# I Matrix Representation of an Operator (from Sakurai)

#### jaurandt

Look, I am sorry for not being able to post any LaTeX. But I am stuck at a place where I feel I should not be stuck. I can not figure out how to correctly do this. I can't seem to recreate the Pauli matrices with that form using the 3 2-dimensional bases representing x, y, and z spin up/down.

Does anyone have any advice on this?

Related Quantum Physics News on Phys.org

#### HomogenousCow

Just to be clear, you're trying to project the spin matrices in an eigenspin bases right? (say in the z-direction) You only need the spin up and spin down vectors in any one direction to form a complete basis.
For $\sigma_z$ this is trivial in its eigenbasis:
$$\sigma_z = \sum_s \sum_r |s\rangle\langle s|\sigma_z|r\rangle\langle r|$$
$$\sigma_z = |\uparrow \rangle \langle \uparrow | - | \downarrow \rangle \langle \downarrow |$$

For $\sigma_x$ and $\sigma_y$ you could use the raising and lowering operators to make the same decomposition.

• jaurandt and vanhees71

#### jaurandt

Just to be clear, you're trying to project the spin matrices in an eigenspin bases right? (say in the z-direction) You only need the spin up and spin down vectors in any one direction to form a complete basis.
For $\sigma_z$ this is trivial in its eigenbasis:
$$\sigma_z = \sum_s \sum_r |s\rangle\langle s|\sigma_z|r\rangle\langle r|$$
$$\sigma_z = |\uparrow \rangle \langle \uparrow | - | \downarrow \rangle \langle \downarrow |$$

For $\sigma_x$ and $\sigma_y$ you could use the raising and lowering operators to make the same decomposition.
Can you please give the same example, but with the $\sigma_y$ operator? What I'm trying to say is that I don't understand how

$$\sigma_z = \sum_s \sum_r |s\rangle\langle s|\sigma_z|r\rangle\langle r|$$

Reveals the entries of the matrix...

#### andresB

The numbers $$\left\langle a'\right|X\left|a''\right\rangle$$
are the entries of the matrix. For the z Pauli matrix we have $$\left\langle \uparrow\right|\sigma_{z}\left|\uparrow\right\rangle =1; \left\langle \downarrow\right|\sigma_{z}\left|\uparrow\right\rangle =0; \left\langle \uparrow\right|\sigma_{z}\left|\downarrow\right\rangle =0; \left\langle \downarrow\right|\sigma_{z}\left|\downarrow\right\rangle =-1$$

• jaurandt

#### HomogenousCow

You don't even have to deal with matrices if you don't want to, all that's happening is we're rewriting the operator by inserting some identity operators. Since $$I = \sum_s |s\rangle\langle s|,$$ we can just stick one in front of and behind an operator to rewrite it in terms of the operator basis $|s \rangle \langle r|$,
$$A = I A I = \sum_s \sum_r |s\rangle\langle s| A |r\rangle\langle r| = \sum_s \sum_r A_{sr} |s\rangle \langle r|,$$ where $A_{sr}$ are the matrix elements.

• jaurandt

#### jaurandt

The numbers $$\left\langle a'\right|X\left|a''\right\rangle$$
are the entries of the matrix. For the z Pauli matrix we have $$\left\langle \uparrow\right|\sigma_{z}\left|\uparrow\right\rangle =1; \left\langle \downarrow\right|\sigma_{z}\left|\uparrow\right\rangle =0; \left\langle \uparrow\right|\sigma_{z}\left|\downarrow\right\rangle =0; \left\langle \downarrow\right|\sigma_{z}\left|\downarrow\right\rangle =-1$$
So then what happens to the rest of the construct if you just pull out

$$\left\langle a'\right|X\left|a''\right\rangle$$

What happened to the summation and what becomes of $$\left|a'\right\rangle\left\langle a''\right|$$

#### andresB

So then what happens to the rest of the construct if you just pull out

$$\left\langle a'\right|X\left|a''\right\rangle$$

What happened to the summation and what becomes of $$\left|a'\right\rangle\left\langle a''\right|$$

Put everything in the formula and you have the representation of the operator in that basis of vectors. HomogenousCow already showed how the z Pauli operator looks like written in terms of its own set of eigenvectors.

Last edited:
• jaurandt

#### kith

The object $|a' \rangle \langle a''|$ is an operator. Every entry in a representation matrix is tied to such an operator.

Consider this rewrite of the third Pauli matrix:
$$\begin{pmatrix} 1 & 0\\ 0 & -1 \end{pmatrix} = 1 \cdot \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix} + 0 \cdot \begin{pmatrix} 0 & 0\\ 1 & 0 \end{pmatrix} + (-1) \cdot \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$$
This is what corresponds to the operator equation
$\sigma_z = 1 \cdot |\!\uparrow_z \rangle \langle \uparrow_z \!| + 0 \cdot |\!\uparrow_z \rangle \langle \downarrow_z\!| + 0 \cdot |\!\downarrow_z \rangle \langle \uparrow_z\!| + (-1) \cdot |\!\downarrow_z \rangle \langle \downarrow_z\!|.$

(Note that the symbol $\sigma_z$ is ofen used to symbolize both the operator and its matrix representation in the z-basis. This is a sloppy but very common notation.)

Last edited:
• jaurandt

"Matrix Representation of an Operator (from Sakurai)"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving