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touqra
Oct22-05, 09:13 PM
Given S=\frac{1}{2}\hbar{\sigma} where \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) , show that
|+> = \left(\begin{array}{cc}1\\0\end{array}\right) and |-> = \left(\begin{array}{cc}0\\1\end{array}\right) are the eigenfunctions for S_z . Obtain the matrix representation for S_y and S_x in the basis (|+>,|->) .

George Jones
Oct23-05, 10:08 AM
S_{z} = \frac{1}{2}\hbar \sigma_z

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

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

Regards,
George

Gokul43201
Oct23-05, 11:41 AM
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
Oct23-05, 11:48 AM
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

Gokul43201
Oct23-05, 02:07 PM
Oh, they've already given you the eigenvectors ! Yes, that would be unnecessary then.

touqra
Oct23-05, 08:43 PM
The first part is just an eigenvalue problem. From the matrix for Sz, find the eigenvalues (and eigenvectors) from the characteristic (and eigenvalue) equation.
S_{z} = \frac{1}{2}\hbar \sigma_z
What happens when S_{z} operates on \left| + \right>?
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.

Gokul43201
Oct24-05, 12:37 AM
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 ?

touqra
Oct24-05, 01:51 AM
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.

George Jones
Oct24-05, 09:50 AM
If A is a linear operator on an n-dimensional vector space V, then the matrix representation of A with respect to the basis \{v_{1}, \dots , v_{n} \} of V is given by


A v_{i} = \sum_{j} A_{ji} v_{j}.
.

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

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

Regards,
George