My textbook says that if $L^2$ is the square of the angular momentum and if it's eigenstate is $|\alpha,\beta>$ then its eigenvalue is $\hbar^2\alpha$ i.e. $$L^2|\alpha,\beta>=\hbar^2\alpha|\alpha,\beta>$$ It then goes on to talk about the $z$ component of the angular momentum ($L_z$) saying that $$L_z|\alpha,\beta>=\hbar\beta|\alpha,\beta>$$ i.e. that $\hbar\beta$ is the eigenvalue of $L_z$.

Here is where I need help later on it says that $\beta_{max}=l$ i.e. the greatest value $\beta$ can take is $l$ and that $\beta=m$ hence: $$L^2|l,m>=\hbar^2l(l-1)|l,m>$$ and that $$L_z|l,m> =\hbar m|l,m>$$ Firstly why is the $\alpha$ in the ket not the same as that outside but the $\beta$ is? and secondly why if $\alpha$ is bigger then or equal to $\beta^2$ (as my book says) does $\alpha := \sqrt{\beta_{max}}=\sqrt{l}$.

Edit: Please look at my comment in neuneck's question

  • $\begingroup$ Are you sure the eigenvalue in equation 1 is $\alpha$ and not a? Also can you post a book title and page? $\endgroup$ – Constandinos Damalas Jul 31 '14 at 10:50
  • $\begingroup$ I have just finishied A-levels hence the book is "Quantum physics for dummies" (nothing to elaborate), and since I have a digitial copy can not give you the page number. $\endgroup$ – user43487 Jul 31 '14 at 14:14

This is along the line of Phonon's response, but I think this can be formulated more clearly.

From the angular momentum algebra $$[L_i, L_j] = i \hbar \epsilon_{ijk} L_k$$ it follows that (for $\vec L^2 = \sum_i L_i^2, L_{\pm} = L_x \pm i L_y$): $$[\vec L^2, L_i] = 0, \quad [\vec L^2, L_\pm] = 0, \quad [L_z, L_{\pm}] = \pm \hbar L_\pm$$ Now $[\vec L^2, L_z] = 0$ implies that there is a common base of eigenvectors of $\vec L^2$ and $L_z$. Thus, there are two labels to an angular momentum eigenstate. These are not necessarily the eigenvalues themselves, but from the labels one should be able to determine the eigenvalues: $$ \vec L^2 \vert \alpha, \beta \rangle = X_{\alpha} \vert \alpha, \beta \rangle, \quad L_z \vert \alpha, \beta \rangle = Y_{\beta} \vert \alpha, \beta \rangle$$ Now, $[L_z, L_\pm] = \pm \hbar L_\pm$ implies that $L_\pm \vert \alpha, \beta \rangle$ is again an eigenstate of $L_z$, but with a different Eigenvalue: $$ L_z (L_\pm \vert \alpha, \beta \rangle) = L_\pm (L_z \vert \alpha, \beta \rangle) + [L_z, L_\pm] \vert \alpha, \beta \rangle = (Y_{\beta} \pm \hbar) (L_\pm \vert \alpha, \beta \rangle) $$ Thus, acting with $L_\pm$ on $\vert \alpha, \beta\rangle$ increases the $L_z$ eigenvalue $Y_\beta$ by $\hbar$. Therefore, let's try to use $m$ instead of $\beta$ and set $Y_m = m \hbar$ (As you can see, even for the $L_z$ operator, the actual eigenvalue is not equal to the label, but carries a factor of $\hbar$!).

Now for the eigenvalue w.r.t. $\vec L^2$ we need to work a bit more. Check that $$ L_+ L_- = L_x^2 + L_y^2 + \hbar L_z, \quad L_- L_+ = L_x^2 + L_y^2 - \hbar L_z, $$$$ \Rightarrow \; \vec L^2 = L_z^2 + \frac{1}{2} (L_+ L_- + L_- L_+)$$ Also, the $\vec L^2$ eigenvalue does not change under operation with $L_\pm$ due to $[\vec L^2, L_\pm] = 0$. Now since $L_- = L_+^\dagger$, we have that $$ \vert L_+ \vert \alpha, m\rangle \vert^2 = \langle \alpha, m \vert L_- L_+ \vert \alpha, m \rangle \geq 0$$ $$\vert L_- \vert \alpha, m \rangle \vert^2 = \langle \alpha, m \vert L_+ L_- \vert m, \alpha\rangle \geq 0 $$ On the other hand we can use the expression of $\vec L^2$ in terms of $L_z$ and $L_\pm$ to see that $$ X_\alpha = \langle \alpha, m \vert \vec L^2 \vert \alpha, m \rangle = \langle \alpha, m \vert L_z^2 + \frac{1}{2} ( L_+ L_- + L_- L_+) \vert \alpha, m \rangle = \\ \hbar^2 m^2 + \text{something positive or zero}$$ Thus, $X_\alpha \geq \hbar^2 m^2$. This has profound implications: $X_\alpha$ is always larger than (or equal to) $\hbar^2 m^2$. But working on a given state $\vert \alpha, m \rangle$ with $L_\pm$ increases $m$ while leaving $\alpha$ constant! If we could increase $m$ indefinitely (by operating with $L_+$ multiple times on a state), then $X_\alpha$ had to be infinite, which does not make sense physically. This means that there must be a state with highest $m$ for which $L_+ \vert \alpha, m_\text{max} \rangle = 0$.

The same chain of though produces the lowest state, for which $L_- \vert \alpha, m_\text{min}\rangle = 0$. If you do not see this, read again what I wrote, think about it, read it in books until you understand it. This is a very important result!

We can use this to construct the eigenvalue $X_\alpha$: Since $$L_+ \vert \alpha, m_\text{max} \rangle = 0$$ we have that $$L_- L_+ \vert \alpha, m_\text{max} \rangle = 0$$ which is nothing but $$(\vec L^2 - L_z^2 - \hbar L_z) \vert \alpha, m_\text{max} \rangle = X_\alpha - \hbar^2 m_\text{max}^2 - \hbar \hbar m_\text{max} = 0$$ This means that $X_\alpha = \hbar^2 (m_\text{max}^2 + m_\text{max})$. The same way you can show that $X_\alpha = \hbar^2 (m_\text{min}^2- m_\text{min})$. You can now characterize the $X_\alpha$ by the largest allowed $m$: $$\vec L^2 \vert m_\text{max}, m\rangle = \hbar^2 (m_\text{max}^2 + m_\text{max}) \vert m_\text{max}, m\rangle$$ or you can just call $m_\text{max} = j$ and have $$ \vec L^2 \vert j, m \rangle = \hbar^2 j (j+1) \vert j, m \rangle, \quad L_z \vert j, m \rangle = \hbar m \vert j, m \rangle $$ Finally, the two expressions for $X_\alpha$ that we derived from the maximal and minimal $m$ can be put together: $$ X_\alpha = X_\alpha$$ $$ \hbar^2 (m_\text{max}^2 + m_\text{max}) = \hbar^2 (m_\text{min}^2 - m_\text{min})$$ This tells us that $m_\text{min} = - m_\text{max}$. Since the eigenvalue of $L_z$ is increased or decreased in steps of $\hbar$, $m_\text{min}$ and $m_\text{max}$ must be separated by an integer number, telling us that $j$ (and $m$) take on either integer or half-integer numbers and it also tells us that a (half)integer $j$ only has (half)integer $m$s.

Now we derived all the basics about angular momentum from nothing but the commutation relations. Make sure you really understand what's going on here, since this is very important. A thorough understanding of these formulas and their meaning will help you understand a lot of things, like angular momentum addition, tensor operators and the Wigner-Eckart-Theorem and finally even group theory, should you ever come across it.

