i formulate the problem of \'optimality\' for such\nsets? or better, what would be a sensible metric function for such\noptimality? \n\nThe same question could apply to obtain a set of N(N-1)/2 2-particle\nwavefunctions that allowed to approximate better to the remainder\ngiven by the expression\n\nN-fermion wavefunction - N 1-particle wavefunction obtained in \n\n\nof course this method could bias against wavefunctions that in\nprincicle are already in a form of a set of K!/(N!(K-N)!) N-particle\nwavefunction, unless of course, my method would give that the\noptimal wavefunctions for M <> N are all zero\n\nComments, ideas, jokes about exportations being made to other\ncountries, all welcome\n\ngreetings\n\n\nCharles J. Quarra\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>disposablemailaccountfornews@yahoo.com.ar (Charles J. Quarra) wrote in message news:<bc979c06.0405151808.583c62b6@posting.google.com>...
> Hi,
>
>
> Usually when one is first brought the problem in Quantum Mechanics to
> deal with systems of more than one particle, one appeals to the notion
> of tensor product spaces. So if i have a system A of wavefunction
>
>
> \psi_{A}(x_{A}) with basis |e_{A}(E_{i})>
> with equivalent formulation in matrix density \rho(x_{A,0} ,
> x_{A,1})=\conjugate \psi_{A}(x_{A,0}) \psi_{A}(x_{A,1})
>
> and a wavefunction of system B
>
> \psi_{B}(x_{B}) with basis |e_{B}(E_{j})>
> with equivalent formulation in matrix density \rho(x_{B,0} ,
> x_{B,1})= \conjugate \psi_{B}(x_{B,0}) \psi_{B}(x_{B,1})
>
> then QM says that the A+B system space is the one spanned by the
> basis
>
> |e_{A}(E_{i})> |e_{B}(E_{j})>
>
> the wavefunction of the coupled system is (assuming fermion fields)
>
> \psi_{A}(x_{A})\psi_{B}(x_{B}) - \psi_{A}(x_{B})\psi_{B}(x_{A})
>
>
> which in the equivalent matrix formalism gives us
>
> \rho_{A}(x_{A,0},x_{A,1})\rho_{B}(x_{B,0},x_{B,1}) -
> \rho_{A}(x_{A,0},x_{B,1})\rho_{B}(x_{B,0},x_{A,1}) -
> \rho_{A}(x_{B,0},x_{A,1})\rho_{B}(x_{A,0},x_{B,1}) +
> \rho_{A}(x_{B,0},x_{B,1})\rho_{B}(x_{A,0},x_{A,1})
>
> which is also totally antisymmetric relative to permutations between
> system A and B
>
>
>
> However, when you want to retrieve the partial systems in when the
> discrepancies between the wavefunction and the matrix density
> formalisms start to arise, because now, if you do a partial trace of
> the A+B matrix density, on for example, you get something like (feel
> free to correct me if i screwd up)
>
> \rho_{A}(x_{A,0},x_{A,1}) + \rho_{B}(x_{A,0},x_{A,1}) - sum ( K ,
> \rho_{A}(x_{A,0} , K)\rho_{B}(K , x_{A,1}) + \rho_{B}(x_{A,0} ,
> K)\rho_{A}(K , x_{A,1}) )
>
> which obviously depends on the wavefunction of both spaces
>
> The problem is that i started with a uncoupled system (one of which
> the wavefunction is factorizable in the tensor space), but now my
> partial trace is not anymore uncoupled
>
> Obviously the problem happened when i assumed the field is symmetric,
> but its likely that happens the same in the boson case. Im confused
> with the relationship between these two steps: (anti)symmetrization
> and partial tracing.
>
> I suppose that i was expecting that the partial trace got me the
> \rho_{A} density alone (which is pure, hence it can be expressed as a
> wavefunction on A).
>
>
> A density/wavefunction that depends upon an antisimetrization step
> before partial trace looks very bad. This should be reversible, since
> we are talking of an uncoupled system. Any ideas out there what is
> wrong here in this reasoning?
>

