contravariant tensor fields, differential operators and quantisation [Archive]

Tobias Fritz

May25-04, 10:01 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\nHere are some thoughts about tensor fields corresponding to differential\noperators acting on the functions on a manifold. The functions on the\nmanifold are interpreted as the state space of a quantum system.\n\nNotation:\nM = n-dim real manifold representing space of the quantum-mechanical system\nT^k(M) = k-times contravariant tensor fields on M\nF(M) = complex-valued functions on M (states)\n\nNow the following is true:\n\n(0) Given an element v_0 in T^0(M), which is just a real scalar function, we\ncan make it into an operator F(M)->F(M) in the obvious way: just multiply\npointwise by v_0; I will interpret this as a differential operator of 0th\norder.\nExample: position operator in QM\n\n(1) Given an element v_1 in T^1(M), which is a vector field, there is also\nan obvious operation on F(M) which is given by the Lie derivative. So an\nelement on T^1(M) gives rise to a differential operator of 1st order.\nExample: (angular) momentum operator in QM\n\n\nNow IMHO we must immediately ask: can we make any tensor field in T^k(M)\ninto a differential operator of k-th order in some natural way? Natural\nshould mean that the map\n\nT^k(M) --> (operators on F(M))\n\nshould have some nice properties, of course R-linearity, but also some other\nkind of morphism. It does not work if we map tensor products to composition\nof operators, because then the wedge product of two vector fields will give\na diff op of 1st order.\nSo probably it is much better to take only symmetric tensors instead of\nT^k(M) (also because we can arbitrarily change the order of partial\nderivatives) and symmetric products; but on the other hand, symmetric\nproducts do commute, while operators do not.\n\nMy first thought also was not to use the tensor algebra, but the inverse\nlimit of the iterated tangent bundles M, TM, TTM, ... but now I think that\ntensors are better suited.\n\nThe goal is to represent differential operators on F(M) directly on the\nmanifold M like in the two examples (0) and (1) above. Then it would be\npossible to expand every sufficiently nice operator into series of diff ops\n(this can be easily seen in a chart: what does the operator do to constant\nfunctions, what happens to linear functions, what to quadratic ones etc.)\nand also represent it somehow geometrically on M.\n\nAnd ideas how to generalize (0) and (1) to higher orders?\n\nTIA,\nTobias\n--\nhang my head drown my fear\ntill you all just disappear\n\nreverse my forename for mail! - saibot\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>Here are some thoughts about tensor fields corresponding to differential
operators acting on the functions on a manifold. The functions on the
manifold are interpreted as the state space of a quantum system.

Notation:
M = n-dim real manifold representing space of the quantum-mechanical system
T^k(M) = k-times contravariant tensor fields on MF(M) = complex-valued functions on M (states)

Now the following is true:

(0) Given an element v_0 in T^0(M), which is just a real scalar function, we
can make it into an operator F(M)->F(M) in the obvious way: just multiply
pointwise by v_0; I will interpret this as a differential operator of 0th
order.
Example: position operator in QM

(1) Given an element v_1 in T^1(M), which is a vector field, there is also
an obvious operation on F(M) which is given by the Lie derivative. So an
element on T^1(M) gives rise to a differential operator of 1st order.
Example: (angular) momentum operator in QM

Now IMHO we must immediately ask: can we make any tensor field in T^k(M)
into a differential operator of k-th order in some natural way? Natural
should mean that the map

T^k(M) --> (operators on F(M))

should have some nice properties, of course R-linearity, but also some other
kind of morphism. It does not work if we map tensor products to composition
of operators, because then the wedge product of two vector fields will give
a diff op of 1st order.
So probably it is much better to take only symmetric tensors instead of
T^k(M) (also because we can arbitrarily change the order of partial
derivatives) and symmetric products; but on the other hand, symmetric
products do commute, while operators do not.

My first thought also was not to use the tensor algebra, but the inverse
limit of the iterated tangent bundles M, TM, TTM, ... but now I think that
tensors are better suited.

The goal is to represent differential operators on F(M) directly on the
manifold M like in the two examples (0) and (1) above. Then it would be
possible to expand every sufficiently nice operator into series of diff ops
(this can be easily seen in a chart: what does the operator do to constant
functions, what happens to linear functions, what to quadratic ones etc.)
and also represent it somehow geometrically on M.

And ideas how to generalize (0) and (1) to higher orders?

