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If we try to find matrix elements of ladder operators ( $J_{\pm}$) for spin when they act on eigenstates of $J^2$ and $J_z$ ( $\newcommand{ket}[1]{\left|#1\right\rangle} \newcommand{avg}[1]{\left\langle#1\right\rangle} \ket{\ j,m}$ ), as for example A.R. Edmonds: Angular Momentum in Quantum Mechanics, or Auletta: Quantum Mechanics, or any other textbook which deals with this issue, we come to a point where we find that:

$$J_{\pm} \ket{\ j,m}=e^{i\phi} \hbar \sqrt{j(j+1)-m(m\pm1)} \ket{\ j,m\pm1}$$

Normally, we set $\phi=0$ for all choices of m for a fixed j. Now, the standard explanation for this is that the choice of phase is arbitrary but must be followed consistently (e.g. Edmonds or Condon, Shortley).

I would like to point out some issues with this explanations and then show what happens if we try to play with the phase choice. The most important point will be that we will allow the choice of a different phase $\phi=\phi(m)$ for different m's as opposed to choice of same phase for every m (the special case of which is the standard choice of $\phi=0$ for all m's). This idea comes from the fact that there is no (at least obvious) reason to claim that it follows from the standard derivation of matrix elements of ladder operators that the phases must be the same for all m's.

We will look at the example of spin=1 matrices. The standard choice of $\phi=0$ for all matrix elements yields the following matrices of ladder operators and then the observables of spin in z,x and y direction:

$$j_z=\begin{bmatrix} 1 & 0&0\\ 0& 0&0\\ 0& 0&-1 \end{bmatrix}, j_+=\sqrt{2} \begin{bmatrix} 0 & 1&0\\ 0& 0&1\\ 0& 0&0 \end{bmatrix}, j_-=\sqrt{2}\begin{bmatrix} 0 & 0&0\\ 1& 0&0\\ 0& 1&0 \end{bmatrix},$$

$$j_x=\frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 1&0\\ 1& 0&1\\ 0& 1&0 \end{bmatrix}, j_y=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & -i&0\\ i& 0&-i\\ 0& i&0 \end{bmatrix}$$

Now, if we make a choice of setting the phase for $j_+\ket{1,0}$ to 0, $\phi(m=0)=0$, but for $j_+\ket{1,-1}$ we set it to a different phase, $\phi(m=-1)=\pi/2$, we construct the following matrices:

$$j_z^{(2)}=j_z=\begin{bmatrix} 1 & 0&0\\ 0& 0&0\\ 0& 0&-1 \end{bmatrix}, j_+^{(2)}=\sqrt{2} \begin{bmatrix} 0 & i&0\\ 0& 0&1\\ 0& 0&0 \end{bmatrix}, j_-^{(2)}=\sqrt{2}\begin{bmatrix} 0 & 0&0\\ -i& 0&0\\ 0& 1&0 \end{bmatrix},$$

$$j_x^{(2)}=\frac{1}{\sqrt{2}} \begin{bmatrix} 0 & i&0\\ -i& 0&1\\ 0& 1&0 \end{bmatrix}, j_y^{(2)}=\frac{1}{\sqrt{2}}\begin{bmatrix} 0 & 1&0\\ 1& 0&-i\\ 0& i&0 \end{bmatrix}$$

Now, to make things clear, I should point out that the new $j_x$ and $j_y$ have the same eigenvalues as the standard ones, and satisfy the same old algebra of angular momentum, though their eigenstates are different. As a result, the expected values of these different observables are different in some superpositions of $\ket{1,m}$, which is interesting, since that is what we observe in experiments.

As an example, here is a graph which shows how the expected values of $j_x$ and $j_y$ on the state $\ket{\psi}=\frac{1}{\sqrt{3}}\left(\ket{1,1}+\ket{1,0}+\ket{1,-1}\right)$ change when we keep the phase $\phi(0)=0$ (the same as above) but we change $\phi(-1)$ (which is for the above matrices set to $\pi/2$). In the graph, $\avg{j_x}$ is in red and $\avg{j_y}$ is in blue.

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As a contrast, here is a graph which shows how the expected values of $j_x$ and $j_y$ on the state $\ket{\psi}=\frac{1}{\sqrt{3}}\left(\ket{1,1}+\ket{1,0}+\ket{1,-1}\right)$ change when we change both phases simultaneously and equally: $\phi(0)=\phi(-1)=\phi$ which is the standardly considered phase choice (in textbooks) and is equivalent to rotating coordinate system around z-axis. That is apparent in the way expected values of $j_x$ and $j_y$ interchange as we change $\phi$. In the following graph, $\avg{j_x}$ is again in red and $\avg{j_y}$ is in blue.