I've read this question again and i've found with my question is
ill-formulated: of course that even if you begin with N 1-particle
wavefunctions , once you form a slater determinant to get a physical
N-fermion wavefunction, its impossible to reobtain the original
1-particle wavefunctions (in the 2-particle case this can be easily
viewed since you would need also the symmetric part to reconstruct the
individual factors, in N>=3 i guess you would need a list of
even-permutation-positive products, but that is another history)

However, Now that i realize i cannot obtain univocally a set of N
1-particle wavefunctions of which slater determinant is equal to my
original N-fermion wavefunction, i do the question: can i pick 'some'
of those sets of N 1-particle wavefunctions and constructs them
algorithmically? can i formulate the problem of 'optimality' for such
sets? or better, what would be a sensible metric function for such
optimality?

The same question could apply to obtain a set of N(N-1)/2 2-particle
wavefunctions that allowed to approximate better to the remainder
given by the expression

N-fermion wavefunction - N 1-particle wavefunction obtained in

of course this method could bias against wavefunctions that in
princicle are already in a form of a set of K!/(N!(K-N)!) N-particle
wavefunction, unless of course, my method would give that the
optimal wavefunctions for M <> N are all zero

Comments, ideas, jokes about exportations being made to other
countries, all welcome

greetings

Charles J. Quarra

Arnold Neumaier

May23-04, 03:15 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>Charles J. Quarra wrote:\n\n> However, Now that i realize i cannot obtain univocally a set of N\n> 1-particle wavefunctions of which slater determinant is equal to my\n> original N-fermion wavefunction,\n\nYou can get it exactly up to an orthogonal transformation in the space spanned\nby these wave functions. There is no need to get more, since electrons are\nindistinguishable - so a single electron in a cloud of N electrons has\nno individuality at all. It is intrinsically entangled with the whole\ncloud.\n\nStrictly speaking, all electrons in the world are entangled with each other\nbecause of the Pauli principle, but at large distance (in units of Bohr radii)\nthe effect is negligible.\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>Charles J. Quarra wrote:

> However, Now that i realize i cannot obtain univocally a set of N
> 1-particle wavefunctions of which slater determinant is equal to my
> original N-fermion wavefunction,

You can get it exactly up to an orthogonal transformation in the space spanned
by these wave functions. There is no need to get more, since electrons are
indistinguishable - so a single electron in a cloud of N electrons has
no individuality at all. It is intrinsically entangled with the whole
cloud.

Strictly speaking, all electrons in the world are entangled with each other
because of the Pauli principle, but at large distance (in units of Bohr radii)
the effect is negligible.

