# Separability of a Hamiltonian with spin

• I
• Salmone
In summary, the Hamiltonian ##\hat{H}=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2+\frac{A}{\hbar^2}(J^2-L^2-S^2)## is separable into two parts ##H_1=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2## and ##H_2=\frac{A}{\hbar^2}(J^2-L^2-S^2)## and ##[H_1,H_2]=0##, where A is a constant. This is because the commutator of ##
Salmone
I'd like to know if this Hamiltonian ##\hat{H}=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2+\frac{A}{\hbar^2}(J^2-L^2-S^2)## is separable into two parts ##H_1=\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2## and ##H_2=\frac{A}{\hbar^2}(J^2-L^2-S^2)## and ##[H_1,H_2]=0##. Here A is a constant. I did so:

##[H_1,H_2]=[\frac{p^2}{2m}+\frac{1}{2}m\omega^2r^2,\frac{A}{\hbar^2}(J^2-L^2-S^2)]=##
##=[\frac{p^2}{2m},\frac{A}{\hbar^2}(J^2-L^2-S^2)]+[\frac{1}{2}m\omega^2r^2,\frac{A}{\hbar^2}(J^2-L^2-S^2)]=##
##=\frac{A}{2m \hbar^2}([p^2,J^2]-[p^2,L^2]-[p^2,S^2])+\frac{a}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])=## ##=\frac{A}{2m \hbar^2}([p^2,L^2]+[p^2,S^2]+2[p^2,L_xS_x]+2[p^2,L_yS_y]+2[p^2,L_zS_z]-[p^2,L_x^2]-[p^2,L_y^2]-[p^2,L_z^2]-## ##-[p^2,S^2])+\frac{A}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])=\frac{A}{2m \hbar^2}([p^2,2L_xS_x]+[p^2,2L_yS_y]+[p^2,2L_zS_z])+\frac{A}{2 \hbar^2}([r^2,J^2]-[r^2,L^2]-[r^2,S^2])## now, since ##p^2## is a scalar and it's a function of spatial coordinates, it commutes with the component ##L_i## of the angular moment and with Spin operators and the same can be said about ##r^2## so the first commutator is ##0## and the hamiltonian is separable in ##H_1+H_2##. Am I right?

Last edited:
vanhees71
That's right!

Salmone and gentzen

## 1. What is the concept of separability in a Hamiltonian with spin?

Separability in a Hamiltonian with spin refers to the ability to separate the spin degrees of freedom from the other degrees of freedom in the system. This means that the Hamiltonian can be written as a sum of two separate Hamiltonians, one for the spin and one for the other degrees of freedom.

## 2. Why is separability important in Hamiltonian systems?

Separability is important because it allows us to study the spin dynamics independently from the other dynamics in the system. This can simplify the analysis and make it easier to understand the behavior of the system.

## 3. How is the separability of a Hamiltonian with spin determined?

The separability of a Hamiltonian with spin can be determined through the use of mathematical techniques such as diagonalization and transformation to a suitable basis. If the Hamiltonian can be written as a sum of two separate Hamiltonians, then it is considered separable.

## 4. What are the implications of a non-separable Hamiltonian with spin?

A non-separable Hamiltonian with spin means that the spin degrees of freedom are coupled to the other degrees of freedom in the system. This can make the analysis and understanding of the system more complex, as the dynamics of the spin are now influenced by the other dynamics in the system.

## 5. Can a non-separable Hamiltonian with spin be simplified or transformed into a separable form?

In some cases, a non-separable Hamiltonian with spin can be transformed into a separable form through the use of mathematical techniques. However, this is not always possible and depends on the specific system and Hamiltonian. In general, it is desirable to have a separable Hamiltonian with spin for easier analysis and understanding.

• Quantum Physics
Replies
1
Views
709
• Quantum Physics
Replies
21
Views
2K
• Quantum Physics
Replies
9
Views
443
• Quantum Physics
Replies
2
Views
1K
• Quantum Physics
Replies
4
Views
786
• Quantum Physics
Replies
4
Views
1K
• Quantum Physics
Replies
1
Views
602
• Quantum Physics
Replies
5
Views
480
• Quantum Physics
Replies
6
Views
1K
• Quantum Physics
Replies
3
Views
804