# Angular momentum in 3-d: expectation values and uncertainty principle

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.3.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

For 3-d angular momentum, we’ve seen that the components ${J_{x}}$ and ${J_{y}}$ can be written in terms of raising and lowering operators

$\displaystyle J_{\pm}\equiv J_{x}\pm iJ_{y} \ \ \ \ \ (1)$

In the basis of eigenvectors of ${J^{2}}$ and ${J_{z}}$ (that is, the states ${\left|jm\right\rangle }$) the raising and lowering operators have the following effects:

$\displaystyle J_{\pm}\left|jm\right\rangle =\hbar\sqrt{\left(j\mp m\right)\left(j\pm m+1\right)}\left|j,m\pm1\right\rangle \ \ \ \ \ (2)$

We can use these relations to construct the matrix elements of ${J_{x}}$ and ${J_{y}}$ in this basis. We can also use these relations to work out expectation values and uncertainties for the angular momentum components in this basis.

First, since diagonals of both the ${J_{x}}$ and ${J_{y}}$ matrices have only zero elements,

 $\displaystyle \left\langle J_{x}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle jm\left|J_{x}\right|jm\right\rangle =0\ \ \ \ \ (3)$ $\displaystyle \left\langle J_{y}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle jm\left|J_{y}\right|jm\right\rangle =0 \ \ \ \ \ (4)$

To work out ${\left\langle J_{x}^{2}\right\rangle }$ and ${\left\langle J_{y}^{2}\right\rangle }$, we can write these operators in terms of the raising and lowering operators:

 $\displaystyle J_{x}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(J_{+}+J_{-}\right)\ \ \ \ \ (5)$ $\displaystyle J_{y}$ $\displaystyle =$ $\displaystyle \frac{1}{2i}\left(J_{+}-J_{-}\right) \ \ \ \ \ (6)$

We can then use the fact that the basis states are orthonormal, so that

$\displaystyle \left\langle j^{\prime}m^{\prime}\left|jm\right.\right\rangle =\delta_{j^{\prime}j}\delta_{m^{\prime}m} \ \ \ \ \ (7)$

The required squares are

 $\displaystyle J_{x}^{2}$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(J_{+}^{2}+J_{+}J_{-}+J_{-}J_{+}+J_{-}^{2}\right)\ \ \ \ \ (8)$ $\displaystyle J_{y}^{2}$ $\displaystyle =$ $\displaystyle -\frac{1}{4}\left(J_{+}^{2}-J_{+}J_{-}-J_{-}J_{+}+J_{-}^{2}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(-J_{+}^{2}+J_{+}J_{-}+J_{-}J_{+}-J_{-}^{2}\right) \ \ \ \ \ (10)$

The diagonal matrix elements ${\left\langle jm\left|J_{x}^{2}\right|jm\right\rangle }$ and ${\left\langle jm\left|J_{y}^{2}\right|jm\right\rangle }$ will get non-zero contributions only from those terms that leave ${j}$ and ${m}$ unchanged when operating on ${\left|jm\right\rangle }$. This means that only the terms that contain an equal number of ${J_{+}}$ and ${J_{-}}$ terms will contribute. We therefore have

 $\displaystyle \left\langle jm\left|J_{x}^{2}\right|jm\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left\langle jm\left|J_{+}J_{-}+J_{-}J_{+}\right|jm\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{4}\sqrt{\left(j+m\right)\left(j-m+1\right)}\left\langle jm\left|J_{+}\right|j,m-1\right\rangle +\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\hbar}{4}\sqrt{\left(j-m\right)\left(j+m+1\right)}\left\langle jm\left|J_{-}\right|j,m+1\right\rangle \ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4}\sqrt{\left(j+m\right)\left(j-m+1\right)}\sqrt{\left(j-m+1\right)\left(j+m\right)}+\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\hbar^{2}}{4}\sqrt{\left(j-m\right)\left(j+m+1\right)}\sqrt{\left(j+m+1\right)\left(j-m\right)}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4}\left(\left(j+m\right)\left(j-m+1\right)+\left(j-m\right)\left(j+m+1\right)\right)\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4}\left(j^{2}-m^{2}+j+m+j^{2}-m^{2}+j-m\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2}\left(j\left(j+1\right)-m^{2}\right) \ \ \ \ \ (18)$

From 10 we see that the only terms that contribute to ${\left\langle jm\left|J_{y}^{2}\right|jm\right\rangle }$ are the same as the corresponding terms in ${\left\langle jm\left|J_{x}^{2}\right|jm\right\rangle }$, so the result is the same:

$\displaystyle \left\langle jm\left|J_{y}^{2}\right|jm\right\rangle =\frac{\hbar^{2}}{2}\left(j\left(j+1\right)-m^{2}\right) \ \ \ \ \ (19)$

We can check that ${J_{x}}$ and ${J_{y}}$ satisfy the uncertainty principle, as derived by Shankar. That is, we want to verify that

$\displaystyle \Delta J_{x}\cdot\Delta J_{y}\ge\left|\left\langle jm\left|\left(J_{x}-\left\langle J_{x}\right\rangle \right)\left(J_{y}-\left\langle J_{y}\right\rangle \right)\right|jm\right\rangle \right| \ \ \ \ \ (20)$

On the LHS

 $\displaystyle \Delta J_{x}$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle J_{x}^{2}\right\rangle -\left\langle J_{x}\right\rangle ^{2}}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left\langle J_{x}^{2}\right\rangle }\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar^{2}}{2}\left(j\left(j+1\right)-m^{2}\right)}\ \ \ \ \ (23)$ $\displaystyle \Delta J_{y}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\hbar^{2}}{2}\left(j\left(j+1\right)-m^{2}\right)}\ \ \ \ \ (24)$ $\displaystyle \Delta J_{x}\cdot\Delta J_{y}$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2}\left(j\left(j+1\right)-m^{2}\right) \ \ \ \ \ (25)$

On the RHS

$\displaystyle \left|\left\langle jm\left|\left(J_{x}-\left\langle J_{x}\right\rangle \right)\left(J_{y}-\left\langle J_{y}\right\rangle \right)\right|jm\right\rangle \right|=\left|\left\langle jm\left|J_{x}J_{y}\right|jm\right\rangle \right| \ \ \ \ \ (26)$

Using the same technique as that above for deriving ${\left\langle jm\left|J_{x}^{2}\right|jm\right\rangle }$ we have

 $\displaystyle \left\langle jm\left|J_{x}J_{y}\right|jm\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{4i}\left\langle jm\left|\left(J_{+}+J_{-}\right)\left(J_{+}-J_{-}\right)\right|jm\right\rangle \ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4i}\left\langle jm\left|J_{-}J_{+}-J_{+}J_{-}\right|jm\right\rangle \ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{4i}\left(\left(j-m\right)\left(j+m+1\right)-\left(j+m\right)\left(j-m+1\right)\right)\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}m}{2i} \ \ \ \ \ (30)$

We therefore need to verify that

$\displaystyle j\left(j+1\right)-m^{2}\ge\left|m\right| \ \ \ \ \ (31)$

for all allowed values of ${m}$. We know that ${-j\le m\le+j}$, so

$\displaystyle j\left(j+1\right)-m^{2}\ge j^{2}+j-j^{2}=j\ge\left|m\right| \ \ \ \ \ (32)$

Thus the inequality is indeed satisfied.

In the case ${\left|m\right|=j}$ we have

$\displaystyle j\left(j+1\right)-j^{2}=j=\left|m\right| \ \ \ \ \ (33)$

so the inequality saturates (becomes an equality) in that case.