TIA,
Tobias
--
hang my head drown my fear
till you all just disappear

reverse my forename for mail! - saibot

Alfred Einstead

May26-04, 01:06 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\nTobias Fritz <tobias@mad.scientist.com> wrote:\n> Notation:\n> M = n-dim real manifold representing space of the quantum-mechanical system\n> T^k(M) = k-times contravariant tensor fields on M\n> F(M) = complex-valued functions on M (states)\n\nF(M) = the space of observables of the *classical* phase space M, and\nforms a C*-algebra.\n\nThe state space Omega(M) of F(M), applying standard definitions and\nrecipes, is the subset of the dual F(M)* consisting of those\nfunctionals W that yield W[1] = 1, and W[f] >= 0, for f >= 0, f in F(M).\n\nThis is a convex space with\nW1, W2 in Omega(M) --> p W1 + q W2 in Omega(M) for p,q>=0;p+q=1\nThe subspace Omega_0(M) being the space of states in Omega(M) that\nare not reducible as above as non-trivial mixtures of other states.\n\nThis is already known to be equivalent to the space of probability\ndistributions over M; the pure states being the singular distributions;\nessentially identified with the phase space M, itself:\nOmega_0(M) = M.\n\nIf you want a *quantum* state space, you need a *non-commutative*\nalgebra somewhere. The (pure) state space of a commutative\nC*-algebra is *always* equivalent to a manifold on which that\nalgebra is realised as a function algebra.\n\nThe Schroedinger picture arises as the standard representation\nof the Heisenberg algebra H_n, given by:\n[q^i,q^j] = [p_i,p_j] = 0; 0 <= i,j < n\n[q^i,p_j] = i h-bar delta^i_j; 0 <= i,j < n.\n\nOne *proves* that the space of pure states Omega_0(H_n) is then\nequivalent to that given by the Schroedinger representation:\nW_{psi}[f(p,q)] = integral psi*(x) f(-i h-bar del, x) psi(x) dx\n(via the Stone von Neumann theorem).\n\nWhat you\'re trying to do is recover the standard Schroedinger\nrepresentation:\np <-> -i h-bar grad\nin a more general context; whereas this is not a suitable\nstarting point, but rather a *theorem* arising from these more\ngeneral considerations above that are universal across all physics\n(classical, or quantum) and hence must be taken as a more\nsuitable starting point.\n\nThe space that f(x) lives over actually arises as a consequence\nof the subalgebra {x^1,x^2,...,x^n} being a maximal Abelian ideal.\nThat means that a state space representation over H_n is parametrized\nby the eigenvalues of the operators in this subalgebra; i.e., is\na function of the x\'s.\n\nThe resemblance to 3-D space is specious and nothing but a\ndistraction.\n\nMore generally, the space f(x) will live in will NOT be a 3-D\nspace, but the *Dirichlet* part of the classical phase space\nof a n-body system, which is a (3n)-D space.\n\nThat\'s where your analogy (and analysis) break down.\n\nThe x\'s are not on a manifold, they\'re living on a phase space.\n\nIn a more general quantum system, given by the standard Hamiltonian\nrecipe, you\'ll always have the split of the algebra of observables\ninto a set of generators of the form p\'s and q\'s with:\n[q^i,q^j] = 0 = [p_i,p_j]; [q^i,p_j] = i h-bar delta^i_j\nthough in the general case, the i\'s and j\'s may range over an\ninfinite (but countable) set.\n\nThe corresponding states would then be parametrized by a suitably\nselected maximal Abelian ideal. There\'s zillions of them running\naround in this algebra, as there were for H_n. For instance, for\nH_3 you had the subalgebras generated by the following sets:\n{q^1,q^2,q^3}\n{p_1,p_2,p_3}\n{p_1,q^2,p_3} \n{q^1,q^2 cos(G) + p_2 k sin(G), q^3 cos(G) - p_3 k sin(G)}\nall of which are Abelian and maximal, and so can each be used as\na parametrization of the state space.\n\nFor H_n a maximal Abelian ideal is *infinite* dimensional (if\nthe algebra above is), with, e.g.:\n{q^1,q^2,q^3,...}\n{p_1,p_2,p_3,...}\nbeing two cases in point.\n\nThe state space would be parametrized by *all* the eigenvalues of\nthe operators in the corresponding set; e.g.\nq^i |x1,x2,x3,...> = x^i |x1,x2,x3,...>; i=1,2,...\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>Tobias Fritz <tobias@mad.scientist.com> wrote:
> Notation:
> M = n-dim real manifold representing space of the quantum-mechanical system
> T^k(M) = k-times contravariant tensor fields on M
> F(M) = complex-valued functions on M (states)

F(M) = the space of observables of the *classical* phase space M, and
forms a C*-algebra.

