A.20 Angular momentum of vector particles

This addendum is concerned with vector particles, particles whose wave functions are vectors. To be sure, the wave function of an electron can also be written as a vector, chapters 3.1 and 5.5.1:

$$\mbox{electron:} \quad \skew{-1}\vec\Psi({\skew0\vec r};... ... r};t) \\ \Psi_-({\skew0\vec r};t) \end{array} \right)$$

But that is not a normal vector. It is a two-di­mensional vector in three-di­mensional space, and is known as a spinor. This addendum is concerned with wave functions that are normal three-di­mensional vectors. That is of importance for under­standing, for example, the spin angular momentum of photons. A photon is a vector particle, though a special one. It will be shown in this addendum that the spin of a vector particle is 1. The parity of such a particle will also be discussed.

To really appreciate this addendum, you may want to read the previous addendum {A.19} first. In any case, according to that addendum angular momentum is related to what happens to the wave function under rotation of the coordinate system. In particular, the angular momentum in the $z$-direction is related to what happens if the coordinate system is rotated around the $z$-axis.

Consider first the simplest possible vector wave function:

$$\vec\Psi = \skew3\vec A' f(r) \qquad \skew3\vec A' = \left(\begin{array}{c} A_x' \\ A_y' \\ A_z' \end{array}\right)$$

Here $\skew3\vec A'$ is a constant vector. Also $r$ is the distance from the origin around which the angular momentum is defined. Finally, $f(r)$ is any arbitrary function of $r$. The big question is now what happens to this wave function if the coordinate system is rotated over some angle $\gamma$ around the $z$-axis.

Figure A.8: Effect of rotation of the coordinate system on a vector. The vector is physi­cally the same, but it has a different mathematical represen­tation, different components, in the two coordinate systems.

The factor $f(r)$ does not change under rotations because the distance from the origin does not. But the vector $\skew3\vec A'$ does change. Figure A.8 shows what happens. The vector in the rotated coordinate system $x,y,z$ has components

$$A_x = \cos(\gamma) A_x' + \sin(\gamma) A_y' \qquad A_y... ...sin(\gamma) A_x' + \cos(\gamma) A_y' \qquad A_z = A_z' %$$ (A.82)

For example, the first relation expresses that $A_x'{\hat\imath}^{\,\prime}$ has a component $\cos(\gamma)A_x'$ in the direction of ${\hat\imath}$, while $A_y'{\hat\jmath}^{\,\prime}$ has a component $\sin(\gamma)A_y'$ in that direction. The second expression follows similarly. The $z$-component does not change under the rotation.

Now consider three very special vectors:

$$\fbox{$\displaystyle \vec Y_1^1 = \left(\begin{array}{... ...-}{\rm i}/\sqrt{2} \\ \phantom{-}0 \end{array}\right) $} %$$ (A.83)

If you plug $\vec{A'}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vec{Y}_1^1$ into the relations (A.82) given above and use the Euler identity (2.5), you get $\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{{\rm i}\gamma}\vec{Y}_1^1$. So the vector $\vec{Y}_1^1$ changes by a mere scalar factor, the exponential $e^{{\rm i}\gamma}$, under the rotation of the coordinate system. According to the relationship between rotations and angular momentum, {A.19}, that makes $\vec{Y}_1^1$ a state of definite angular momentum in the $z$-direction. Also, the magnetic quantum number $m_s$ of the momentum in the $z$-direction is by definition the coefficient of ${\rm i}\gamma$ in the exponential factor $e^{{\rm i}\gamma}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^{1{\rm i}\gamma}$. Here that is 1, so $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. This value is shown as the super­script of the state $\vec{Y}_1^1$. The actual angular momentum in the $z$-direction is $m_s\hbar$, so it is $\hbar$ for this state. This angular momentum should be called spin, in analogy to the case for the electron. It is due to the fact that the wave function is a vector.