# Total angular momentum – matrix elements and commutation relations

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.2.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

In Shankar’s Chapter 12 treatment of the eigenvalues of the angular momentum operators ${L^{2}}$ and ${L_{z}}$, he retraces much of what we’ve already covered as a result of working through Griffiths’s book. He defines raising and lowering operators for angular momentum as

$\displaystyle L_{\pm}\equiv L_{x}\pm iL_{y} \ \ \ \ \ (1)$

These operators can be used to discover the eigenvalues of ${L^{2}}$ to be ${\ell\left(\ell+1\right)\hbar^{2}}$, where ${\ell=0,\frac{1}{2},1,\frac{3}{2},\ldots}$ and the eigenvalues of ${L_{z}}$ are ${m\hbar}$ where ${m}$ ranges from ${-\ell}$ to ${+\ell}$ in integer steps. The eigenvalues of ${L_{\pm}}$ can also be found to satisfy

$\displaystyle L_{\pm}\left|\ell m\right\rangle =\hbar\sqrt{\left(\ell\mp m\right)\left(\ell\pm m+1\right)}\left|\ell,m\pm1\right\rangle \ \ \ \ \ (2)$

When dealing with vector wave functions (as opposed to scalar ones) in two dimensions, we found that a quantity ${J_{z}}$ is the generator of infinitesimal rotations about the ${z}$ axis, where

$\displaystyle J_{z}=L_{z}+S_{z} \ \ \ \ \ (3)$

and the operator producing the rotation by ${\varepsilon_{z}}$ is

$\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]=I-\frac{i\varepsilon_{z}}{\hbar}J_{z} \ \ \ \ \ (4)$

For a scalar wave function in three dimensions, we found that the properties of two successive rotations by ${\varepsilon_{x}}$ about the ${x}$ axis and ${\varepsilon_{y}}$ about the ${y}$ axis led to the commutations

 $\displaystyle \left[L_{x},L_{y}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{z}\ \ \ \ \ (5)$ $\displaystyle \left[L_{y},L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{x}\ \ \ \ \ (6)$ $\displaystyle \left[L_{z},L_{x}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{y} \ \ \ \ \ (7)$

For a vector wave function, the rotation is generated by ${J_{i}}$ rather than ${L_{i}}$ but because the effects of rotations are the same, the ${J_{i}}$ must have the same commutation relations, so that

 $\displaystyle \left[J_{x},J_{y}\right]$ $\displaystyle =$ $\displaystyle i\hbar J_{z}\ \ \ \ \ (8)$ $\displaystyle \left[J_{y},J_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar J_{x}\ \ \ \ \ (9)$ $\displaystyle \left[J_{z},J_{x}\right]$ $\displaystyle =$ $\displaystyle i\hbar J_{y} \ \ \ \ \ (10)$

We can do the same analysis on ${J}$ as we did above with ${L}$ to define the raising and lowering operators

$\displaystyle J_{\pm}\equiv J_{x}\pm iJ_{y} \ \ \ \ \ (11)$

and get the same eigenvalue relations

$\displaystyle J_{\pm}\left|jm\right\rangle =\hbar\sqrt{\left(j\mp m\right)\left(j\pm m+1\right)}\left|j,m\pm1\right\rangle \ \ \ \ \ (12)$

The three components of ${\mathbf{J}}$ are then ${J_{z}}$ and

 $\displaystyle J_{x}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(J_{+}+J_{-}\right)\ \ \ \ \ (13)$ $\displaystyle J_{y}$ $\displaystyle =$ $\displaystyle \frac{1}{2i}\left(J_{+}-J_{-}\right) \ \ \ \ \ (14)$

Using these three equations, we can generate the matrix elements of the components of ${\mathbf{J}}$ in the orthonormal basis ${\left|jm\right\rangle }$ (that is, the basis consisting of eigenfunctions with total angular momentum number ${j}$ and ${J_{z}}$ number ${m}$). These matrix elements are

 $\displaystyle \left\langle j^{\prime}m^{\prime}\left|J_{x}\right|jm\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left\langle j^{\prime}m^{\prime}\left|J_{+}+J_{-}\right|jm\right\rangle \ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sqrt{\left(j-m\right)\left(j+m+1\right)}\left\langle j^{\prime}m^{\prime}\left|j,m+1\right.\right\rangle +\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\hbar}{2}\sqrt{\left(j+m\right)\left(j-m+1\right)}\left\langle j^{\prime}m^{\prime}\left|j,m-1\right.\right\rangle \ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left[\sqrt{\left(j-m\right)\left(j+m+1\right)}\delta_{j^{\prime}j}\delta_{m^{\prime},m+1}+\sqrt{\left(j+m\right)\left(j-m+1\right)}\delta_{j^{\prime}j}\delta_{m^{\prime},m-1}\right] \ \ \ \ \ (18)$
 $\displaystyle \left\langle j^{\prime}m^{\prime}\left|J_{y}\right|jm\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2i}\left\langle j^{\prime}m^{\prime}\left|J_{+}-J_{-}\right|jm\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2i}\sqrt{\left(j-m\right)\left(j+m+1\right)}\left\langle j^{\prime}m^{\prime}\left|j,m+1\right.\right\rangle -\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{\hbar}{2i}\sqrt{\left(j+m\right)\left(j-m+1\right)}\left\langle j^{\prime}m^{\prime}\left|j,m-1\right.\right\rangle \ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2i}\left[\sqrt{\left(j-m\right)\left(j+m+1\right)}\delta_{j^{\prime}j}\delta_{m^{\prime},m+1}-\sqrt{\left(j+m\right)\left(j-m+1\right)}\delta_{j^{\prime}j}\delta_{m^{\prime},m-1}\right] \ \ \ \ \ (22)$

$\displaystyle \left\langle j^{\prime}m^{\prime}\left|J_{z}\right|jm\right\rangle =m\hbar\delta_{j^{\prime}j}\delta_{m^{\prime},m} \ \ \ \ \ (23)$

The full matrix for each component ${J_{i}}$ is actually infinite-dimensional, since ${j}$ can be any half-integer from 0 up to infinity. However, the sub-matrix for each value of ${j}$ is completely orthogonal to all other sub-matrices with different ${j}$ values, so the complete matrix for each ${J_{i}}$ is block-diagonal. Shankar gives the matrices for ${J_{x}}$ and ${J_{y}}$ up to ${j=1}$ in his equations 12.5.23 and 12.5.24. This means that the commutation relations 9 should be obeyed for each set of sub-matrices corresponding to a particular ${j}$ value.

For ${j=\frac{1}{2}}$ we have for the 3 sub-matrices (we can copy these from Shankar or use the above formulas to work them out). The values of ${m}$ are ${+\frac{1}{2}}$ and ${-\frac{1}{2}}$ in that order, from top to bottom and left to right.

 $\displaystyle J_{x}^{\left(1/2\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (24)$ $\displaystyle J_{y}^{\left(1/2\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2i}\left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2}\left[\begin{array}{cc} 0 & -1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (26)$ $\displaystyle \left[J_{x}^{\left(1/2\right)},J_{y}^{\left(1/2\right)}\right]$ $\displaystyle =$ $\displaystyle \frac{i\hbar^{2}}{4}\left(\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]-\left[\begin{array}{cc} -1 & 0\\ 0 & 1 \end{array}\right]\right)\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{i\hbar^{2}}{2}\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar J_{z}^{\left(1/2\right)} \ \ \ \ \ (29)$

For ${j=1}$ we have for the 3 sub-matrices

 $\displaystyle J_{x}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{array}\right]\ \ \ \ \ (30)$ $\displaystyle J_{y}^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{\sqrt{2}}\left[\begin{array}{ccc} 0 & -1 & 0\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{array}\right]\ \ \ \ \ (31)$ $\displaystyle \left[J_{x}^{\left(1\right)},J_{y}^{\left(1\right)}\right]$ $\displaystyle =$ $\displaystyle \frac{i\hbar^{2}}{2}\left(\left[\begin{array}{ccc} 1 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & -1 \end{array}\right]-\left[\begin{array}{ccc} -1 & 0 & -1\\ 0 & 0 & 0\\ 1 & 0 & 1 \end{array}\right]\right)\ \ \ \ \ (32)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar^{2}\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{array}\right]\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar J_{z}^{\left(1\right)} \ \ \ \ \ (34)$