The state space \Omega(M) of F(M), applying standard definitions and
recipes, is the subset of the dual F(M)* consisting of those
functionals W that yield W[1] = 1, and W[f] >= 0, for f >= 0, f in F(M).

This is a convex space with
W1, W2 in \Omega(M) --> p W1 + q W2 in \Omega(M) for p,q>=0;p+q=1
The subspace \Omega_0(M) being the space of states in \Omega(M) that
are not reducible as above as non-trivial mixtures of other states.

This is already known to be equivalent to the space of probability
distributions over M; the pure states being the singular distributions;
essentially identified with the phase space M, itself:
\Omega_0(M) = M.

If you want a *quantum* state space, you need a *non-commutative*
algebra somewhere. The (pure) state space of a commutative
C*-algebra is *always* equivalent to a manifold on which that
algebra is realised as a function algebra.

The Schroedinger picture arises as the standard representation
of the Heisenberg algebra H_n, given by:
[q^i,q^j] = [p_i,p_j] = 0; <= i,j < n
[q^i,p_j] = i h-bar \delta^i_j; <= i,j < n.

One *proves* that the space of pure states \Omega_0(H_n) is then
equivalent to that given by the Schroedinger representation:
W_{\psi}[f(p,q)] = integral \psi*(x) f(-i h-bar del, x) \psi(x) dx
(via the Stone von Neumann theorem).

What you're trying to do is recover the standard Schroedinger
representation:
p <-> -i h-bar grad
in a more general context; whereas this is not a suitable
starting point, but rather a *theorem* arising from these more
general considerations above that are universal across all physics
(classical, or quantum) and hence must be taken as a more
suitable starting point.

The space that f(x) lives over actually arises as a consequence
of the subalgebra {x^1,x^2,...,x^n} being a maximal Abelian ideal.
That means that a state space representation over H_n is parametrized
by the eigenvalues of the operators in this subalgebra; i.e., is
a function of the x's.

The resemblance to 3-D space is specious and nothing but a
distraction.

More generally, the space f(x) will live in will NOT be a 3-D
space, but the *Dirichlet* part of the classical phase space
of a n-body system, which is a (3n)-D space.

That's where your analogy (and analysis) break down.

The x's are not on a manifold, they're living on a phase space.

In a more general quantum system, given by the standard Hamiltonian
recipe, you'll always have the split of the algebra of observables
into a set of generators of the form p's and q's with:
[q^i,q^j] == [p_i,p_j]; [q^i,p_j] = i h-bar \delta^i_j
though in the general case, the i's and j's may range over an
infinite (but countable) set.

The corresponding states would then be parametrized by a suitably
selected maximal Abelian ideal. There's zillions of them running
around in this algebra, as there were for H_n. For instance, for
H_3 you had the subalgebras generated by the following sets:
{q^1,q^2,q^3}{p_1,p_2,p_3}{p_1,q^2,p_3}{q^1,q^2 cos(G) + p_2 k sin(G), q^3 cos(G) - p_3 k sin(G)}
all of which are Abelian and maximal, and so can each be used as
a parametrization of the state space.

For H_n a maximal Abelian ideal is *infinite* dimensional (if
the algebra above is), with, e.g.:
{q^1,q^2,q^3,...}
{p_1,p_2,p_3,...}
being two cases in point.

The state space would be parametrized by *all* the eigenvalues of
the operators in the corresponding set; e.g.
q^i |x1,x2,x3,...> = x^i |x1,x2,x3,...>; i=1,2,...

