Natural Eigenbasis composite 2 X Spin 1/2 system

In summary, the way I am being taught in my course is that instead of s=1/2, m_1=1/2; s_2=1/2, m_2=1/2 > = !+ + >, m_1=-1/2; s_2=1/2, m_2=-1/2 > = !- + >, m_1=0; s_2=0,m_2=0 > = !+ + >.
  • #1
cathalcummins
46
0
The way I am being taught in my course is that instead of


!s_1= 1/2, m_1 =1/2 ; s_2= 1/2, m_2 =1/2 > = !+ + >
!s_1= 1/2, m_1 =1/2 ; s_2= 1/2, m_2 =-1/2 > = !+ - >
!s_1= 1/2, m_1 =-1/2 ; s_2= 1/2, m_2 =1/2 > = !- + >
!s_1= 1/2, m_1 =-1/2 ; s_2= 1/2, m_2 =-1/2 > = !- - >

We have


¦s=1,m_z=1>= ¦++>
¦s=1,m_z=0>= 1/sqrt(2) (¦+->+¦-+>)
¦s=1,m_z=-1>= ¦++>

¦s=0,m_z=0>= 1/sqrt(2) (¦+->-¦-+>)

I understand the spin-1 triplet. My question is "how do you compute ¦+->-¦-+> to get composite s=0,m=0". Is it essentially

1/sqrt(2) (¦1 0> - ¦1 0> )= ¦0 0>


?

Thanks.
 
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  • #2
The idea is that you derive this new basis by requiring that it be eigenstates of total angular momentum. However, if you are just given the states, you should still be able to show that it is an eigenstate.

For example, show that the state is an eigenstate of [tex]S_1^z + S_2^z[/tex] and also [tex]S_1^2 + S_2^2[/tex], and what are the eigenvalues?
 
  • #3
The spin zero state is chosen to be orthogonal to the spin one state.
 
  • #4
Looks like I misunderstood the question. Orthogonality is usually the way to go but you can do the Clebsch-Gordan decomposition more systematically I believe.
 
  • #5
Of course! orthogonality, thank you Pam. When reading David J Griffiths Introduction to quantum mechanics - (Prentice Hall, 1995) p166, the way it is presented almost seems to insinuate a "nice consequence" as opposed to a necessary condition.

[tex]
\left( S_{1z} \otimes \mathbb{I} + \mathbb{I} \otimes S_{2z} \right)\frac{1}{\sqrt{2}}\{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\}=\frac{\hbar}{2\sqrt{2}} \left( \ket{\uparrow\downarrow}+\ket{\downarrow\uparrow}-\ket{\uparrow\downarrow}-\ket{\downarrow\uparrow} \right)[/tex]

So that

[tex]
\left( S_{1z} \otimes \mathbb{I} + \mathbb{I} \otimes S_{2z} \right)\frac{1}{\sqrt{2}}\{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\}=\left(0 \right)\frac{1}{\sqrt{2}} \left( \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\right)
[/tex]

Meaning that

[tex]
\{ \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow}\}
[/tex]

is an eigenvector of the [tex]S_z=\left{ S_{1z} \otimes \mathbb{I} + \mathbb{I} \otimes S_{2z}\right}[/tex] operator with corresponding eigenvalue [tex]0[/tex]. Similarly, but with more algebra, we should arrive with the corresponding eigenvalue [tex]s(s+1)\hbar^2=0(0+1)\hbar^2=0[/tex] for the [tex]S^2[/tex] operator.

Technically, the zero spin state is orthogonal to every other eigenvector yes?
 
Last edited:
  • #6
All non-degenerate evectors are orthogonal.
 
  • #7
To motivate the Clebsch-Gordon, and prove [tex]m=m_1+m_2[/tex] my lecturer done the obvious exploitation of the fact:

[tex]
J_z=J_{1z}+J_{2z}
[/tex]

so that the expectation value of [tex]J_z-J_{1z}-J_{2z}[/tex] is zero. That makes sense to a certain extent. Though, and here is the problem: the operators above do not posess simultaneous eigenkets. But, regardless, he goes on to calculate it wrt two separate eigenbases; [tex]| \phi \rangle[/tex] and [tex]| \psi \rangle[/tex] where

[tex]
| \phi \rangle = | j_1 j_2 m_1 m_2\rangle
[/tex]

and

[tex]
| \psi \rangle = | j_1 j_2 j m \rangle
[/tex]

This formalism is that of Modern Quantum Mechanics - J.J Sakurai (Addison-Wesley Rev .ed, 1994). The Expectation value is found by (according to my lecturers notes)

[tex]
\langle \phi | J_z-J_{1z}-J_{2z} | \psi \rangle
[/tex]

which must return zero. Okay I'm going to stop there because I simply don't agree that this is the expectation value.

You can still prove [tex]m=m_1+m_2[/tex] by the elementary steps outlined in Modern Quantum Mechanics - J.J Sakurai (Addison-Wesley Rev .ed, 1994) p 208 . This way makes sense to me and it mentions nothing of expectation value. Sorry to post such a trivial question but I'm just trying to weed out any mistakes in my lecturers notes.
 
Last edited:

1. What is a Natural Eigenbasis composite 2 X Spin 1/2 system?

A Natural Eigenbasis composite 2 X Spin 1/2 system is a quantum mechanical system made up of two particles, each with a spin of 1/2. The system has a total of four basis states, which are the combinations of the individual spin states of the two particles. These basis states are referred to as the "natural eigenbasis" because they are the eigenvectors of the system's Hamiltonian operator.

2. How is the Natural Eigenbasis determined for a composite 2 X Spin 1/2 system?

The Natural Eigenbasis for a composite 2 X Spin 1/2 system is determined by finding the eigenvectors of the system's Hamiltonian operator. The Hamiltonian operator describes the total energy of the system and its eigenvectors represent the possible states the system can occupy.

3. What are the physical implications of a composite 2 X Spin 1/2 system?

A composite 2 X Spin 1/2 system has physical implications in various fields, such as quantum computing and quantum information processing. It also has implications in the study of entanglement, where the states of the two particles are correlated and cannot be described independently.

4. How is a composite 2 X Spin 1/2 system different from a single particle spin system?

A composite 2 X Spin 1/2 system is different from a single particle spin system because it involves two particles, each with a spin of 1/2, instead of just one particle. This means that the composite system has a larger set of basis states and a more complex Hamiltonian operator.

5. What are the applications of a Natural Eigenbasis composite 2 X Spin 1/2 system?

The Natural Eigenbasis composite 2 X Spin 1/2 system has applications in quantum information processing and quantum computing, where it can be used to perform operations on quantum bits (qubits). It also has applications in quantum cryptography and quantum communication, where entanglement of the two particles can be used for secure communication.

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