For ${j=\frac{3}{2}}$ we need to work out the matrices from the formulas above for the matrix elements. Ordering the values of ${m=\frac{3}{2},\frac{1}{2},-\frac{1}{2},\frac{3}{2}}$ from left to right (columns) and top to bottom (rows), we get

 $\displaystyle J_{x}^{\left(3/2\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left[\begin{array}{cccc} 0 & \sqrt{3} & 0 & 0\\ \sqrt{3} & 0 & 2 & 0\\ 0 & 2 & 0 & \sqrt{3}\\ 0 & 0 & \sqrt{3} & 0 \end{array}\right]\ \ \ \ \ (35)$ $\displaystyle J_{y}^{\left(3/2\right)}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2i}\left[\begin{array}{cccc} 0 & \sqrt{3} & 0 & 0\\ -\sqrt{3} & 0 & 2 & 0\\ 0 & -2 & 0 & \sqrt{3}\\ 0 & 0 & -\sqrt{3} & 0 \end{array}\right]\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{i\hbar}{2}\left[\begin{array}{cccc} 0 & -\sqrt{3} & 0 & 0\\ \sqrt{3} & 0 & -2 & 0\\ 0 & 2 & 0 & -\sqrt{3}\\ 0 & 0 & \sqrt{3} & 0 \end{array}\right]\ \ \ \ \ (37)$ $\displaystyle \left[J_{x}^{\left(3/2\right)},J_{y}^{\left(3/2\right)}\right]$ $\displaystyle =$ $\displaystyle \frac{i\hbar^{2}}{4}\left(\left[\begin{array}{cccc} 3 & 0 & -2\sqrt{3} & 0\\ 0 & 1 & 0 & -2\sqrt{3}\\ 2\sqrt{3} & 0 & -1 & 0\\ 0 & 2\sqrt{3} & 0 & -3 \end{array}\right]-\left[\begin{array}{cccc} -3 & 0 & -2\sqrt{3} & 0\\ 0 & -1 & 0 & -2\sqrt{3}\\ 2\sqrt{3} & 0 & 1 & 0\\ 0 & 2\sqrt{3} & 0 & 3 \end{array}\right]\right)\ \ \ \ \ (38)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar^{2}\left[\begin{array}{cccc} \frac{3}{2} & 0 & 0 & 0\\ 0 & \frac{1}{2} & 0 & 0\\ 0 & 0 & -\frac{1}{2} & 0\\ 0 & 0 & 0 & -\frac{3}{2} \end{array}\right]\ \ \ \ \ (39)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar J_{z}^{\left(3/2\right)} \ \ \ \ \ (40)$

Thus the commutation relation ${\left[J_{x},J_{y}\right]=i\hbar J_{z}}$ is satisfied for these three sets of sub-matrices.

# Rotation of a vector wave function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.5.1.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We’ve seen that, for a rotation by an infinitesimal angle ${\varepsilon_{z}}$ about the ${z}$ axis, a scalar wave function transforms according to

$\displaystyle \psi\left(x,y\right)\rightarrow\psi\left(x+\varepsilon_{z}y,y-\varepsilon_{z}x\right) \ \ \ \ \ (1)$

The meaning of this transformation can be seen in the figure:

The physical system represented by the wave function ${\Psi}$ is rigidly rotated by the angle ${\varepsilon_{z}}$, so that the value of ${\Psi}$ at point ${A}$ is now sitting over the point ${B}$. However, in the primed (rotated) coordinate system, the numerical value of the coordinates of the point ${B}$ in the figure are the same as the numerical values that the point ${A}$ had in the original, unrotated coordinates. That is

$\displaystyle \left(x_{B}^{\prime},y_{B}^{\prime}\right)=\left(x_{A},y_{A}\right) \ \ \ \ \ (2)$

Just as ${B}$ is obtained from ${A}$ by rotating ${A}$ by ${+\varepsilon_{z}}$, we can obtain ${A}$ from ${B}$ by rotating by ${-\varepsilon_{z}}$. For any given point, the primed (rotated) and unprimed (unrotated) coordinates are related by (all relations are to first order in ${\varepsilon_{z}}$):

 $\displaystyle x^{\prime}$ $\displaystyle =$ $\displaystyle x-y\varepsilon_{z}\ \ \ \ \ (3)$ $\displaystyle y^{\prime}$ $\displaystyle =$ $\displaystyle y+x\varepsilon_{z} \ \ \ \ \ (4)$

The inverse relations are obtained by a rotation by ${-\varepsilon_{z}}$:

 $\displaystyle x$ $\displaystyle =$ $\displaystyle x^{\prime}+y^{\prime}\varepsilon_{z}\ \ \ \ \ (5)$ $\displaystyle y$ $\displaystyle =$ $\displaystyle y^{\prime}-x^{\prime}\varepsilon_{z} \ \ \ \ \ (6)$

After rotation, the values of ${\Psi^{\prime}}$ are related to the values ${\Psi}$ before rotation by rotating through the angle ${-\varepsilon_{z}}$, so that

$\displaystyle \Psi^{\prime}\left(x,y\right)=\Psi\left(x+y\varepsilon_{z},y-x\varepsilon_{z}\right) \ \ \ \ \ (7)$

Now suppose the wave function is a vector ${\mathbf{V}=V_{x}\hat{\mathbf{x}}+V_{y}\hat{\mathbf{y}}}$. The situation is as shown:

The initial unrotated vector ${\mathbf{V}}$ is the value of the wave function at point ${A}$ (and is entirely in the ${x}$ direction for convenience). After rotation, the vector gets moved to ${B}$ and is also rotated so that it now makes an angle ${\varepsilon_{z}}$ with the original ${x}$ axis. However, its direction is now along the ${x^{\prime}}$ axis, which makes an angle of ${\varepsilon_{z}}$ with the original ${x}$ axis.

In this case, each component of ${\mathbf{V}}$ still gets transformed in the same way as the scalar function above, but the vector itself is also rotated. If the components ${V_{x}}$ and ${V_{y}}$ of the vector were constants, then the rotated vector is given by applying the 2-d rotation matrix

$\displaystyle R=\left[\begin{array}{cc} 1 & -\varepsilon_{z}\\ \varepsilon_{z} & 1 \end{array}\right] \ \ \ \ \ (8)$

so we get ${\mathbf{V}^{\prime}=R\mathbf{V}}$, or, in components:

 $\displaystyle V_{x}^{\prime}$ $\displaystyle =$ $\displaystyle V_{x}-V_{y}\varepsilon_{z}\ \ \ \ \ (9)$ $\displaystyle V_{y}^{\prime}$ $\displaystyle =$ $\displaystyle V_{y}+V_{x}\varepsilon_{z} \ \ \ \ \ (10)$

If ${V_{x}}$ and ${V_{y}}$ vary from point to point, then we must apply the transformation 1 to each component, so that the overall transformation is

 $\displaystyle V_{x}^{\prime}$ $\displaystyle =$ $\displaystyle V_{x}\left(x+\varepsilon_{z}y,y-\varepsilon_{z}x\right)-V_{y}\left(x+\varepsilon_{z}y,y-\varepsilon_{z}x\right)\varepsilon_{z}\ \ \ \ \ (11)$ $\displaystyle V_{y}^{\prime}$ $\displaystyle =$ $\displaystyle V_{y}\left(x+\varepsilon_{z}y,y-\varepsilon_{z}x\right)+V_{x}\left(x+\varepsilon_{z}y,y-\varepsilon_{z}x\right)\varepsilon_{z} \ \ \ \ \ (12)$

The operator that generates the transformation of a scalar function by an infinitesimal angle ${\delta\boldsymbol{\theta}}$ is