The vector $\vec{Y}_1^0$ has only a $z$-component, so it does not change under the rotation. Phrased differen­tly, it changes by a unit factor $e^{0{\rm i}\gamma}$. That makes its magnetic quantum number $m_s$ zero, as the super­script in $\vec{Y}_1^0$ says. Then the angular momentum in the $z$-direction of this state is zero too. Finally, the vector $\vec{Y}_1^{-1}$ changes by a factor $e^{-1{\rm i}\gamma}$ under rotation, so it has $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1, as its super­script says, and the angular momentum in the $z$-direction is $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar$.

To get at square angular momentum, first the operator ${\widehat S}_z$ of spin angular momentum around the $z$-axis is needed. The relation between angular momentum and rotations shows that this operator takes the general form, {A.19} (A.76),

$${\widehat S}_z = \frac{\hbar}{{\rm i}} \lim_{\gamma\to 0}\frac{{\cal R}_{z,\gamma}-1}{\gamma}$$

Here ${\cal R}_{z,\gamma}$ is the operator that describes the effect of the rotation of the coordinate system on the wave function. Applied on the vector $\skew3\vec A'$ in the unrotated coordinate system, that means

$${\widehat S}_z \skew3\vec A' = \frac{\hbar}{{\rm i}} \lim_{\gamma\to 0}\frac{\skew3\vec A- \skew3\vec A'}{\gamma}$$

Plugging in the components of $\skew3\vec A$ as given earlier, (A.82), and taking the limits using l’Hôpital, that produces

$${\widehat S}_z \left(\begin{array}{c} A_x' \\ A_y' \\ A_z'... ...\phantom{-}A_y' \\ {-}A_x' \\ \phantom{-}0 \end{array}\right)$$

So the operator ${\widehat S}_z$ drops the $z$-component and swaps the other two components, changing the sign of the first, and then adds a factor $\hbar$$\raisebox{.5pt}{$/$}$${\rm i}$. If the same operations are performed another time, the net result is:

$${\widehat S}_z^2 \left(\begin{array}{c} A_x' \\ A_y' \\ A_... ... \left(\begin{array}{c} A_x' \\ A_y' \\ 0 \end{array}\right)$$

So the square operator just drops the $z$-component and adds a factor $\hbar^2$.

Of course, the operators ${\widehat S}_x^2$ and ${\widehat S}_y^2$ are defined similarly. There is nothing special about the $z$-axis. The operator of square spin angular momentum is defined as

$${\widehat S}^2 \equiv {\widehat S}_x^2 + {\widehat S}_y^2 + {\widehat S}_z^2$$

Since each of the operators in the right hand side drops a different component and adds a factor $\hbar^2$ to the other two, the total for any vector $\skew3\vec A'$ is,

$${\widehat S}^2 \skew3\vec A' = 2 \hbar^2 \skew3\vec A'$$

So the square spin operator always produces a simple multiple of the original vector. That makes any vector an eigen­vector of square spin angular momentum. Also, the azimuthal quantum number $s$, the spin, can by definition be found from equating the coefficient $2\hbar^2$ of $\skew3\vec A'$ in the right hand side above to $s(s+1)\hbar^2$. The only non­negative value $s$ that can satisfy this condition is $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

That then means that the spin $s$ of vector particles is equal to 1. So a vector particle is a boson of spin 1. The subscript on the special vectors $\vec{Y}_1^{m_s}$ indicates their spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1.

You can write the most general vector wave function in the form

$$\skew{-1}\vec\Psi({\skew0\vec r};t) = \Psi_1({\skew0\vec r... ...ec r};t)\vec Y_1^1 + \Psi_{-1}({\skew0\vec r};t)\vec Y_1^{-1}$$

Then you can put the coefficients in a vector much like the wave function of the electron, but now three-di­mensional:

$$\mbox{vector boson:}\quad \skew{-1}\vec\Psi({\skew0\vec ... ...;t) \\ \Psi_{-1}({\skew0\vec r};t) \end{array} \right)$$

Like the electron, the vector particle can of course also have orbital angular momentum. That is due to the coefficients $\Psi_{m_s}$ in the wave function above. So far it has been assumed that these coefficients only depended on the distance $r$ from the origin. However, consider the following more general component of a vector wave function:

$$\vec Y_1^{m_s} f(r) Y_l^{m_l}(\theta,\phi) %$$ (A.84)

Here $\theta$ and $\phi$ are the position angles in spherical coordinates, and $Y_l^{m_l}$ is a so-called spherical harmonic of orbital angular momentum, chapter 4.2.3. For the above wave function to be properly normalized,

$$\int_{r=0}^\infty r^2 f(r){\,\rm d}r = 1$$

(To check this, take a dot, or rather inner, product of the wave function with itself. Then integrate over all space using spherical coordinates and the ortho­normality of the spherical harmonics.)