Tobias Fritz

May29-04, 11:58 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>>> Notation:\n>> M = n-dim real manifold representing space of the quantum-mechanical\n>> system T^k(M) = k-times contravariant tensor fields on M\n>> F(M) = complex-valued functions on M (states)\n>\n> F(M) = the space of observables of the *classical* phase space M, and\n> forms a C*-algebra.\n>\n> The state space Omega(M) of F(M), applying standard definitions and\n> recipes, is the subset of the dual F(M)* consisting of those\n> functionals W that yield W[1] = 1, and W[f] >= 0, for f >= 0, f in F(M).\n>\nWhat is the physical meaning of this? I mean, why do we take the dual space?\nIntuitively, shouldn\'t the algebra of observables in some sense be "larger"\nthan the state space?\nIs this the same as passing from ordinary wave functions L^2(M) (hmm... what\nmeasure do we use here?) to the distributions (with the delta distribution\nwe get pure states).\n\n>\n> This is a convex space with\n> W1, W2 in Omega(M) --> p W1 + q W2 in Omega(M) for p,q>=0;p+q=1\n> The subspace Omega_0(M) being the space of states in Omega(M) that\n> are not reducible as above as non-trivial mixtures of other states.\n>\n> This is already known to be equivalent to the space of probability\n> distributions over M; the pure states being the singular distributions;\n> essentially identified with the phase space M, itself:\n> Omega_0(M) = M.\n>\nDo you really mean *phase* space here? This would imply that we have an\neigenstate of position *and* momentum operator. OK, probably it has to do\nwith the commutativity of the next section, so that real quantum systems\ncoming from non-abelian operator algebras cannot be represented this way.\n\n>\n> If you want a *quantum* state space, you need a *non-commutative*\n> algebra somewhere. The (pure) state space of a commutative\n> C*-algebra is *always* equivalent to a manifold on which that\n> algebra is realised as a function algebra.\n>\n> The Schroedinger picture arises as the standard representation\n> of the Heisenberg algebra H_n, given by:\n> [q^i,q^j] = [p_i,p_j] = 0; 0 <= i,j < n\n> [q^i,p_j] = i h-bar delta^i_j; 0 <= i,j < n.\n>\n> One *proves* that the space of pure states Omega_0(H_n) is then\n> equivalent to that given by the Schroedinger representation:\n> W_{psi}[f(p,q)] = integral psi*(x) f(-i h-bar del, x) psi(x) dx\n> (via the Stone von Neumann theorem).\n>\n> What you\'re trying to do is recover the standard Schroedinger\n> representation:\n> p <-> -i h-bar grad\n> in a more general context; whereas this is not a suitable\n> starting point, but rather a *theorem* arising from these more\n> general considerations above that are universal across all physics\n> (classical, or quantum) and hence must be taken as a more\n> suitable starting point.\n>\nYes, I am more familiar with this algebraic approach than with the\ndefinitions above you gave for phase space. Could you elaborate a bit in\nthem?\n\n> The space that f(x) lives over actually arises as a consequence\n> of the subalgebra {x^1,x^2,...,x^n} being a maximal Abelian ideal.\n> That means that a state space representation over H_n is parametrized\n> by the eigenvalues of the operators in this subalgebra; i.e., is\n> a function of the x\'s.\n>\n> The resemblance to 3-D space is specious and nothing but a\n> distraction.\n>\n> More generally, the space f(x) will live in will NOT be a 3-D\n> space, but the *Dirichlet* part of the classical phase space\n> of a n-body system, which is a (3n)-D space.\n>\n> That\'s where your analogy (and analysis) break down.\n>\n> The x\'s are not on a manifold, they\'re living on a phase space.\n>\n> In a more general quantum system, given by the standard Hamiltonian\n> recipe, you\'ll always have the split of the algebra of observables\n> into a set of generators of the form p\'s and q\'s with:\n> [q^i,q^j] = 0 = [p_i,p_j]; [q^i,p_j] = i h-bar delta^i_j\n> though in the general case, the i\'s and j\'s may range over an\n> infinite (but countable) set.\n>\n> The corresponding states would then be parametrized by a suitably\n> selected maximal Abelian ideal. There\'s zillions of them running\n> around in this algebra, as there were for H_n. For instance, for\n> H_3 you had the subalgebras generated by the following sets:\n> {q^1,q^2,q^3}\n> {p_1,p_2,p_3}\n> {p_1,q^2,p_3}\n> {q^1,q^2 cos(G) + p_2 k sin(G), q^3 cos(G) - p_3 k sin(G)}\n> all of which are Abelian and maximal, and so can each be used as\n> a parametrization of the state space.\n>\n> For H_n a maximal Abelian ideal is *infinite* dimensional (if\n> the algebra above is), with, e.g.:\n> {q^1,q^2,q^3,...}\n> {p_1,p_2,p_3,...}\n> being two cases in point.\n>\n> The state space would be parametrized by *all* the eigenvalues of\n> the operators in the corresponding set; e.g.\n> q^i |x1,x2,x3,...> = x^i |x1,x2,x3,...>; i=1,2,...\n\n\nThank you for your explanations!\nTobias\n--\nhang my head drown my fear\ntill you all just disappear\n\nreverse my forename for mail! - saibot\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>>> Notation:
>> M = n-dim real manifold representing space of the quantum-mechanical
>> system T^k(M) = k-times contravariant tensor fields on M>> F(M) = complex-valued functions on M (states)
>
> F(M) = the space of observables of the *classical* phase space M, and
> forms a C*-algebra.
>
> The state space \Omega(M) of F(M), applying standard definitions and
> recipes, is the subset of the dual F(M)* consisting of those
> functionals W that yield W[1] = 1, and W[f] >= 0, for f >= 0, f in F(M).
>
What is the physical meaning of this? I mean, why do we take the dual space?
Intuitively, shouldn't the algebra of observables in some sense be "larger"
than the state space?
Is this the same as passing from ordinary wave functions L^2(M) (hmm... what
measure do we use here?) to the distributions (with the \delta distribution
we get pure states).

