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  quantum mechanics current operators

+ 4 like - 0 dislike
1188 views

How to derive the charge current and the energy current operators in second quantized form in Quantum mechanics ? Also if you could comment in a similar way on the entropy current operator, that will be nice.

The way I am proceeding is simply taking the time derivative of the number operator and the Hamiltonian operator. What is wrong in this approach and what is a more appropriate approach ?

Consider a non-relativistic fermionic/bosonic Hamiltonian given in second-quantized form. For example, we can take the Hubbard Hamiltonian. But please start with a general Hamiltonian H to explain the process in general.

This post imported from StackExchange Physics at 2015-06-15 19:45 (UTC), posted by SE-user cleanplay
asked Jul 5, 2013 in Theoretical Physics by cleanplay (80 points) [ no revision ]
Would you mind including some physical context? About which charge and energy currents are you asking? Is this a non-relativistic setup about which you're asking?

This post imported from StackExchange Physics at 2015-06-15 19:45 (UTC), posted by SE-user joshphysics
@joshphysics I have edited the question to make it more clear.

This post imported from StackExchange Physics at 2015-06-15 19:45 (UTC), posted by SE-user cleanplay

1 Answer

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Whether you use second quantization formalism or whether you are even talking about classical or quantum systems current is defined via the continuity equation for some quantity, $\hat{O}$,

$$\frac{\partial \hat{O}}{\partial t} + \nabla \cdot \mathbf{\hat{J}} = 0,$$

where I have used hat to denote we are talking about quantum mechanical observables.

The question is can we find a pair of observables for which the above equation holds. For integrable models, such as the Hubbard model, Heisenberg spin chain model, free fermions the answer is yes. We can identify local conserved charges for which the above equation holds.

Now, in the Heisenberg picture we have, $$\frac{d}{dt}\hat{O}(t)=\frac{i}{\hbar}[H,\hat{O}(t)]$$

So if you have some Hamiltonian and some corresponding local conserved charge you compute it's commutator with the Hamiltonian and use that to find the current operator. For instance, for the Hubbard model,

$$ H = -t \sum_{\langle i,j \rangle,\sigma}( c^{\dagger}_{i,\sigma} c^{}_{j,\sigma}+ h.c.) + U \sum_{i=1}^{N} n_{i\uparrow} n_{i\downarrow}$$

the number density current at site $i$, $n_i=c^\dagger_i c_i$ can be easily found by a discretized version of the continuity equation, $$\frac{i}{\hbar}[H,n_i(t)]=- (J_{i+1}-J_{i}) = -t(-i c^\dagger_{i+1} c_i+ h.c.)+t(-i c^\dagger_{i} c_{i-1}+ h.c.),$$ which allows us to identify, $$J_i=t(-i c^\dagger_{i} c_{i-1}+ h.c.) $$ as the particle number current density at site $i$. Please note that the total (sum for all sites) particle number density commutes with the Hamiltonian, assuming periodic boundary conditions. This is very important for integrability.

For more examples (e.g., energy current in the Hubbard model), see this paper: http://arxiv.org/abs/cond-mat/9611007

This post imported from StackExchange Physics at 2015-06-15 19:45 (UTC), posted by SE-user Bubble
answered Jul 13, 2013 by Bubble (210 points) [ no revision ]
Nice answer, thanks for the very useful paper. I found equivalent approach which defines the current in terms of discretized form of polarization.

This post imported from StackExchange Physics at 2015-06-15 19:45 (UTC), posted by SE-user cleanplay

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