How Is the Particle Density Operator Defined?

Click For Summary
SUMMARY

The particle density operator, denoted as Ï(r), is defined through the wavefunction's modulus squared, represented as |ψ(r)|². The discussion highlights a reformulation where |ψ(r)|² is expressed as an integral involving the delta function: |ψ(r)|² = ∫dr' ψ*(r')δ(r-r')ψ(r'). This formulation is crucial for understanding the expectation value of the density operator, , which is derived from the density matrix rather than a simple average value. The confusion arises from the distinction between the density matrix elements and the delta function representation.

PREREQUISITES
  • Understanding of quantum mechanics and wavefunctions
  • Familiarity with the concept of density operators
  • Knowledge of delta functions in mathematical physics
  • Experience with expectation values in quantum mechanics
NEXT STEPS
  • Study the derivation of the density matrix in quantum mechanics
  • Explore the role of delta functions in quantum field theory
  • Learn about the implications of particle density operators in many-body physics
  • Investigate the relationship between wavefunctions and density operators in quantum statistical mechanics
USEFUL FOR

Quantum physicists, graduate students in physics, and researchers focusing on quantum mechanics and many-body theory will benefit from this discussion.

aaaa202
Messages
1,144
Reaction score
2
There is something I do not understand. One way to define the current density operator is through the particle density operator Ï(r). From the fundamental interpretation of the wavefunction we have:

Ï(r)= lψ(r)l2

But my book takes this a step further by rewriting the equality above:

lψ(r)l2 = ∫dr' ψ*(r')δ(r-r')ψ(r')

And thus identifies the particle density operator as the delta function above. How does this make sense in any way?
 
Physics news on Phys.org
We have <r|Ï|r> = ∫dr' ψ*(r')δ(r-r')ψ(r')
where do you see a possible identification?
|ψ(r)|² is found on the diagonal of rho not on delta.
 
In general, we write expectation of operator A as ##\displaystyle \langle \psi |A|\psi \rangle = \int \psi^*(r') A \psi(r') dr'##. Substituting ##A = \delta(r-r')## gives you the correct expression for expectation of ##\rho(r)##. So what's the problem?
 
Yes but
<r|ψ><ψ|r'> = <r|Ï|r'> is not identified to δ(r-r') like aaa202 said.
it is not here an average value but an element of a density matrix.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
0
Views
795
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
4K