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This may be obvious but I have limited experience in physics , The generators of Spatial translation symmetry commutes with each other i.e [P(i),P(j)] = 0 but if Spacetime is a curved manifolds then the value of the commutator should not be zero but some invariant property related to curvature i.e a Function of the curvature tensor If this is false then what should the commutator be like e.g in the vicinity of a gravitational source according to GR , I'm sorry I do not know much in relativity nor differential geometry

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The momentum of a particle is a global construction, it is defined for wavefunctions extending over the whole manifold. Most manifolds don't have symmetry generators corresponding to translations, but highly symmetric spaces, like dS and AdS do. In these case, you can replace the commutators with the dS version, which includes the cosmological constant curvature contribution. You might have a local relation if you can make a scaling limit, i.e. only for classical particle motion, where you would give the Poisson bracket in terms of the curvature tensor, but this wouldn't work in QM. –  Ron Maimon Feb 29 '12 at 15:56
    
In dS spacetime in units in which the de Sitter radius is unity, an example momentum commutation relation is $[P_{x},P_{y}]=iJ_{z}$. –  Stephen Blake Feb 29 '12 at 21:05
    
Wouldn't the momentum commutation relations be $[\nabla_\alpha ,\nabla_\beta ]$ with $\nabla_\gamma$ the covariant derivative? With this acting on a scalar field you get back the normal relations, and with it acting on a vector the curvature tensor shows up? –  kηives May 16 '12 at 17:22

2 Answers 2

The commutator you are interested in is non-trivial if you generalize the translations to curvilinear coordinates. For a vector function $A^{\alpha }\left( x\right) $, a «translation» along $dx^{\mu}$ is the following transformation: $$ A^{\alpha}\rightarrow A^{\alpha}-dx^{\mu}D_{\mu}A^{\alpha} $$ (where $D_{\mu}$ is the so called covariant derivative) instead of $A^{\alpha }\rightarrow A^{\alpha}-dx^{\mu}\partial_{\mu}A^{\alpha}$.

First of all, you should understand one basic fact from the group theory: a commutator of generators corresponds to a generator of some transformation. This transformation is a superposition of transformations which form a infinitesimal closed contour in the parametric space of a group. It sounds complicated but the idea is very simple. Let's imagine you have a group element: $$ T\left( \mathbf{a}\right) =\exp\left( i\mathbf{g}\cdot\mathbf{a}\right) =1+i\mathbf{g}\cdot\mathbf{a+}\frac{1}{2}\left( i\mathbf{\mathbf{g} \cdot\mathbf{a}}\right) ^{2}+\mathbf{\ldots,} $$ where $a^{n}$ are parameters of the group, $g_{n}$ are generators of the group and $\mathbf{g}\cdot\mathbf{a}=g_{n}a^{n}$. Let's now consider the following sequence of transformations: $T\left( \mathbf{a}\right) $ then $T\left( \mathbf{b}\right) $, so that $\mathbf{b}\neq\mathbf{a}$, then $T\left( -\mathbf{a}\right) $, \ so that $T\left( -\mathbf{a}\right) T\left( \mathbf{a}\right) =1,$ and finally $T\left( -\mathbf{b}\right) $. The parameters of these transformations form a rectangle in the group parameter space (see the picture below). Therefore, the total composite transformation has the form: \begin{align*} & T\left( -\mathbf{b}\right) T\left( -\mathbf{a}\right) T\left( \mathbf{b}\right) T\left( \mathbf{a}\right) =\\ & = \left( 1-i\mathbf{g} \cdot\mathbf{b+\ldots}\right) \left( 1-i\mathbf{g}\cdot\mathbf{a+\ldots }\right) \left( 1+i\mathbf{g}\cdot\mathbf{b+\ldots}\right) \left( 1+i\mathbf{g}\cdot\mathbf{a+\ldots}\right) . \end{align*} Let's now assume that $\mathbf{a}$ and $\mathbf{b}$ are infinitesimal small, so that the expansion of the composite transformation has the form: \begin{align*} T\left( -\mathbf{b}\right) T\left( -\mathbf{a}\right) T\left( \mathbf{b}\right) T\left( \mathbf{a}\right) & \approx1+\left( \mathbf{g}\cdot\mathbf{a}\right) \left( \mathbf{g}\cdot\mathbf{b}\right) -\left( \mathbf{g}\cdot\mathbf{b}\right) \left( \mathbf{g}\cdot \mathbf{a}\right) \\ & =1+\left[ g_{m},g_{n}\right] a^{m}b^{n}=1+\frac{1}{2}\left[ g_{m} ,g_{n}\right] f^{mn},\qquad(1) \end{align*} where $$ \left[ g_{m},g_{n}\right] =g_{m}g_{n}-g_{n}g_{m} $$ and $f^{mn}$ is the so called oriented area element (or directed area measure): $$ f^{mn}=a^{m}b^{n}-a^{n}b^{m}. $$ For example, if the parameter space of the group is three dimensional (as it is for 3D translations) then the vector $$ s^{k}=\frac{1}{2}\epsilon^{kmn}f^{mn}=\left[ \mathbf{a}\times\mathbf{b} \right] ^{k} $$ is transverse to $\mathbf{a}$ and $\mathbf{b}$, so that its length squared is the area of the rectangle with the sides $\mathbf{a}$ and $\mathbf{b}$ (see the figure below): $$ s^{2}=\mathbf{a}^{2}\mathbf{b}^{2}-\left( \mathbf{a}\cdot\mathbf{b}\right) ^{2}. $$

