The Eigenvectors of $L_3$ (for spin 1) are $\left| m \right>$ with $m=1,0,-1$. One can compute the matrix

$D_i=\begin{pmatrix}\left< 1 \middle| L_i \middle| 1 \right> & \left< 1 \middle| L_i \middle| 0 \right> & \left< 1 \middle| L_i \middle| -1 \right>\\\left< 0 \middle| L_i \middle| 1 \right> & \left< 0 \middle| L_i \middle| 0 \right> & \left< 0 \middle| L_i \middle| -1 \right> \\\left< -1 \middle| L_i \middle| 1 \right> & \left< -1 \middle| L_i \middle| 0 \right> & \left< -1 \middle| L_i \middle| -1 \right>\end{pmatrix}$

for $i=+,-,1,2,3$.

As an exercise I was supposed to calculate

$D_i'=\begin{pmatrix}\left< x \middle| L_i \middle| x \right> & \left< x \middle| L_i \middle| y \right> & \left< x \middle| L_i \middle| z \right>\\\left< y \middle| L_i \middle| x \right> & \left< y \middle| L_i \middle| y \right> & \left< y \middle| L_i \middle| z \right> \\\left< z \middle| L_i \middle| x \right> & \left< z \middle| L_i \middle| y \right> & \left< z \middle| L_i \middle| z \right>\end{pmatrix}$

In the basis $\left| x \right>=\frac{1}{\sqrt{2}}(\left| -1 \right>-\left| 1 \right>)$, $\left| y \right>=\frac{i}{\sqrt{2}}(\left| -1 \right>+\left| 1 \right>)$, $\left| z \right>=\left| 0 \right>$.

I did that by writing

$\left< x \middle| L_i \middle| x \right> = \frac{1}{\sqrt{2}}(\left< -1 \right| - \left< 1 \right|)L_i(\left| -1 \right> - \left| 1 \right>)=\frac{1}{\sqrt{2}}(\left< -1 \middle| L_i \middle| -1 \right>-\left< -1 \middle| L_i \middle| 1 \right>-\left< 1 \middle| L_i \middle| -1 \right> + \left< 1 \middle| L_i \middle| 1 \right>)$

For all elements of $D_i'$.

What I am wondering is, what does each base physicly mean? I guess in $\left| x \right>,\left| y \right>,\left| z \right>$ base the corresponding matrix $D_i'$ is the angular momentum operator in the position space while $D_i$ is the angular momentum in the Eigenbasis - but how can I imagine it/get an intuitive idea of the difference of using two bases?

  • $\begingroup$ but how can I imagine it/get an intuitive idea of the difference of using two bases? I am not sure if this question is clear enough. Can you elaborate? What type of difference are you looking for? Can your question be boiled down to "What are the reasons we choose to work in some particular basis?"? $\endgroup$ Commented Jul 2, 2019 at 18:49
  • 1
    $\begingroup$ The closest you can probably get is en.wikipedia.org/wiki/Vector_model_of_the_atom , but this is quite an outdated approach. $\endgroup$
    – Cryo
    Commented Jul 3, 2019 at 0:45
  • $\begingroup$ @AaronStevens Yes, "What are the reasons we choose to work in some particular basis?" might be better phrasing of my question, sorry. $\endgroup$
    – Kekks
    Commented Jul 3, 2019 at 5:45

1 Answer 1


In QM it is best to think of basis as eigenbasis of some operator (or multiple commuting operators). That is, operators come first; they are defined independently from any basis or even dimension of Hilbert space. The angular momentum operators for example are defined by how they commute with each other (which can be intuitively derived from how rotations in 3D space commute with each other). Once we define the operators we ask how can they be represented. The answer is usually: there are many ways to do it with matrices of different dimension. Then we say, ok, how about 3 dimensional complex matrices; this is the spin 1 representation you mentioned. Once representation is chosen we can diagonalize one of the matrices, say i=3, and represent the rest in the eigenbasis of that matrix. That is how basis are choses. In this case their meaning is: eigenbasis of i=3 angular momentum operator.

Regarding you question about arbitrary changes of basis. In general it may be tough to assign meaning to basis that you get by just changing some meaningful basis to something else. Usually when we change basis we do it with purpose so the meaning comes built in and once again, it is something like: this are the eigenbasis of that other operator where that other operator has some meaning. In general, if you have a state (basis is a special case), you can get some meaning out of it by asking what are the expectation values of some physically meaningful operators. It is also a good idea to ask what are the variances of those operators. In your case you can use the 3 angular momentum operators on those new basis and see what are the expectation values and what are the variances.

By the way, if those new basis diagonalize another angular momentum operator then there you have it. It's just a different choice of which operator you choose to be diagonal, i.e, it does not really matter.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.