Arnold Neumaier

Charles J. Quarra

May25-04, 04:55 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>\nArnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40AF86A2.7030207@univie.ac.at>...\n> Charles J. Quarra wrote:\n>\n> > However, Now that i realize i cannot obtain univocally a set of N\n> > 1-particle wavefunctions of which slater determinant is equal to my\n> > original N-fermion wavefunction,\n>\n> You can get it exactly up to an orthogonal transformation in the space spanned\n> by these wave functions. There is no need to get more, since electrons are\n> indistinguishable - so a single electron in a cloud of N electrons has\n> no individuality at all. It is intrinsically entangled with the whole\n> cloud.\n>\n> Strictly speaking, all electrons in the world are entangled with each other\n> because of the Pauli principle, but at large distance (in units of Bohr radii)\n> the effect is negligible.\n>\n\nBut, how would you do that? I mean, in principle you can compose\n3-electron wavefunctions from a 2-electron entangled wavefunction and\n1-electron wavefunctions doing\n\nW1(x) 1-electron\nW2(y,z) 2-electron (antisymmetric but entangled)\n\nW3(x,y,z) = W1(x)W2(y,z) + W1(y)W2(z,x) + W1(z)W2(x,y)\n- W1(x)W2(z,y) - W1(z)W2(y,x) - W1(y)W2(x,z)\n\n(however the last three terms are not needed based on the antisymmetry\nof W2)\n\nnote that in principle a decomposition in 3 wavefunctions cannot be\ncomplete (i.e: W3 - the slater determinant of such "decomposition"\nwould be different from zero)\n\na entangled 2-electron wavefunction cannot be decomposed in\n1-electron wavefunctions since that would lose the entanglement;\nhowever in principle you could go on and ask for such a decomposition\nin terms of a 2-electron slater determinant + traceless density matrix\n(lets suppose that such decomposition of W2 is W2_A(y)*W2_B(z) -\nW2_A(z)*W2_B(y) + p_0(y,z) where Tr(p_0)=0 )\n\nbtw, adding density matrices can be interpreted as adding quantum\n(pure) states to a mixture, but i dont understand how to interpret a\ntraceless density matrix.\n\nQ: can a traceless matrix be written in a special way?\n\nOn the former question, a sensible decomposition of W3 should be\nconsistent with the W2 decomposition giving\n\nW1(x)W2_A(y)W2_B(z) + W1(y)W2_A(z)W2_B(x) + W1(z)W2_A(x)W2_B(y)\n- W1(x)W2_A(z)W2_B(y) - W1(z)W2_A(y)W2_B(x) - W1(y)W2_A(x)W2_B(z)\n\n+ W1(x)p_0(y,z) + W1(y)p_0(z,x) + W1(z)p_0(x,y)\n\n\n\nBut i have no clue _how_ we can obtain such a decomposition from an\narbitrary N-electron wavefunction\n\n\n\n\nbut a decomposition of the 3-electron wavefunction\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40AF86A2.7030207@univie.ac.at>...
> Charles J. Quarra wrote:
>
> > However, Now that i realize i cannot obtain univocally a set of N
> > 1-particle wavefunctions of which slater determinant is equal to my
> > original N-fermion wavefunction,
>
> You can get it exactly up to an orthogonal transformation in the space spanned
> by these wave functions. There is no need to get more, since electrons are
> indistinguishable - so a single electron in a cloud of N electrons has
> no individuality at all. It is intrinsically entangled with the whole
> cloud.
>
> Strictly speaking, all electrons in the world are entangled with each other
> because of the Pauli principle, but at large distance (in units of Bohr radii)
> the effect is negligible.
>

But, how would you do that? I mean, in principle you can compose
3-electron wavefunctions from a 2-electron entangled wavefunction and
1-electron wavefunctions doing

W1(x) 1-electron
W2(y,z) 2-electron (antisymmetric but entangled)

W3(x,y,z) = W1(x)W2(y,z) + W1(y)W2(z,x) + W1(z)W2(x,y)- W1(x)W2(z,y) - W1(z)W2(y,x) - W1(y)W2(x,z)

(however the last three terms are not needed based on the antisymmetry
of W2)

note that in principle a decomposition in 3 wavefunctions cannot be
complete (i.e: W3 - the slater determinant of such "decomposition"
would be different from zero)

a entangled 2-electron wavefunction cannot be decomposed in
1-electron wavefunctions since that would lose the entanglement;
however in principle you could go on and ask for such a decomposition
in terms of a 2-electron slater determinant + traceless density matrix
(lets suppose that such decomposition of W2 is W2_A(y)*W2_B(z) -W2_A(z)*W2_B(y) + p_0(y,z) where Tr(p_0)=0 )

btw, adding density matrices can be interpreted as adding quantum
(pure) states to a mixture, but i dont understand how to interpret a
traceless density matrix.

Q: can a traceless matrix be written in a special way?