$\displaystyle U\left[R\left(\delta\boldsymbol{\theta}\right)\right]=I-\frac{i}{\hbar}\delta\boldsymbol{\theta}\cdot\mathbf{L} \ \ \ \ \ (13)$

In this case, the rotation is about the ${z}$ axis so

 $\displaystyle \delta\boldsymbol{\theta}$ $\displaystyle =$ $\displaystyle \varepsilon_{z}\hat{\mathbf{z}}\ \ \ \ \ (14)$ $\displaystyle \delta\boldsymbol{\theta}\cdot\mathbf{L}$ $\displaystyle =$ $\displaystyle \varepsilon_{z}L_{z} \ \ \ \ \ (15)$

Thus we have

$\displaystyle V_{x,y}\left(x+\varepsilon_{z}y,y-\varepsilon_{z}x\right)=\left(I-\frac{i}{\hbar}\varepsilon_{z}L_{z}\right)V_{x,y}\left(x,y\right) \ \ \ \ \ (16)$

Plugging this into 11 and keeping terms only up to order ${\varepsilon_{z}}$ we have

 $\displaystyle V_{x}^{\prime}$ $\displaystyle =$ $\displaystyle \left(I-\frac{i}{\hbar}\varepsilon_{z}L_{z}\right)V_{x}-V_{y}\varepsilon_{z}\ \ \ \ \ (17)$ $\displaystyle V_{y}^{\prime}$ $\displaystyle =$ $\displaystyle \left(I-\frac{i}{\hbar}\varepsilon_{z}L_{z}\right)V_{y}+V_{x}\varepsilon_{z} \ \ \ \ \ (18)$

In matrix form, this is

 $\displaystyle \left[\begin{array}{c} V_{x}^{\prime}\\ V_{y}^{\prime} \end{array}\right]$ $\displaystyle =$ $\displaystyle \left(\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]-\frac{i\varepsilon_{z}}{\hbar}\left[\begin{array}{cc} L_{z} & 0\\ 0 & L_{z} \end{array}\right]-\varepsilon_{z}\left[\begin{array}{cc} 0 & 1\\ -1 & 0 \end{array}\right]\right)\left[\begin{array}{c} V_{x}\\ V_{y} \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]-\frac{i\varepsilon_{z}}{\hbar}\left[\begin{array}{cc} L_{z} & 0\\ 0 & L_{z} \end{array}\right]-\frac{i\varepsilon_{z}}{\hbar}\left[\begin{array}{cc} 0 & -i\hbar\\ i\hbar & 0 \end{array}\right]\right)\left[\begin{array}{c} V_{x}\\ V_{y} \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(I-\frac{i\varepsilon_{z}}{\hbar}J_{z}\right)\left[\begin{array}{c} V_{x}\\ V_{y} \end{array}\right] \ \ \ \ \ (21)$

This has the same form as 13, except that the angular momentum generator is now the sum of ${L_{z}}$ and the final matrix on the RHS above, which Shankar calls suggestively ${S_{z}}$, in anticipation of spin which at this stage he hasn’t considered. That is,

 $\displaystyle J_{z}$ $\displaystyle =$ $\displaystyle L_{z}+S_{z}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} L_{z} & 0\\ 0 & L_{z} \end{array}\right]+\left[\begin{array}{cc} 0 & -i\hbar\\ i\hbar & 0 \end{array}\right] \ \ \ \ \ (23)$

The eigenvalues of the second matrix here are just ${\pm\hbar}$, so we haven’t yet encountered half-integral values of angular momentum.

# Finite rotations about an arbitrary axis in three dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.4.3.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The operators for an infinitesimal rotation in 3-d are

 $\displaystyle U\left[R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{x}L_{x}}{\hbar}\ \ \ \ \ (1)$ $\displaystyle U\left[R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{y}L_{y}}{\hbar}\ \ \ \ \ (2)$ $\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (3)$

If we have a finite (larger than infinitesimal) rotation about one of the coordinate axes, we can create the operator by dividing up the finite rotation angle ${\theta}$ into ${N}$ small increments and take the limit as ${N\rightarrow\infty}$, just as we did with finite translations. For example, for a finite rotation about the ${x}$ axis, we have

$\displaystyle U\left[R\left(\theta\hat{\mathbf{x}}\right)\right]=\lim_{N\rightarrow\infty}\left(I-\frac{i\theta L_{x}}{N\hbar}\right)^{N}=e^{-i\theta L_{x}/\hbar} \ \ \ \ \ (4)$

What if we have a finite rotation about some arbitrarily directed axis? Suppose we have a vector ${\mathbf{r}}$ as shown in the figure:

The vector ${\mathbf{r}}$ makes an angle ${\alpha}$ with the ${z}$ axis, and we wish to rotate ${\mathbf{r}}$ about the ${z}$ axis by an angle ${\delta\theta}$. Note that this argument is completely general, since if the axis of rotation is not the ${z}$ axis, we can rotate the entire coordinate system so that the axis of rotation is the ${z}$ axis. The generality enters through the fact that we’re keeping the angle ${\alpha}$ arbitrary.

The rotation by ${\delta\theta\hat{\mathbf{z}}\equiv\delta\boldsymbol{\theta}}$ shifts the tip of ${\mathbf{r}}$ along the circle shown by a distance ${\left(r\sin\alpha\right)\delta\theta}$ in a counterclockwise direction (looking down the ${z}$ axis). This shift is in a direction that is perpendicular to both ${\hat{\mathbf{z}}}$ and ${\mathbf{r}}$, so the little vector representing the shift in ${\mathbf{r}}$ is

$\displaystyle \delta\mathbf{r}=\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r} \ \ \ \ \ (5)$

Thus under the rotation ${\delta\boldsymbol{\theta}}$, a vector transforms as

$\displaystyle \mathbf{r}\rightarrow\mathbf{r}+\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r} \ \ \ \ \ (6)$

Just as with translations, if we rotate the coordinate system by an amount ${\delta\boldsymbol{\theta}}$, this is equivalent to rotating the wave function ${\psi\left(\mathbf{r}\right)}$ by the same angle, but in the opposite direction, so we require

$\displaystyle \psi\left(\mathbf{r}\right)\rightarrow\psi\left(\mathbf{r}-\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r}\right) \ \ \ \ \ (7)$

A first order Taylor expansion of the quantity on the RHS gives

$\displaystyle \psi\left(\mathbf{r}-\left(\delta\boldsymbol{\theta}\right)\times\mathbf{r}\right)=\psi\left(\mathbf{r}\right)-\left(\delta\boldsymbol{\theta}\times\mathbf{r}\right)\cdot\nabla\psi \ \ \ \ \ (8)$

The operator generating this rotation will have the form (in analogy with the forms for the coordinate axes above):

$\displaystyle U\left[R\left(\delta\boldsymbol{\theta}\right)\right]=I-\frac{i\delta\theta}{\hbar}L_{\hat{\theta}} \ \ \ \ \ (9)$

where ${L_{\hat{\theta}}}$ is an angular momentum operator to be determined.