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One interesting point is that the observables we construct by choosing different phases for different matrix elements of ladder operators, i.e. $\phi(0) \neq \phi(-1)$, can't be written as linear combination of the standard operators (even with complex coefficients).

Another interesting thing to note is that these different observables arise only for spin>=1, since for spin=1/2 only one matrix element of ladder operators is different from 0.

My question is, does anyone know of an interpretation of these "$j_x$" and "$j_y$" we construct by different phase choices? Further on, does anyone know of an argument I am not aware of, for choosing the same phase for every matrix element as opposed to my choice of different phases (to me it seems that the standard choice is just a special case of a more general choice, but I don't yet understand how to interpret the results)?

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  • $\begingroup$ You may want to use Joined->True option for ListPlot in your plots since it doesn't really make any sense to show discreteness of your sampling points, and it just clutters the plots. $\endgroup$ – Ruslan Jan 19 '15 at 11:55
  • $\begingroup$ As for the question itself: did you change the $\ket{\psi}$ accordingly when you changed the $\phi$'s (especially important for the first plot)? $\endgroup$ – Ruslan Jan 19 '15 at 11:57
  • $\begingroup$ If I do that, the graph again doesn't look similar to the standard expected values graph. Then again, I don't understand why would I have to change $|\psi>$, am I missing something very important here? $\endgroup$ – bsoda Jan 19 '15 at 19:55
  • $\begingroup$ Indeed, it shouldn't make any difference. BTW, why do you plot $\avg{j_i}^2$ instead of just $\avg{j_i}$? $\endgroup$ – Ruslan Jan 20 '15 at 7:32
  • $\begingroup$ Fair point, it's fixed now. $\endgroup$ – bsoda Jan 20 '15 at 8:02
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There must be an error in the way you calculate the expectation values.

The fact that your transformed operators have the same spectrum of eigenvalues and the same algebra comes from the fact that redefinitions of the phase lead to unitarily equivalent operators, i.e. they are related by $$\tilde j_+ = U^\dagger j_+ U,\quad \tilde j_- = U^\dagger j_- U,\quad \tilde j_z = U^\dagger j_z U$$ with a unitary matrix $U$ and the states in the two bases (different phase choices) are related by $$\langle \tilde \phi \vert = \langle \phi \vert U, \qquad \vert \tilde \phi \rangle = U^\dagger \vert \phi \rangle.$$ Since $j_z$ and $j_\pm$ span the entire angular momentum algebra, any operator leaving the angular momentum $\vec J^2$ invariant can be written as combination of $j_z$ and $j_\pm$. Since any product of these will transform by getting a $U^\dagger$ on the left and a $U$ on the right, we have $$ \tilde O = U^\dagger O U.$$

Now the expectation value of any operator can be computed in both bases and we find $$ \langle \tilde \phi \vert \tilde O \vert \tilde \phi \rangle = \langle \phi \vert U U^\dagger O U U^\dagger \vert \phi \rangle = \langle \phi \vert O \vert \phi \rangle.$$ This is without choosing any specific phase convention, but making sure we follow the convention consistently with the $U$ matrices.

Now since you find a difference between different conventions there should be a detail you overlooked. Did you remember that if the state $\vert 1, -1\rangle$ gets a phase $\phi$, the state $\langle 1, -1 \vert$ gets the phase $-\phi$?

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  • $\begingroup$ Thank you for the answer, Neuneck. I now understand what is the fault in my logic up there. The problem is that I (mistakenly) thought of the states as independently constructed before finding the matrix elements of ladder operators. This lead to the expected values shown above. But they are by def. constructed through ladder operators. This means that, to consistently build a set of operators and states, I need to do exactly as it's written in your answer. $\endgroup$ – bsoda Jan 20 '15 at 16:12
  • $\begingroup$ Exactly, (for the second comment) $j_z$ and U commute and so $j_z$ stays the same. $\endgroup$ – bsoda Jan 20 '15 at 16:13
  • $\begingroup$ @ Neuneck OK, thanks to both of you and bsoda, I now seem to understand what's wrong. Thank you for the answer and @bsoda for an interesting question. $\endgroup$ – Ruslan Jan 20 '15 at 16:20

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