  • $\begingroup$ Thank you for also taking the time to further clarify some of the details that I may have left out, I just wanted leave some steps for him to go through on his own. But anyway I think now Joseph has enough to mull over as it is :-). $\endgroup$ – Ellie Jul 31 '14 at 13:55
  • $\begingroup$ Thanks for you help, there is just one bit I do not understand, this is when you change $\alpha$ into $j$ please could you explain $\endgroup$ – user43487 Jul 31 '14 at 15:45
  • $\begingroup$ @Joseph I just renamed the variable. You could also say that $\vec L^2$ has eigenvalues of $\hbar^2 \alpha (\alpha + 1)$. By renaming I wanted to get closer to the canonical nomenclature. Before that I used $\alpha$ to stress that this is just a label. $\endgroup$ – Neuneck Aug 1 '14 at 8:03

Use the link below for a complete derivation of what you're looking for.

Second page of link

EDIT: as the link may have been unclear to you, I will elaborate a bit further myself:

First thing: Operator $L^2$ is of interest as it commutes with either components of $L$: $$[L^2,L_x] = [L^2,L_z]=[L^2,L_y]=0 $$ Consequently it means that it is possible to find simultaneous eigenstates(let's call the eigenstate $f$ to shorten the notations) of $L^2$ and of any of the three components, e.g. $L_z$ (the usual conventional choice): $$L^2 f = \alpha f, L_z f=\beta f $$ Now by introducing the ladder operators $L_+$ and $L_-$, we can obtain eigenvalues of $L^2$ and $L_z$ algebraically. $$L_{\pm} =L_x \pm iL_y $$ Which gives the following commutators: $[L_z,L_{\pm}]=\pm \hbar L_{\pm}$ and $[L^2,L_{\pm}]=0$

Now about your first question:

Using the 2nd commutation condition below, one obtains: $L^2(L_{\pm}f)=\alpha (L_{\pm})$ (if any step is unclear, do ask!) and using the first relation: $L_z(L_{\pm}f)=\pm\hbar L_{\pm}f+L_{\pm}(\beta f)=(\beta \pm \hbar)(L_{\pm} f)$

The ladder operators raise and lower the eigenvalues of $L_z$ by $\hbar$ while eigenvalues of $L^2$ unchanged, and for each value of $\alpha$ there's a set of discrete values of $\beta$ separated by $\hbar$, (simply stating what we just calculated). Finally about your second question: As $f$ is an eigenfunction of $L^2$ and $L_{\pm}$ then it is obvious that $\beta \leq \alpha$. So let us choose the state with the largest value of $\beta$ and call it $f'$: $$ L_z f' = \hbar l f' ; L^2 f' = \alpha f'$$ With $l$ the max of $L_z$ eigenvalues, we can obtain an expression for $\alpha$: Using $$L_{\pm} L_{\mp} =(L_x \pm iL_y)(L_x \mp iL_y) = L^2 - L_z^2 \mp i(i\hbar L_z) $$ Extracting $L^2$ from it and applying it on $f'$: $$L^2 f' = (L_- L_+ + L^2 + \hbar L_z)f'=(\hbar^2 l^2 + \hbar^2 l) f'$$ Finally the expression for $\alpha$ : $\alpha=\hbar^2 l(l+1)$

Now starting from the bottom of the ladder of eigenvalues (denoted $f_0$): $L_z f_0 =\hbar \bar{l}f_0$ In terms of $\alpha$: $$\alpha=\hbar^2 \bar{l}(\bar{l}-1) \leftrightarrow \bar{l}=-l $$ To conclude, now we know that $L_z$ has for eigenvalues: $m\hbar$ (step: $\hbar$) where $m$ varies between $-l$ and $l$, i.e. : $m=-l,-l+1,...,l$

$$L^2 f_l^m = \hbar^2 l(l+1)f_l^m$$ $$L_z f_l^m = \hbar m f_l^m $$ Sources: 1,2,3,4

  • $\begingroup$ I have looked at this link and it does not seem to answer my question $\endgroup$ – user43487 Jul 31 '14 at 8:49
  • $\begingroup$ There you go, I just took the time, and detailed it step by step for you. Hope it is clearer now. $\endgroup$ – Ellie Jul 31 '14 at 11:21

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