I Matrix Representation of an Operator (from Sakurai)

  • Thread starter jaurandt
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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.

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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?
 
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.
 
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...
 
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44
The numbers $$\left\langle a'\right|X\left|a''\right\rangle $$
are the entries of the matrix. For the z Pauli matrix we have $$
\left\langle \uparrow\right|\sigma_{z}\left|\uparrow\right\rangle =1;
\left\langle \downarrow\right|\sigma_{z}\left|\uparrow\right\rangle =0;
\left\langle \uparrow\right|\sigma_{z}\left|\downarrow\right\rangle =0;
\left\langle \downarrow\right|\sigma_{z}\left|\downarrow\right\rangle =-1$$
 
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.
 
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 $$\left|a'\right\rangle\left\langle a''\right| $$
 
232
44
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:

kith

Science Advisor
1,263
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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:
[tex]
\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}
[/tex]
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:

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