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While dealing with General Sigma models (See e.g. Ref. 1)

$$\tag{10.67} S ~=~ \frac{1}{2}\int \! dt ~g_{ij}(X) \dot{X^i} \dot{X^j}, $$

where the Riemann metric can be expanded as,

$$\tag{10.68} g_{ij}(X) ~=~ \delta_{ij} + C_{ijkl}X^kX^l+ \ldots $$ The Hamiltonian is given by,

$$ H ~=~ \frac{1}{2} g^{ij}(X) P_{i}P_{j}.$$

The authors say that in quantum theory the above expression is ambiguous, because $X$ and $P$ don't commute. Hence there are many nonequivalent quantum choices for $H$ reduces to the same classical object. I am not able to figure this out.

Also this Hamiltonian is related to Laplacian, which I am not able to understand, why ? This Hamiltonian can be related to Laplacian if $g^{ij}$ is the usual $\eta^{ij}$. Do the authors want to say that in some atlas we can always find a local coordinates which reduces to $\eta^{ij}$ or is there a general definition of Laplacian which I am unaware of?


  1. K. Hori, S. Katz, A. Klemm, R. Pandharipande, R. Thomas, C. Vafa, R. Vakil, and E. Zaslow, Mirror Symmetry, 2003, chapter 10, eqs. 10.67-10.68. The pdf file is available here or here.
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The operator is clearly "equal" to a Laplacian only if the metric $g$ is flat and positively definite. Otherwise it's just similar, that's why they say it's "related". Also, there are ordering ambiguities because $g^{ij}$ are functions of $X$ which don't commute with $P$. Does it answer all your questions? – Luboš Motl Sep 27 '12 at 15:33
@LubošMotl : Ok I get the "related" part now. Thanks for that interpretation, but is there an example of showing that two different definitions of Hamiltonian leads to same classical Hamiltonian. May be considering different definitions of momenta leading to same classical Hamiltonian. Earlier in the text, conjugate momentum was defined as $P_i = \dfrac{\delta S}{\delta \dot{X}^i} = g_{ij} \partial_t X^j $ – Jaswin Sep 27 '12 at 17:17
Dear Jaswin, differently ordered products of operators (those that exist classically) always differ by terms proportional to $\hbar$ or its positive powers, so in the classical $\hbar\to 0$ limit, they're the same. I would have to go to higher, 5th order polynomials for a good example. – Luboš Motl Sep 28 '12 at 4:28
@LubošMotl : Yes, now I could figure it out, probably he meant that $\dfrac{1}{2} g^{ij} P_iP_j \neq \dfrac{1}{2} P_i P_j g^{ij} $ quantum mechanically, but classically it is true. – Jaswin Sep 28 '12 at 5:23
@Qmechanic : I was reading from $Mirror$ $Symmetry$, chapter 10, equation 10.68. It is available freely in library. – Jaswin Sep 28 '12 at 5:24
up vote 2 down vote accepted

Well, the metric on the target space (not to be confused with the spacetime metric) $g_{ij}$ looks like $$g_{ij}\sim \delta_{ij}+(C_{ijkl}X^{k}X^{l})+\mathcal{O}(X^{4}).$$ We can invert this, obtaining ("for small $X$") $$g^{ij}\sim \delta^{ij}-{D^{ij}}_{kl}X^{k}X^{l}+\mathcal{O}(X^{4})$$ where ${D^{ij}}_{kl}$ are "some coefficients" we could figure out if forced to.

Really, to prove operator ordering ambiguity in the Hamiltonian, you just have to show that $$H\approx g^{ij}P_{i}P_{j} = \delta^{ij}P_{i}P_{j}-{D^{ij}}_{kl}X^{k}X^{l}P_{i}P_{k}+\mathcal{O}(X^{4}P^{2})$$ has ambiguities when quantized.