>
> This is a convex space with
> W1, W2 in \Omega(M) --> p W1 + q W2 in \Omega(M) for p,q>=0;p+q=1
> The subspace \Omega_0(M) being the space of states in \Omega(M) that
> are not reducible as above as non-trivial mixtures of other states.
>
> This is already known to be equivalent to the space of probability
> distributions over M; the pure states being the singular distributions;
> essentially identified with the phase space M, itself:
> \Omega_0(M) = M.
>
Do you really mean *phase* space here? This would imply that we have an
eigenstate of position *and* momentum operator. OK, probably it has to do
with the commutativity of the next section, so that real quantum systems
coming from non-abelian operator algebras cannot be represented this way.

>
> If you want a *quantum* state space, you need a *non-commutative*
> algebra somewhere. The (pure) state space of a commutative
> C*-algebra is *always* equivalent to a manifold on which that
> algebra is realised as a function algebra.
>
> The Schroedinger picture arises as the standard representation
> of the Heisenberg algebra H_n, given by:
> [q^i,q^j] = [p_i,p_j] = 0; <= i,j < n
> [q^i,p_j] = i h-bar \delta^i_j; <= i,j < n.
>
> One *proves* that the space of pure states \Omega_0(H_n) is then
> equivalent to that given by the Schroedinger representation:
> W_{\psi}[f(p,q)] = integral \psi*(x) f(-i h-bar del, x) \psi(x) dx
> (via the Stone von Neumann theorem).
>
> What you're trying to do is recover the standard Schroedinger
> representation:
> p <-> -i h-bar grad
> in a more general context; whereas this is not a suitable
> starting point, but rather a *theorem* arising from these more
> general considerations above that are universal across all physics
> (classical, or quantum) and hence must be taken as a more
> suitable starting point.
>
Yes, I am more familiar with this algebraic approach than with the
definitions above you gave for phase space. Could you elaborate a bit in
them?

> The space that f(x) lives over actually arises as a consequence
> of the subalgebra {x^1,x^2,...,x^n} being a maximal Abelian ideal.
> That means that a state space representation over H_n is parametrized
> by the eigenvalues of the operators in this subalgebra; i.e., is
> a function of the x's.
>
> The resemblance to 3-D space is specious and nothing but a
> distraction.
>
> More generally, the space f(x) will live in will NOT be a 3-D
> space, but the *Dirichlet* part of the classical phase space
> of a n-body system, which is a (3n)-D space.
>
> That's where your analogy (and analysis) break down.
>
> The x's are not on a manifold, they're living on a phase space.
>
> In a more general quantum system, given by the standard Hamiltonian
> recipe, you'll always have the split of the algebra of observables
> into a set of generators of the form p's and q's with:
> [q^i,q^j] == [p_i,p_j]; [q^i,p_j] = i h-bar \delta^i_j
> though in the general case, the i's and j's may range over an
> infinite (but countable) set.
>
> The corresponding states would then be parametrized by a suitably
> selected maximal Abelian ideal. There's zillions of them running
> around in this algebra, as there were for H_n. For instance, for
> H_3 you had the subalgebras generated by the following sets:
> {q^1,q^2,q^3}
> {p_1,p_2,p_3}
> {p_1,q^2,p_3}
> {q^1,q^2 cos(G) + p_2 k sin(G), q^3 cos(G) - p_3 k sin(G)}
> all of which are Abelian and maximal, and so can each be used as
> a parametrization of the state space.
>
> For H_n a maximal Abelian ideal is *infinite* dimensional (if
> the algebra above is), with, e.g.:
> {q^1,q^2,q^3,...}
> {p_1,p_2,p_3,...}
> being two cases in point.
>
> The state space would be parametrized by *all* the eigenvalues of
> the operators in the corresponding set; e.g.
> q^i |x1,x2,x3,...> = x^i |x1,x2,x3,...>; i=1,2,...

Thank you for your explanations!
Tobias
--
hang my head drown my fear
till you all just disappear

reverse my forename for mail! - saibot