Writing out the RHS of 8, we have

 $\displaystyle \psi\left(\mathbf{r}\right)-\left(\delta\boldsymbol{\theta}\times\mathbf{r}\right)\cdot\nabla\psi$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\left(\delta\theta_{y}z-\delta\theta_{z}y\right)\frac{\partial\psi}{\partial x}+\left(\delta\theta_{x}z-\delta\theta_{z}x\right)\frac{\partial\psi}{\partial y}-\left(\delta\theta_{x}y-\delta\theta_{y}x\right)\frac{\partial\psi}{\partial z}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\delta\theta_{x}\left(y\frac{\partial\psi}{\partial z}-z\frac{\partial\psi}{\partial y}\right)-\delta\theta_{y}\left(z\frac{\partial\psi}{\partial x}-x\frac{\partial\psi}{\partial z}\right)-\delta\theta_{z}\left(x\frac{\partial\psi}{\partial y}-y\frac{\partial\psi}{\partial x}\right)\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\delta\boldsymbol{\theta}\cdot\frac{i}{\hbar}\mathbf{r}\times\mathbf{p}\psi\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \psi\left(\mathbf{r}\right)-\frac{i}{\hbar}\delta\boldsymbol{\theta}\cdot\mathbf{L}\psi\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle U\left[R\left(\delta\boldsymbol{\theta}\right)\right]\psi \ \ \ \ \ (14)$

Comparing this with 9, we see that

$\displaystyle L_{\hat{\theta}}=\hat{\boldsymbol{\theta}}\cdot\mathbf{L} \ \ \ \ \ (15)$

where ${\hat{\boldsymbol{\theta}}}$ is the unit vector along the axis of rotation. Since all rotations about the same axis commute, we can use the same procedure as above to generate a finite rotation ${\boldsymbol{\theta}}$ about an arbitrary axis and get

$\displaystyle U\left[R\left(\boldsymbol{\theta}\right)\right]=e^{-i\boldsymbol{\theta}\cdot\mathbf{L}/\hbar} \ \ \ \ \ (16)$

# Angular momentum in three dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.4.2.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We can now generalize our treatment of rotation, originally studied in two dimensions, to three dimensions. We’ll view a 3-d rotation as a combination of rotations about the ${x}$, ${y}$ and ${z}$ axes, each of which can be represented by a ${3\times3}$ matrix. These matrices are as follows:

 $\displaystyle R\left(\theta\hat{\mathbf{x}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & \cos\theta & -\sin\theta\\ 0 & \sin\theta & \cos\theta \end{array}\right]\ \ \ \ \ (1)$ $\displaystyle R\left(\theta\hat{\mathbf{y}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\theta & 0 & \sin\theta\\ 0 & 1 & 0\\ -\sin\theta & 0 & \cos\theta \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle R\left(\theta\hat{\mathbf{z}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (3)$

We’re interested in infinitesimal rotations, for which we retain terms up to first order in the rotation angle ${\varepsilon_{i}}$, so that ${\cos\varepsilon_{i}=1}$ and ${\sin\varepsilon_{i}=\varepsilon_{i}}$. This gives the infinitesimal rotation matrices as

 $\displaystyle R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & -\varepsilon_{x}\\ 0 & \varepsilon_{x} & 1 \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & \varepsilon_{y}\\ 0 & 1 & 0\\ -\varepsilon_{y} & 0 & 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & -\varepsilon_{z} & 0\\ \varepsilon_{z} & 1 & 0\\ 0 & 0 & 1 \end{array}\right] \ \ \ \ \ (6)$

We now consider the series of rotations as follows: first, a rotation by ${\varepsilon_{x}\hat{\mathbf{x}}}$, then by ${\varepsilon_{y}\hat{\mathbf{y}}}$, then by ${-\varepsilon_{x}\hat{\mathbf{x}}}$ and finally by ${-\varepsilon_{y}\hat{\mathbf{y}}}$. Because the various rotations don’t commute, we don’t end up back where we started. We can calculate the matrix products to find the final rotation.

 $\displaystyle R$ $\displaystyle =$ $\displaystyle R\left(-\varepsilon_{y}\hat{\mathbf{y}}\right)R\left(-\varepsilon_{x}\hat{\mathbf{x}}\right)R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & 0 & -\varepsilon_{y}\\ 0 & 1 & 0\\ \varepsilon_{y} & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & \varepsilon_{x}\\ 0 & -\varepsilon_{x} & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & \varepsilon_{y}\\ 0 & 1 & 0\\ -\varepsilon_{y} & 0 & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & 1 & -\varepsilon_{x}\\ 0 & \varepsilon_{x} & 1 \end{array}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1 & \varepsilon_{x}\varepsilon_{y} & -\varepsilon_{y}\\ 0 & 1 & \varepsilon_{x}\\ \varepsilon_{y} & -\varepsilon_{x} & 1 \end{array}\right]\left[\begin{array}{ccc} 1 & \varepsilon_{x}\varepsilon_{y} & \varepsilon_{y}\\ 0 & 1 & -\varepsilon_{x}\\ -\varepsilon_{y} & \varepsilon_{x} & 1 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{ccc} 1+\varepsilon_{y}^{2} & \varepsilon_{x}\varepsilon_{y} & -\varepsilon_{x}^{2}\varepsilon_{y}\\ -\varepsilon_{x}\varepsilon_{y} & 1+\varepsilon_{x}^{2} & 0\\ 0 & \varepsilon_{x}\varepsilon_{y}^{2} & 1+\varepsilon_{x}^{2}+\varepsilon_{y}^{2} \end{array}\right] \ \ \ \ \ (10)$

To get the third line, we multiplied the first two matrices in the second line, and the last two matrices in the second line. In the final result, we can discard terms containing ${\varepsilon_{x}^{2}}$ or ${\varepsilon_{y}^{2}}$ to get

$\displaystyle R=\left[\begin{array}{ccc} 1 & \varepsilon_{x}\varepsilon_{y} & 0\\ -\varepsilon_{x}\varepsilon_{y} & 1 & 0\\ 0 & 0 & 1 \end{array}\right]=R\left(-\varepsilon_{x}\varepsilon_{y}\hat{\mathbf{z}}\right) \ \ \ \ \ (11)$

Thus the result of the four rotations about the ${x}$ and ${y}$ axes is a single rotation about the ${z}$ axis.

To convert this to quantum operators, we define the operator ${U\left[R\right]}$ by comparison with the procedure we used for 2-d rotations. That is, the operator ${U}$ is given by the corresponding angular momentum operator ${L_{x}}$, ${L_{y}}$ or ${L_{z}}$ as

 $\displaystyle U\left[R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{x}L_{x}}{\hbar}\ \ \ \ \ (12)$ $\displaystyle U\left[R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{y}L_{y}}{\hbar}\ \ \ \ \ (13)$ $\displaystyle U\left[R\left(\varepsilon_{z}\hat{\mathbf{z}}\right)\right]$ $\displaystyle =$ $\displaystyle I-\frac{i\varepsilon_{z}L_{z}}{\hbar} \ \ \ \ \ (14)$

By comparing 7 and 11 we thus require these ${U}$ operators to satisfy

$\displaystyle U\left[R\left(-\varepsilon_{y}\hat{\mathbf{y}}\right)\right]U\left[R\left(-\varepsilon_{x}\hat{\mathbf{x}}\right)\right]U\left[R\left(\varepsilon_{y}\hat{\mathbf{y}}\right)\right]U\left[R\left(\varepsilon_{x}\hat{\mathbf{x}}\right)\right]=U\left[R\left(-\varepsilon_{x}\varepsilon_{y}\hat{\mathbf{z}}\right)\right] \ \ \ \ \ (15)$

We can get the commutation relation ${\left[L_{x},L_{y}\right]}$ by matching coefficients of ${\varepsilon_{x}\varepsilon_{y}}$ on each side of this equation. On the RHS, the coefficient is ${\frac{iL_{z}}{\hbar}}$. On the LHS, we can pick out the terms involving ${\varepsilon_{x}\varepsilon_{y}}$ to get

$\displaystyle -\frac{1}{\hbar^{2}}\left(L_{y}L_{x}-L_{y}L_{x}-L_{x}L_{y}+L_{y}L_{x}\right)=\frac{1}{\hbar^{2}}\left[L_{x},L_{y}\right] \ \ \ \ \ (16)$

The first term on the LHS comes from the ${\varepsilon_{x}}$ term in the first ${U}$ in 15 multiplied by the ${\varepsilon_{y}}$ term in the second ${U}$ (with the ${I}$ term in the other two ${U}$s); the second term on the LHS comes from the ${\varepsilon_{x}}$ term in the first ${U}$ in 15 multiplied by the ${\varepsilon_{y}}$ term in the fourth ${U}$, and so on.

