# Quantum Mechanics - question about spin

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 $$(|+>,|->)$$ .

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## Answers and Replies

George Jones
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$$S_{z} = \frac{1}{2}\hbar \sigma_z$$

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
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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
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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

Gokul43201
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Oh, they've already given you the eigenvectors ! Yes, that would be unnecessary then.

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:
$$S_{z} = \frac{1}{2}\hbar \sigma_z$$
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.

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Gokul43201
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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 ?

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.

George Jones
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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

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