On the former question, a sensible decomposition of W3 should be
consistent with the W2 decomposition giving

W1(x)W2_A(y)W2_B(z) + W1(y)W2_A(z)W2_B(x) + W1(z)W2_A(x)W2_B(y)- W1(x)W2_A(z)W2_B(y) - W1(z)W2_A(y)W2_B(x) - W1(y)W2_A(x)W2_B(z)+ W1(x)p_0(y,z) + W1(y)p_0(z,x) + W1(z)p_0(x,y)

But i have no clue _how_ we can obtain such a decomposition from an
arbitrary N-electron wavefunction

but a decomposition of the 3-electron wavefunction

Arnold Neumaier

May25-04, 08:26 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>\n\nCharles J. Quarra wrote:\n> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40AF86A2.7030207@univie.ac.at>...\n>\n>>Char les J. Quarra wrote:\n>>\n>>\n>>> However, Now that i realize i cannot obtain univocally a set of N\n>>>1-particle wavefunctions of which slater determinant is equal to my\n>>>original N-fermion wavefunction,\n>>\n>>You can get it exactly up to an orthogonal transformation in the space spanned\n>>by these wave functions. There is no need to get more, since electrons are\n>>indistinguishable - so a single electron in a cloud of N electrons has\n>>no individuality at all. It is intrinsically entangled with the whole\n>>cloud.\n>>\n>>Strictly speaking, all electrons in the world are entangled with each other\n>>because of the Pauli principle, but at large distance (in units of Bohr radii)\n>>the effect is negligible.\n>>\n>\n>\n> But, how would you do that?\n\nI don\'t understand what precisely you want. I just meant, a 2-particle\nSlater state is already entangled.\n\n.. I mean, in principle you can compose\n> 3-electron wavefunctions from a 2-electron entangled wavefunction and\n> 1-electron wavefunctions doing\n>\n> W1(x) 1-electron\n> W2(y,z) 2-electron (antisymmetric but entangled)\n>\n> W3(x,y,z) = W1(x)W2(y,z) + W1(y)W2(z,x) + W1(z)W2(x,y)\n> - W1(x)W2(z,y) - W1(z)W2(y,x) - W1(y)W2(x,z)\n>\n> (however the last three terms are not needed based on the antisymmetry\n> of W2)\n\nYes.\n\n> a entangled 2-electron wavefunction cannot be decomposed in\n> 1-electron wavefunctions\n\nYou can write it as a superposition of Slater stes.\n\n\n> btw, adding density matrices can be interpreted as adding quantum\n> (pure) states to a mixture, but i dont understand how to interpret a\n> traceless density matrix.\n\nWhat do you mean??? A density matrix always has trace 1.\n\n\nArnold Neumaier\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>Charles J. Quarra wrote:
> Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40AF86A2.7030207@univie.ac.at>...
>
>>Charles J. Quarra wrote:
>>
>>
>>> However, Now that i realize i cannot obtain univocally a set of N
>>>1-particle wavefunctions of which slater determinant is equal to my
>>>original N-fermion wavefunction,
>>
>>You can get it exactly up to an orthogonal transformation in the space spanned
>>by these wave functions. There is no need to get more, since electrons are
>>indistinguishable - so a single electron in a cloud of N electrons has
>>no individuality at all. It is intrinsically entangled with the whole
>>cloud.
>>
>>Strictly speaking, all electrons in the world are entangled with each other
>>because of the Pauli principle, but at large distance (in units of Bohr radii)
>>the effect is negligible.
>>
>
>
> But, how would you do that?

I don't understand what precisely you want. I just meant, a 2-particle
Slater state is already entangled.

.. I mean, in principle you can compose
> 3-electron wavefunctions from a 2-electron entangled wavefunction and
> 1-electron wavefunctions doing
>
> W1(x) 1-electron
> W2(y,z) 2-electron (antisymmetric but entangled)
>
> W3(x,y,z) = W1(x)W2(y,z) + W1(y)W2(z,x) + W1(z)W2(x,y)
> - W1(x)W2(z,y) - W1(z)W2(y,x) - W1(y)W2(x,z)
>
> (however the last three terms are not needed based on the antisymmetry
> of W2)

Yes.