Matching the two sides, we get

$\displaystyle \left[L_{x},L_{y}\right]=i\hbar L_{z} \ \ \ \ \ (17)$

By comparison with the classical definitions of the three components of ${\mathbf{L}}$, we can write the quantum operators in terms of position and momentum operators as

 $\displaystyle L_{x}$ $\displaystyle =$ $\displaystyle YP_{z}-ZP_{y}\ \ \ \ \ (18)$ $\displaystyle L_{y}$ $\displaystyle =$ $\displaystyle ZP_{x}-XP_{z}\ \ \ \ \ (19)$ $\displaystyle L_{z}$ $\displaystyle =$ $\displaystyle XP_{y}-YP_{x} \ \ \ \ \ (20)$

From the commutators of position and momentum ${\left[X,P_{x}\right]=i\hbar}$ and so on, we can verify 17 from these relations as well.

 $\displaystyle \left[L_{x},L_{y}\right]$ $\displaystyle =$ $\displaystyle \left[YP_{z}-ZP_{y},ZP_{x}-XP_{z}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[YP_{z},ZP_{x}-XP_{z}\right]-\left[ZP_{y},ZP_{x}-XP_{z}\right]\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -i\hbar YP_{x}+i\hbar P_{y}X\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(XP_{y}-YP_{x}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar L_{z} \ \ \ \ \ (25)$

The third line follows because ${\left[YP_{z},XP_{z}\right]=\left[ZP_{y},ZP_{x}\right]=0}$. The other two commutation relations follow by cyclic permutation of ${x}$, ${y}$ and ${z}$:

 $\displaystyle \left[L_{y},L_{z}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{x}\ \ \ \ \ (26)$ $\displaystyle \left[L_{z},L_{x}\right]$ $\displaystyle =$ $\displaystyle i\hbar L_{y} \ \ \ \ \ (27)$

# Angular momentum of circular motion

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.6.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

A particle of mass ${\mu}$ constrained to move (at constant speed ${v}$, we assume) on a circle of radius ${a}$ centred at the origin in the ${xy}$ plane has a constant kinetic energy of ${\frac{1}{2}\mu v^{2}}$. As its momentum ${\mathbf{p}}$ is always perpendicular to the radius vector ${\mathbf{r}}$, the angular momentum is given by

$\displaystyle \mathbf{L}=\mathbf{r}\times\mathbf{p}=\mu av\hat{\mathbf{z}}=L_{z}\hat{\mathbf{z}} \ \ \ \ \ (1)$

The energy can thus be written as

$\displaystyle H=\frac{1}{2}\mu v^{2}=\frac{L_{z}^{2}}{2\mu a^{2}} \ \ \ \ \ (2)$

In polar coordinates, the angular momentum operator is

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (3)$

The eigenvalue problem for this system is therefore

 $\displaystyle H\psi$ $\displaystyle =$ $\displaystyle E\psi\ \ \ \ \ (4)$ $\displaystyle -\frac{\hbar^{2}}{2\mu a^{2}}\frac{\partial^{2}\psi}{\partial\phi^{2}}$ $\displaystyle =$ $\displaystyle E\psi \ \ \ \ \ (5)$

The eigenvalues of ${L_{z}}$ are the solutions of

 $\displaystyle -\hbar^{2}\frac{\partial^{2}\psi}{\partial\phi^{2}}$ $\displaystyle =$ $\displaystyle \ell_{z}\psi \ \ \ \ \ (6)$

which are

$\displaystyle \psi=Ae^{i\ell_{z}\phi/\hbar}=Ae^{im\phi} \ \ \ \ \ (7)$

for some constant ${A}$, with the quantization condition (arising from the requirement that ${\psi\left(\phi+2\pi\right)=\psi\left(\phi\right)}$)

$\displaystyle \ell_{z}=m\hbar \ \ \ \ \ (8)$

where ${m}$ is an integer (positive, negative or zero). Plugging this into 5 we find

$\displaystyle E=\frac{\hbar^{2}m^{2}}{2\mu a^{2}} \ \ \ \ \ (9)$

Each energy is two-fold degenerate since ${\pm m}$ both give the same energy. This corresponds to the particle moving round the circle in the clockwise or counterclockwise direction.

# Radially symmetric potentials, angular momentum and centrifugal force

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.5.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We’ve seen that the eigenfunctions of two-dimensional angular momentum have the form

$\displaystyle \psi\left(\rho,\phi\right)=R\left(\rho\right)\Phi_{m}\left(\phi\right) \ \ \ \ \ (1)$

where

$\displaystyle \Phi_{m}\left(\phi\right)=\frac{1}{\sqrt{2\pi}}e^{im\phi} \ \ \ \ \ (2)$

In 2 dimensions and polar coordinates, the hamiltonian can be written as

$\displaystyle H=-\frac{\hbar^{2}}{2\mu}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+V\left(\rho,\phi\right) \ \ \ \ \ (3)$

If the potential is radially symmetric, that is, it doesn’t depend on ${\phi}$, then

$\displaystyle H=-\frac{\hbar^{2}}{2\mu}\left(\frac{\partial^{2}}{\partial\rho^{2}}+\frac{1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}\right)+V\left(\rho\right) \ \ \ \ \ (4)$

In polar coordinates, the angular momentum operator has the form

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (5)$

Thus ${L_{z}}$ commutes with every term in the hamiltonian 4, so for ${V=V\left(\rho\right)}$, we find

$\displaystyle \left[H,L_{z}\right]=0 \ \ \ \ \ (6)$

meaning that we can find a set of functions that are simultaneously eigenfunctions of both ${H}$ and ${L_{z}}$. Since we already know what the most general eigenfunctions of ${L_{z}}$ are (eqn 1), the problem is then to find the radial function ${R\left(\rho\right)}$ so that

$\displaystyle H\left[R\left(\rho\right)\Phi_{m}\left(\phi\right)\right]=ER\left(\rho\right)\Phi_{m}\left(\phi\right) \ \ \ \ \ (7)$

If we use 4 for ${H}$ and 2 for ${\Phi}$ we find that we must solve the differential equation

$\displaystyle -\frac{\hbar^{2}}{2\mu}\left(\frac{d^{2}R}{d\rho^{2}}+\frac{1}{\rho}\frac{dR}{d\rho}-\frac{m^{2}}{\rho^{2}}R\right)+V\left(\rho\right)R=ER \ \ \ \ \ (8)$

We’ve replaced the partial derivatives in 4 by ordinary derivatives, since we now have an ODE in one independent variable, namely ${\rho}$.

The term arising from the ${\frac{1}{\rho^{2}}\frac{\partial^{2}}{\partial\phi^{2}}}$ term in 4 is similar to a potential term, since it doesn’t involve any derivatives of ${R}$. The potential term is

$\displaystyle V_{c}=\frac{\hbar^{2}}{2\mu}\frac{m^{2}}{\rho^{2}} \ \ \ \ \ (9)$

We can find the force corresponding to ${V_{c}}$ by taking the gradient, which in this case amounts to

$\displaystyle F_{c}=\frac{\partial V_{c}}{\partial\rho}=-\frac{\hbar^{2}m^{2}}{\mu\rho^{3}} \ \ \ \ \ (10)$

Since the quantum angular momentum is ${\ell_{z}=m\hbar}$, this can be written as

$\displaystyle F_{c}=-\frac{\ell_{z}^{2}}{\mu\rho^{3}} \ \ \ \ \ (11)$

If the particle is in a circular orbit, then ${\ell_{z}=\rho p}$ where ${p}$ is its momentum, so this becomes

$\displaystyle F_{c}=-\frac{p^{2}}{\mu\rho} \ \ \ \ \ (12)$

Classically, ${p=\mu v^{2}}$ so this is equivalent to

$\displaystyle F_{c}=-\frac{\mu v^{2}}{\rho} \ \ \ \ \ (13)$

which is the formula for centripetal force in Newtonian physics. (Shankar calls it the centrifugal force, but the minus sign indicates it acts towards the centre of rotation rather than outwards, and of course as we well know, the centrifugal force is a fictitious force anyway.)