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For a flat space, the parallel translation of a vector $\mathbf{v}$ along a vector $\mathbf{a}$ doesn't change the direction of the vector $\mathbf{v}$. Therefore, the parallel translation around the infinitesimal closed contour $(\mathbf{a},\mathbf{b},-\mathbf{a},-\mathbf{b})$ equals to the identical transformation, hence from the equation (1) we obtain $\left[ p_{i} ,p_{j}\right] =0$. For curved space, the parallel displacement along a small 4-vector $dx^{\nu}$ is non-trivial: $$ \delta A^{\alpha}=-\Gamma_{\mu\nu}^{\alpha}A^{\mu}dx^{\nu}, $$ where $\Gamma_{\mu\nu}^{\alpha}$ are the so called Christoffel symbols. Therefore the parallel translation around an infinitesimal closed contour $C$ is the contour integral: $$ \Delta A^{\alpha}=- {\displaystyle\oint\limits_{C}} \Gamma_{\mu\nu}^{\alpha}A^{\mu}dx^{\nu}. $$ Applying Stokes' theorem to this integral and assuming that the area enclosed by the contour $C$ has the infinitesimal small value $\Delta f^{\mu\nu}$, one can show that $$ \Delta A^{\alpha}=-\frac{1}{2}R^{\alpha}{}_{\beta\mu\nu}A^{\beta}\Delta f^{\mu\nu}, $$ where $$ R^{\alpha}{}_{\beta\mu\nu}=\partial_{\mu}\Gamma_{\beta\nu}^{\alpha} -\partial_{\nu}\Gamma_{\beta\mu}^{\alpha}+\Gamma_{\mu\rho}^{\alpha} \Gamma_{\beta\nu}^{\rho}-\Gamma_{\nu\rho}^{\alpha}\Gamma_{\beta\mu}^{\rho}, $$ is the well known Riemann tensor. Now you should remember that the transformation around an infinitesimal closed contour is a commutator, see (1). Hence for a vector function $A^{\alpha}\left( x\right) $ we have: $$ \left[ D_{\mu},D_{\nu}\right] A^{\alpha}=R^{\alpha}{}_{\beta\mu\nu}A^{\beta }. $$

Therefore you are right saying that the commutator of translations in a curved space is non zero, it is in fact Riemann tensor. Although the form of the covariant derivative depends on the geometric type of the field it acts on, e.g., for a co-variant tensor field: $$ \left[ D_{\mu},D_{\nu}\right] A_{\alpha\beta}=A_{\alpha\rho}R^{\rho} {}_{\beta\nu\mu}+A_{\rho\beta}R^{\rho}{}_{\alpha\nu\mu}. $$

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Before discussing commutators in a curved space-time, you should specify rigorously the quantum framework you are going to use. In particular, you should specify what a quantum state is, how exactly do you construct a Hilbert space, and how exactly do you define operators.

As a possible way to do this, you may introduce a 3+1 decomposition of your space-time, define a coordinate basis for your Hilbert space in this composition, and define 3-momenta operators as operators of infinitesimal translations, canonically conjugated to the operators of basis, and commuting with themselves by definition, that is $[\hat p_i,\hat p_j]=0$. Note, that with such a difinition $\hat p_i\neq\dfrac{\hbar}{i}\dfrac{\partial}{\partial x_i}$.

However, as there is no universal way so far to construct a classical quantum theory on a relativistic background, and one may come up with some other construction. In any case, the relation $[\hat p_i,\hat p_j]=0$ is 1) achievable, 2) desirable. Desirable - becuase you would actually want to work with momenta defined in such a way that you can measure their components independantly.

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This is not the question--- it isn't GR. How do you formulate the quantum theory of a particle restricted to a sphere. This can be done completely precisely, and it can be realized experimentally in surface states of electrons. The question is if you define the translation operators on the sphere as rotations in two perpendicular directions, then the commutator of $p_i$ and $p_j$ is nonzero. The general case is more difficult, because there is no translation symmetry. Still one can say certain things about the asymptotic p operators and their asymptotic commutation relations. –  Ron Maimon Apr 17 '12 at 0:12
    
Infinitely small rotations commute, so there is no problem here. Another point, which you probably implied, is that for a sphere there must be a point where the operator is divergent (equivalent to saying that a continuous vector field should have a zero value at some point). In itself it is not a problem either, as the operator proper values are not observables in GR, as only scalars can be measured. –  Alexey Bobrick Apr 17 '12 at 21:08
    
Except that the generator is defined by taking out the leading infintesimal order, so that they wouldn't commute. On a sphere, $[p_x,p_y]=L$ where p_x is one rotation, p_y is another rotation, and L a third. This only becomes zero when scaling to act on states of extremely high momentum, that is, at very small distances, which restores commutativity of momenta. Your comments about vector fields with zeros is not apropos--- the generators are defined as those which generate the motion corresponding to the rotations, and the two zeros are irrelevant. –  Ron Maimon Apr 18 '12 at 3:13
    
I was wrong about rotations. But: You do not have to define $\hat p_\mu$ through rotations. As I said, you may generalize the definition of momenta to them being the generators of displacements along any vector field you like. –  Alexey Bobrick Apr 18 '12 at 19:35

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