Translation operator on a sphere

Click For Summary
SUMMARY

The discussion focuses on deriving translation operators for an electron in the presence of a Dirac monopole's magnetic field and on a sphere. The translation operator is defined as &hat;T(a) = e^{-ia&hat;p}, with &hat;p = -i\nabla. It is established that in spherical coordinates, the momentum operator must be replaced by angular momentum &hat;L due to the constraints of the sphere. The covariant derivative &nabla \to &nabla - ieA is suggested for use in the presence of a field, leading to the conclusion that translation operators on a sphere can be treated similarly to those on a flat surface.

PREREQUISITES
  • Understanding of quantum mechanics, specifically operators and their algebra.
  • Familiarity with angular momentum operators and their properties.
  • Knowledge of covariant derivatives and gauge transformations.
  • Basic concepts of spherical coordinates and constraints in physics.
NEXT STEPS
  • Research the application of covariant derivatives in quantum mechanics.
  • Study the properties and algebra of angular momentum operators.
  • Explore the concept of Landau levels and their physical implications.
  • Investigate constraint quantization methods in quantum systems.
USEFUL FOR

Physicists, particularly those specializing in quantum mechanics, gauge theories, and geometric approaches to particle dynamics on curved spaces.

kashokjayaram
Messages
15
Reaction score
0
I'm considering a system where an electron is subjected to magnetic field which is produced by dirac monopole. Here I'm interested in looking for a translation operator. Now how can I get a translation operator in presence of field and in absence of field.?? I need both the operators. Can anybody help me..? Thank you...
 
Physics news on Phys.org
The translation operator is

\hat{T}(a) = e^{-ia\hat{p}}

In position-space this is

\hat{p} = -i\nabla

Are you thinking about using the covariant derivative instead?

\nabla \to \nabla - ieA
 
is this applicable to sphere also..?? Why both are same...? Since this operator is used for translation in plane, how come it is same in sphere?
 
You haven't said anything about a particle on a sphere. Is your particle subject to a constraint r2=const.? Or equivalently, are you using spherical coordinates with r being a constant?

Then the flat-space translation operator is no longer valid. On the sphere only operators respecting the symmetry of the sphere (and the constraint) are allowed. A simple approach is to use spherical coordinates, to fix r=const. and to drop all ∂r terms. I am not absolutely sure whether this approach is valid for all operators. Perhaps an approach like constraint quantization is required.

But in principle it's obvious that instead of momentum p as a generator of translation you have to consider angular momentum L as a generator of rotation.
 
tom.stoer said:
You haven't said anything about a particle on a sphere.
Have a look at the title.

The problem is that with a constant field, the hamiltonian is not directly translation invariant. But the translated hamiltonian can be brought back to its original form applying an additional gauge transformation. Effectively, the translation operator becomes the covariant derivative. Physically, this leads to the appearance of Landau levels.
On a sphere, the translation operators are the same as that for a flat surface tangential to the point you want to rotate. So it should be possible to use a covariant version of the angular momentum operators.
 
Ah, I see, it's in the title of the thread ;-)

What you have to do is to use

\hat{D}(n,\phi) = e^{-i\,n^a\,\hat{L}^a\,\phi}

which generates rotations with angle phi and axis n; n is the unit vector in direction n.

Note that formally L is an angular momentum operator, but not necessarily an orbital angular momentum operator. That means it satisfies the algebra

[\hat{L}^a,\hat{L}^b] = i\epsilon^{abc}\,\hat{L}^c

but it need not be represented using r, theta and phi. Of course it has the usual eigenstates ##|lm\rangle##.

This allows you treat the particle on a sphere as a rigid rotor.

But I am not sure whether this helps for your problem with a external field.
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

Similar threads

  • · Replies 1 ·
Replies
1
Views
662
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 10 ·
Replies
10
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
  • · Replies 15 ·
Replies
15
Views
820
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 31 ·
2
Replies
31
Views
3K