# Angular momentum: probabilities of eigenvalues in two dimensions

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercises 12.3.3 – 12.3.4.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

We’ve seen that the eigenfunctions of two-dimensional angular momentum have the form

$\displaystyle \psi\left(\rho,\phi\right)=R\left(\rho\right)e^{i\ell_{z}\phi/\hbar} \ \ \ \ \ (1)$

where ${\ell_{z}}$ (the eigenvalue) is an integral multiple of ${\hbar}$ and ${R\left(\rho\right)}$ is some function of the radial coordinate ${\rho}$ which depends on the particular potential function in the hamiltonian. It’s more convenient to write the angular function as

$\displaystyle \Phi_{m}\left(\phi\right)=\frac{1}{\sqrt{2\pi}}e^{im\phi} \ \ \ \ \ (2)$

This set of functions is orthonormal over the interval ${\phi\in\left[0,2\pi\right]}$, that is

$\displaystyle \int_{0}^{2\pi}\Phi_{m}^*\left(\phi\right)\Phi_{m^{\prime}}\left(\phi\right)d\phi=\delta_{mm^{\prime}} \ \ \ \ \ (3)$

This set of functions forms the angular part of the eigenfunctions of ${L_{z}}$, which in some cases allows us to determine the probabilities of a system being in a particular eigenstate of ${L_{z}}$. Here are a couple of examples.

Example 1 A particle is described by the wave function

$\displaystyle \psi\left(\rho,\phi\right)=Ae^{-\rho^{2}/2\Delta^{2}}\cos^{2}\phi \ \ \ \ \ (4)$

where ${A}$ is a normalization constant, and ${\Delta}$ is another constant.

We can use the trig identity

$\displaystyle \cos^{2}\phi=\frac{1}{2}\left(1+\cos2\phi\right) \ \ \ \ \ (5)$

to write this wave function as

 $\displaystyle \psi\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle \frac{A}{2}e^{-\rho^{2}/2\Delta^{2}}\left[1+\cos2\phi\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A}{2}e^{-\rho^{2}/2\Delta^{2}}\left(1+\frac{e^{2i\phi}+e^{-2i\phi}}{2}\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{A\sqrt{2\pi}}{2}e^{-\rho^{2}/2\Delta^{2}}\left(\Phi_{0}+\frac{1}{2}\left(\Phi_{2}+\Phi_{-2}\right)\right) \ \ \ \ \ (8)$

Thus the wave function has the form

$\displaystyle \psi\left(\rho,\phi\right)=c_{0}\Phi_{0}+c_{2}\Phi_{2}+c_{-2}\Phi_{-2} \ \ \ \ \ (9)$

where the coefficients ${c_{m}}$ can be found by comparison with 8. Since the ${\Phi_{m}}$ are orthonormal functions, the probability of the particle being in state ${i}$ is

$\displaystyle P\left(\ell_{z}=m\hbar\right)=\frac{\left|c_{m}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}} \ \ \ \ \ (10)$

We can see from this formula that the factor of ${\frac{A\sqrt{2\pi}}{2}e^{-\rho^{2}/2\Delta^{2}}}$ cancels out of the probability formula, so we have

 $\displaystyle P\left(\ell_{z}=0\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{0}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{1+\frac{1}{4}+\frac{1}{4}}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{3}\ \ \ \ \ (13)$ $\displaystyle P\left(\ell_{z}=2\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{2}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{4}}{1+\frac{1}{4}+\frac{1}{4}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{6}\ \ \ \ \ (16)$ $\displaystyle P\left(\ell_{z}=-2\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{-2}\right|^{2}}{\sum_{j}\left|c_{j}\right|^{2}}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\frac{1}{4}}{1+\frac{1}{4}+\frac{1}{4}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{6} \ \ \ \ \ (19)$

Example 2 Now we have the wave function

$\displaystyle \psi\left(\rho,\phi\right)=Ae^{-\rho^{2}/2\Delta^{2}}\left(\frac{\rho}{\Delta}\cos\phi+\sin\phi\right) \ \ \ \ \ (20)$

Again, we write the trig functions in terms of ${\Phi_{m}}$ to get

 $\displaystyle \psi\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle Ae^{-\rho^{2}/2\Delta^{2}}\left(\frac{\rho}{\Delta}\frac{e^{i\phi}+e^{-i\phi}}{2}+\frac{e^{i\phi}-e^{-i\phi}}{2i}\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\sqrt{2\pi}e^{-\rho^{2}/2\Delta^{2}}\left[\left(\frac{\rho}{2\Delta}+\frac{1}{2i}\right)\Phi_{1}+\left(\frac{\rho}{2\Delta}-\frac{1}{2i}\right)\Phi_{-1}\right] \ \ \ \ \ (22)$

As above, the factor of ${A\sqrt{2\pi}e^{-\rho^{2}/2\Delta^{2}}}$ cancels out when calculating probabilities, so we have

 $\displaystyle P\left(\ell_{z}=\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{1}\right|^{2}}{\left|c_{1}\right|^{2}+\left|c_{-1}\right|^{2}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\frac{\rho}{2\Delta}+\frac{1}{2i}\right|^{2}}{\left|\frac{\rho}{2\Delta}+\frac{1}{2i}\right|^{2}+\left|\frac{\rho}{2\Delta}-\frac{1}{2i}\right|^{2}}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}}{2\left[\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}\right]}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\ \ \ \ \ (26)$ $\displaystyle P\left(\ell_{z}=-\hbar\right)$ $\displaystyle =$ $\displaystyle \frac{\left|c_{-1}\right|^{2}}{\left|c_{1}\right|^{2}+\left|c_{-1}\right|^{2}}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\frac{\rho}{2\Delta}-\frac{1}{2i}\right|^{2}}{\left|\frac{\rho}{2\Delta}+\frac{1}{2i}\right|^{2}+\left|\frac{\rho}{2\Delta}-\frac{1}{2i}\right|^{2}}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}}{2\left[\left(\frac{\rho}{2\Delta}\right)^{2}+\frac{1}{4}\right]}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2} \ \ \ \ \ (30)$

Thus in this case, the ${\rho}$ dependence cancels out when calculating the probabilities, although we can’t expect this to be true in general.

# Eigenvalues of angular momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.2.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

One consequence of requiring the angular momentum operator ${L_{z}}$ to be hermitian is that the eigenvalues must be integral multiples of ${\hbar}$, so that ${\ell_{z}=m\hbar}$ for ${m=0,\pm1,\pm2,\ldots}$. Shankar proposes another method by which we might try to obtain this restriction on ${\ell_{z}}$. We start with a superposition of two eigenstates of ${L_{z}}$, so that

 $\displaystyle \psi\left(\rho,\phi\right)$ $\displaystyle =$ $\displaystyle A\left(\rho\right)e^{i\phi\ell_{z}/\hbar}+B\left(\rho\right)e^{i\phi\ell_{z}^{\prime}/\hbar}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{i\phi\ell_{z}^{\prime}/\hbar}\left[A\left(\rho\right)e^{i\phi\left(\ell_{z}-\ell_{z}^{\prime}\right)/\hbar}+B\left(\rho\right)\right] \ \ \ \ \ (2)$

where ${A}$ and ${B}$ are two unknown functions of the radial coordinate ${\rho}$, and ${\ell_{z}}$ and ${\ell_{z}^{\prime}}$ are two eigenvalues of ${L_{z}}$. If we rotate the system by a complete circle, so that ${\phi\rightarrow\phi+2\pi}$, the physical state should remain unchanged. This means that