> a entangled 2-electron wavefunction cannot be decomposed in
> 1-electron wavefunctions

You can write it as a superposition of Slater stes.

> btw, adding density matrices can be interpreted as adding quantum
> (pure) states to a mixture, but i dont understand how to interpret a
> traceless density matrix.

What do you mean??? A density matrix always has trace 1.

Arnold Neumaier

Charles J. Quarra

May29-04, 11:54 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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40B342D3.2050008@univie.ac.at>...\n\n> I don\'t understand what precisely you want. I just meant, a 2-particle\n> Slater state is already entangled.\n>\n\nWell, going back to my original question; i want to get a set _S_ of\n1-particle wavefunctions (with possibly K-particle wavefunction\nantisymmetrized corrections with 2<=K<=N) from a N-particle\n(antisymmetric) wavefunction.\n\nThe idea is precisely that the set _S_ of 1-particle wavefunctions\nshould reconstruct the original N-particle wavefunction from a Slater\ndeterminant operation.\n\nI also talk about "possible K-particle wavefunction antisymmetrized\ncorrections" because its not clear to me at this moment that any\nantisymmetric N-dimensional function can be written as a Slater\ndeterminant of 1-dim functions\n\n\n\n> > a entangled 2-electron wavefunction cannot be decomposed in\n> > 1-electron wavefunctions\n>\n> You can write it as a superposition of Slater stes.\n>\n\nOk, this being true based on the completeness in the antisymmetric\nsubspace of E^n (E being a 1-particle Hillbert space)\n\nI guess i was thinking into breaking N-particle wavefunctions into a\nsum of N 1-particle uncoupled states plus K-particle amplitude\ncorrelations, but i guess all those K-correlations can also be broken\ninto sums of slater determinants of 1-particle states\n\n>\n> > btw, adding density matrices can be interpreted as adding quantum\n> > (pure) states to a mixture, but i dont understand how to interpret a\n> > traceless density matrix.\n>\n> What do you mean??? A density matrix always has trace 1.\n>\n\n\nWhen you have a density matrix that is not pure, how would you break\nit into sums of pure parts?\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>Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote in message news:<40B342D3.2050008@univie.ac.at>...

> I don't understand what precisely you want. I just meant, a 2-particle
> Slater state is already entangled.
>

Well, going back to my original question; i want to get a set _S_ of
1-particle wavefunctions (with possibly K-particle wavefunction
antisymmetrized corrections with 2<=K<=N) from a N-particle
(antisymmetric) wavefunction.

The idea is precisely that the set _S_ of 1-particle wavefunctions
should reconstruct the original N-particle wavefunction from a Slater
determinant operation.

I also talk about "possible K-particle wavefunction antisymmetrized
corrections" because its not clear to me at this moment that any
antisymmetric N-dimensional function can be written as a Slater
determinant of 1-dim functions

> > a entangled 2-electron wavefunction cannot be decomposed in
> > 1-electron wavefunctions
>
> You can write it as a superposition of Slater stes.
>

Ok, this being true based on the completeness in the antisymmetric
subspace of E^n (E being a 1-particle Hillbert space)

I guess i was thinking into breaking N-particle wavefunctions into a
sum of N 1-particle uncoupled states plus K-particle amplitude
correlations, but i guess all those K-correlations can also be broken
into sums of slater determinants of 1-particle states

>
> > btw, adding density matrices can be interpreted as adding quantum
> > (pure) states to a mixture, but i dont understand how to interpret a
> > traceless density matrix.
>
> What do you mean??? A density matrix always has trace 1.
>

When you have a density matrix that is not pure, how would you break
it into sums of pure parts?