How? Well, consider the simpler case of a one-dimensional particle. We see that the Poisson brackets satisfy $$\tag{1}p^{2}x^{2}=(px)^{2}=\{x^{3},p^{3}\}-p^{2}x\{x^{2},p\}-\{x^{3},p\}p^{2}.$$ Woah, how did we get this equality? Well we use the property $$ \{fg,h\}=f\{g,h\}+g\{f,h\},\quad\mbox{and}\quad\{f,gh\}=g\{f,h\}+h\{f,g\}.$$ Then we consider $\{x^{3},p^{3}\}$ and do some algebra. But when (1) is quantized, these equalities fails badly. It's unclear (or ambiguous) what's important to quantize, and how to do it.

In other words, if we have quantization as a map $$Q:\mathrm{classical}\to\mathrm{quantum}$$ satisfying:

  1. quantization "puts hats" on position and momentum: $Q(x)=\widehat{x}$ and $Q(p)=\widehat{p}$, and are "represented irreducibly" (this is a technical condition, don't worry too much about it!);
  2. $Q$ is linear, so $Q(c_{1}f+c_{2}g)=c_{1}Q(f)+c_{2}Q(g)$ where $f,g$ are functions of momentum and position;
  3. Poisson brackets become $\displaystyle Q(\{f,g\})=\frac{1}{\mathrm{i}\hbar}[Q(f),Q(g)]$;
  4. The number 1 is mapped to the identity operator $Q(1)=\mathrm{id}$.

We have problems trying to evaluate $Q(x^{2}p^{2})$. Do we have $$Q(x^{2}p^{2})\stackrel{??}{=}Q(x)^{2}Q(p)^{2}\stackrel{??}{=}Q(xp)^{2}?$$ What happens to equation (1)? It's ambiguous :(

For more on operator ordering ambiguities, see S. Twareque Ali, Miroslav Engliš "Quantization Methods: A Guide for Physicists and Analysts" arXiv:math-ph/0405065.

Also this Hamiltonian is related to Laplacian, which I am not able to understand, why ?

When we work with a linear sigma model, we have $g_{ij}=\delta_{ij}$ and we recover the usual Hamiltonian as the Laplacian (up to some constant).

This can be seen from the formula, and noting in this particular case $g^{ij}=\delta^{ij}$ so we find $$ H = \frac{1}{2}\delta^{ij}P_{i}P_{j} = \frac{1}{2}P^{i}P_{i}$$ Again, up to some constant. (See equation (10.70) of the book you're reading, and you find $P_{i}=\mathrm{i}\partial/\partial X^{i}$)

And again do not confuse the "target space metric" $g_{ij}$ with the "spacetime metric" which I think you denote by $\eta_{ij}$ (later on in the book, I think the authors use $h_{ij}$ for the "spacetime metric").

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Thanks a lot Nelson – Jaswin Oct 1 '12 at 14:25

The authors say that in quantum theory the above expression is ambiguous, because X and P don't commute. Hence there are many nonequivalent quantum choices for H reduces to the same classical object. I am not able to figure this out.

If you review the text, you'll see the authors have provided the commutation relation in equation 10.70 as:

$$[X^i,P_j] = i\delta_j^i$$

which tells you X and P are non-commutative and the commutation operation produce an imaginary number (and this equation might be understood as $\dfrac{1}{X}P-P\dfrac{1}{X} = i$).

The equation you reference is actually written as:

$$H = \dfrac{1}{2}g^{ij}(X)P_iP_j$$

The position variable X is important since momentum is classically understood as $mass \times velocity$ so $p^2 = m^2v^2$and kinetic energy is $\dfrac{1}{2}mv^2$. So the $g^{ij}(X)$ is taking place of an inverse mass term in order to stay consistent with the classically defined energy equation.

The problem arises if someone is trying to solve for an unknown value in the equation. Imagine that you know H and X and want to solve for P, you might get some answer for P and assume everything is hunky dory, however, if for some reason you have the value of P you just calculated as well as the value of H, and then try to solve for X, you run into problems, because the X and P are non-commutative, you will not get the same value of X as you started out with, and as defined you have to include an imaginary component.

This is what is meant by ambiguity, knowing two components of the equation will not give you a definite value for the third, only a range of values.

As far as the relationship to the Laplacian, if one looks at the Schrodinger equation one can see that the kinetic energy operator term is:


so if one compares this to the equation you reference, it should be clear that the momentum symbols (P) are taking the place of the Laplacian operator.

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