$\displaystyle \left|\psi\left(\rho,\phi+2\pi\right)\right|=\left|\psi\left(\rho,\phi\right)\right| \ \ \ \ \ (3)$

so that ${\psi\left(\rho,\phi+2\pi\right)}$ may differ from ${\psi\left(\rho,\phi\right)}$ by a phase factor. From 2

$\displaystyle \psi\left(\rho,\phi+2\pi\right)=e^{i\left(\phi+2\pi\right)\ell_{z}^{\prime}/\hbar}\left[A\left(\rho\right)e^{i\left(\phi+2\pi\right)\left(\ell_{z}-\ell_{z}^{\prime}\right)/\hbar}+B\left(\rho\right)\right] \ \ \ \ \ (4)$

The phase factor of ${e^{i\left(\phi+2\pi\right)\ell_{z}^{\prime}/\hbar}}$ on the RHS can be anything (provided the exponent is purely imaginary), but the quantity in the square brackets must be numerically the same as the corresponding quantity in 2. This means that

$\displaystyle \frac{\left(\phi+2\pi\right)\left(\ell_{z}-\ell_{z}^{\prime}\right)}{\hbar}=\frac{\phi\left(\ell_{z}-\ell_{z}^{\prime}\right)}{\hbar}+2m\pi \ \ \ \ \ (5)$

where ${m}$ is an integer. This gives the condition

$\displaystyle \ell_{z}-\ell_{z}^{\prime}=m\hbar \ \ \ \ \ (6)$

To proceed further, we need to argue that ${\ell_{z}}$ is symmetric about zero, that is, if ${\ell_{z}}$ is an eigenvalue, then so is ${-\ell_{z}}$. I’m not sure if Shankar expects us to prove this rigorously, but it seems plausible, since the only difference between ${+\ell_{z}}$ and ${-\ell_{z}}$ is (classically, anyway) that the direction of rotation is reversed. Given this condition, ${\ell_{z}}$ must be a multiple of ${\frac{1}{2}\hbar}$, since any other value doesn’t satisfy both the conditions of symmetry about zero, and 6. (For example, if we try ${\ell_{z}=\frac{1}{4}\hbar}$, then the symmetry requirement means we must also allow ${\ell_{z}=-\frac{1}{4}\hbar}$, but this violates 6.) If ${\ell_{z}}$ is an odd multiple of ${\frac{1}{2}\hbar}$, then we get the sequence ${\ldots,-\frac{3}{2}\hbar}$,-${\frac{1}{2}\hbar,+\frac{1}{2}\hbar,+\frac{3}{2}\hbar,\ldots}$ while if ${\ell_{z}}$ is an even multiple of ${\frac{1}{2}\hbar}$ we get the sequence ${\ldots,-2\hbar,-\hbar,0,+\hbar,+2\hbar,\ldots}$. In reality, only the latter sequence is correct, but we can’t show that from this argument.

# Eigenvalues of two-dimensional angular momentum

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.3.1.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

The angular momentum operator ${L_{z}}$ for rotations in two dimensions has the form, in polar coordinates, of

$\displaystyle L_{z}=-i\hbar\frac{\partial}{\partial\phi} \ \ \ \ \ (1)$

To find the eigenvalues and eigenfunctions, we need to solve

$\displaystyle L_{z}\left|\ell_{z}\right\rangle =\ell_{z}\left|\ell_{z}\right\rangle \ \ \ \ \ (2)$

where ${\left|\ell_{z}\right\rangle }$ is the eigenfunction and ${\ell_{z}}$ is the corresponding eigenvalue. Using polar coordinates, we must solve

$\displaystyle -i\hbar\frac{\partial}{\partial\phi}\psi_{\ell_{z}}\left(\rho,\phi\right)=\ell_{z}\psi_{\ell_{z}}\left(\rho,\phi\right) \ \ \ \ \ (3)$

where ${\rho}$ is the radial coordinate. As the only derivative here is with respect to ${\phi}$, we can solve this using separation of variables by proposing a solution of form

$\displaystyle \psi_{\ell_{z}}\left(\rho,\phi\right)=R\left(\rho\right)\Phi\left(\phi\right) \ \ \ \ \ (4)$

Substituting this and cancelling off ${R\left(\rho\right)}$ we get

$\displaystyle -i\hbar\frac{\partial}{\partial\phi}\Phi\left(\phi\right)=\ell_{z}\Phi\left(\phi\right) \ \ \ \ \ (5)$

which has the solution

$\displaystyle \Phi\left(\phi\right)=Ae^{i\ell_{z}\phi/\hbar} \ \ \ \ \ (6)$

for some constant ${A}$, which we can absorb into ${R\left(\rho\right)}$ to give the general solution

$\displaystyle \psi_{\ell_{z}}\left(\rho,\phi\right)=R\left(\rho\right)e^{i\ell_{z}\phi/\hbar} \ \ \ \ \ (7)$

[This is actually the two-dimensional version of the more general 3-d case, in which the solution involved a radial function multiplied by a spherical harmonic.]

At this stage, the eigenvalue ${\ell_{z}}$ could be any number, real or complex, since they all satisfy 3. However, since ${L_{z}}$ is an observable, it must be hermitian, which implies that ${L_{z}^{\dagger}=L_{z}}$, so that

$\displaystyle \left\langle \psi_{1}\left|L_{z}\right|\psi_{2}\right\rangle =\left\langle \psi_{2}\left|L_{z}\right|\psi_{1}\right\rangle ^* \ \ \ \ \ (8)$

In the coordinate basis, we have

$\displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\psi_{1}^*\left(-i\hbar\frac{\partial}{\partial\phi}\right)\psi_{2}d\phi\;d\rho=\left[\int_{0}^{\infty}\int_{0}^{2\pi}\psi_{2}^*\left(-i\hbar\frac{\partial}{\partial\phi}\right)\psi_{1}d\phi\;d\rho\right]^* \ \ \ \ \ (9)$

Integrating the LHS by parts, we have

$\displaystyle \int_{0}^{\infty}\int_{0}^{2\pi}\psi_{1}^*\left(-i\hbar\frac{\partial}{\partial\phi}\right)\psi_{2}d\phi\;d\rho=-i\hbar\int_{0}^{\infty}\left.\psi_{1}^*\psi_{2}\right|_{0}^{2\pi}d\rho+i\hbar\int_{0}^{\infty}\int_{0}^{2\pi}\frac{\partial\psi_{1}^*}{\partial\phi}\psi_{2}d\phi\;d\rho \ \ \ \ \ (10)$

The second term on the RHS is seen to be equal to the RHS of 9, so in order for 9 to be true, we must have

$\displaystyle \int_{0}^{\infty}\left.\psi_{1}^*\psi_{2}\right|_{0}^{2\pi}d\rho=0 \ \ \ \ \ (11)$

Although two different eigenfunctions ${\psi_{1}}$ and ${\psi_{2}}$ are orthogonal and thus would satisfy this condition automatically, the condition must also be true when ${\psi_{1}=\psi_{2}}$. This gives us the condition that

$\displaystyle \psi_{\ell_{z}}\left(2\pi\right)=\psi_{\ell_{z}}\left(0\right) \ \ \ \ \ (12)$

That is, the eigenfunctions must be periodic with period ${2\pi}$. Looking back at 7, we see that this forces the eigenvalues ${\ell_{z}}$ to be integral multiples of ${\hbar}$:

 $\displaystyle \ell_{z}$ $\displaystyle =$ $\displaystyle m\hbar\ \ \ \ \ (13)$ $\displaystyle m$ $\displaystyle =$ $\displaystyle 0,\pm1,\pm2,\ldots \ \ \ \ \ (14)$

Here ${m}$ is the magnetic quantum number, not the mass.