Quantum Mechanics - question about spin

In summary, the matrix representation for an operator with respect to a basis is found by writing its own eigenkets as suitable linear combinations of the eigenkets of the operator.
  • #1
touqra
287
0
Given [tex]S=\frac{1}{2}\hbar{\sigma} [/tex] where [tex]\sigma = \left(\left(\begin{array}{cc}0&1\\1&0\end{array}\right),\left(\begin{array}{cc}0&-i\\i&0\end{array}\right),\left(\begin{array}{cc}1&0\\0&-1\end{array}\right)\right) [/tex], show that
[tex] |+> = \left(\begin{array}{cc}1\\0\end{array}\right) [/tex] and [tex] |-> = \left(\begin{array}{cc}0\\1\end{array}\right) [/tex] are the eigenfunctions for [tex] S_z [/tex] . Obtain the matrix representation for [tex] S_y [/tex] and [tex] S_x [/tex] in the basis [tex] (|+>,|->)[/tex] .
 
Last edited:
Physics news on Phys.org
  • #2
[tex]S_{z} = \frac{1}{2}\hbar \sigma_z[/tex]

What happens when [tex]S_{z}[/tex] operates on [tex]\left| + \right>[/tex]?

How is a matrix representation (with respect to a basis) of an operator found? You've basically written down the final answer.

Regards,
George
 
  • #3
The first part is just an eigenvalue problem. From the matrix for Sz, find the eigenvalues (and eigenvectors) from the characteristic (and eigenvalue) equation.
 
  • #4
Gokul43201 said:
The first part is just an eigenvalue problem. From the matrix for Sz, find the eigenvalues (and eigenvectors) from the characteristic (and eigenvalue) equation.

The is the comprehensive way to do things, which is probably a good idea, but it not strictly necessary to do this in order to show that |+> and |-> are eignevectors of S_z.

Regards,
George
 
  • #5
Oh, they've already given you the eigenvectors ! Yes, that would be unnecessary then.
 
  • #6
Gokul43201 said:
The first part is just an eigenvalue problem. From the matrix for Sz, find the eigenvalues (and eigenvectors) from the characteristic (and eigenvalue) equation.
George Jones said:
[tex]S_{z} = \frac{1}{2}\hbar \sigma_z[/tex]
What happens when [tex]S_{z}[/tex] operates on [tex]\left| + \right>[/tex]?
How is a matrix representation (with respect to a basis) of an operator found? You've basically written down the final answer.
The original question was exactly the same as I've written here. Initially when I look at the question, I was dumbfounded.
The first part of the question, I would agree with Gokul43201 on finding the characteristic equation first.
How about the second part of the question? How do you find the matrix representation? I really have no idea.
 
Last edited:
  • #7
touqra said:
The original question was exactly the same as I've written here. Initially when I look at the question, I was dumbfounded.
I'm a little puzzled too...
The first part of the question, I would agree with Gokul43201 on finding the characteristic equation first.
As George said above, you can do it this way, if you wish, but it's not necessary. All you have to do is operate Sz on |+> and |-> and show what happens.

How about the second part of the question? How do you find the matrix representation? I really have no idea.
This is the weird part. What are you supposed to start from ? The matrix representation is what you're actually given.

The matrix form comes from the outer-product representation of those operators (by writing their own eigenkets as suitable linear combinations of the eigenkets of Sz).

Is this from a textbook ? If so, which one ?
 
  • #8
Gokul43201 said:
This is the weird part. What are you supposed to start from ? The matrix representation is what you're actually given.

The matrix form comes from the outer-product representation of those operators (by writing their own eigenkets as suitable linear combinations of the eigenkets of Sz).

Is this from a textbook ? If so, which one ?

It's a past year exam question. I wonder how the students did for this question. IMO, if the question was phrased in another way, like, prove that the matrix representation for ... is ... Then, it should be ok.
 
  • #9
If [tex]A[/tex] is a linear operator on an n-dimensional vector space [tex]V[/tex], then the matrix representation of [tex]A[/tex] with respect to the basis [tex]\{v_{1}, \dots , v_{n} \}[/tex] of [tex]V[/tex] is given by

[tex]
A v_{i} = \sum_{j} A_{ji} v_{j}.
[/tex].

In this case, [tex]V = \mathbb{R}^{2}[/tex], [tex]v_{1} = |+> = \left(\begin{array}{cc}1\\0\end{array}\right)[/tex] and [tex]v_{2} = |-> = \left(\begin{array}{cc}0\\1\end{array}\right)[/tex].

The case where [tex]v_{1} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc}1\\1\end{array}\right)[/tex] and [tex]v_{2} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc}-1\\1\end{array}\right)[/tex] is more interesting.

Regards,
George
 
Last edited:

Related to Quantum Mechanics - question about spin

1. What is spin in quantum mechanics?

Spin is a fundamental property of elementary particles in quantum mechanics. It is an intrinsic angular momentum that describes the rotation of a particle around its axis. It is a quantum mechanical property and cannot be fully understood in classical terms.

2. How is spin measured in quantum mechanics?

Spin is measured in units of angular momentum called "spin quantum number" and denoted by the symbol "s". It can have both positive and negative values, and the magnitude of the spin determines the type of particle it belongs to.

3. What are the implications of spin in quantum mechanics?

Spin has many implications in quantum mechanics, including determining the behavior of particles in magnetic fields, explaining the stability of atoms, and influencing the properties of materials. Spin also plays a crucial role in quantum computing and quantum information processing.

4. Can particles have fractional spin in quantum mechanics?

Yes, according to quantum mechanics, particles can have fractional spin values, such as 1/2, 3/2, and so on. These particles are called fermions and are subject to the Pauli exclusion principle, which states that no two identical fermions can occupy the same quantum state simultaneously.

5. How does spin differ from orbital angular momentum in quantum mechanics?

Spin is a type of intrinsic angular momentum that is independent of a particle's motion, while orbital angular momentum is associated with a particle's motion around a central point. Spin has discrete values and is not affected by external forces, whereas orbital angular momentum can have continuous values and can be influenced by external forces.

Similar threads

  • Advanced Physics Homework Help
Replies
28
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
446
  • Advanced Physics Homework Help
Replies
7
Views
1K
Replies
19
Views
1K
  • Advanced Physics Homework Help
Replies
8
Views
2K
  • Advanced Physics Homework Help
Replies
6
Views
956
  • Advanced Physics Homework Help
2
Replies
36
Views
2K
Replies
1
Views
426
  • Advanced Physics Homework Help
Replies
5
Views
824
  • Advanced Physics Homework Help
Replies
10
Views
2K
Back
Top