The wave function (A.84) above has orbital angular momentum in the $z$-direction equal to $m_l\hbar$ in addition to the spin angular momentum $m_s\hbar$. So the total angular momentum in the $z$-direction is $(m_s+m_l)\hbar$. To check that, note that under rotations of the coordinate system, the vector changes by a factor $e^{m_s{\rm i}\gamma}$ while the spherical harmonic changes by an additional factor $e^{m_l{\rm i}\gamma}$. That makes the total change a factor $e^{(m_s+m_l){\rm i}\gamma}$.

In general, then, the magnetic quantum number $m_j$ of the net angular momentum is simply the sum of the spin and orbital ones,

$$\fbox{$\displaystyle m_j = m_s + m_l $} %$$ (A.85)

However, the situation for the azimuthal quantum number $j$ of the net angular momentum is not so simple. In general the wave function (A.84) above will have uncertainty in the value of $j$. Combin­ations of wave functions of the form (A.84) are usually needed to get states of definite $j$.

That is a complicated issue best left to chapter 12. But a couple of special cases are worth mentioning already. First, if $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 and $m_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $l$, or alternatively, if $m_s$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$1 and $m_l$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$l$, then $j$ is simply the sum of the spin and orbital azimuthal quantum numbers $1+l$.

The other special case is that there is zero net angular momentum. Zero net angular momentum means that the wave function is exactly the same regardless how the coordinate system is rotated. And that only happens for a vector wave function if it is purely radial:

$${\hat\imath}_r f(r) \frac{1}{\sqrt{4\pi}}$$

Here ${\hat\imath}_r$ is the unit vector sticking radially outward away from the origin. The final constant is the spherical harmonic $Y_0^0$. It is needed the satisfy the normal­ization requirement unless you change the one on $f$.

The above state has zero net angular momentum. The question of interest is what can be said about its spin and orbital angular momentum. To answer that, it must be rewritten in terms of Cartesian components. Now the unit vector ${\hat\imath}_r$ has Cartesian components

$${\hat\imath}_r = \frac{x}{r} {\hat\imath}+ \frac{y}{r} {\hat\jmath}+ \frac{z}{r} {\hat k}$$

The spatial factors in this expression can be written in terms of the spherical harmonics $Y_1^{m_l}$, chapter 4.2.3. That gives the state of zero net angular momentum as

$${\hat\imath}_r f(r) Y_0^0 = \left( {\textstyle\sqrt{\l... ...scriptfont0 3}\kern.05em}} \vec Y_1^{-1} Y_1^1 \right) f(r)$$

To check this, just plug in the expressions for the $\vec{Y}_1^{m_s}$ of (A.83), and for the $Y_1^{m_l}$ of table 4.3.

The bottom line is that by combining states of unit spin $s$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, and unit orbital angular momentum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, you can create a state of zero net angular momentum, $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. Note also that in each of the three terms in the right hand side above, $m_s$ and $m_l$ add up to zero. A state of zero angular momentum $j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 must have $m_j$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 without uncertainty. Further note that the values of both the spin and orbital angular momentum in the $z$-direction are uncertain. Each of the two has measurable values $\hbar$, 0, or $\vphantom0\raisebox{1.5pt}{$-$}$$\hbar$ with equal proba­bility $\frac13$.

The above relation may be written more neatly in terms of “ket notation.” In ket notation, an angular momentum state with azimuthal quantum number $j$ and magnetic quantum number $m_j$ is indicated as $\big\vert j\:m_j\big\rangle$. Using this notation, and dropping the common factor $f(r)$, the above relation can be written as

$${\big\vert\:0\big\rangle }_j = {\textstyle\sqrt{\leavevm... ...pt]{5pt}{.5pt}1\big\rangle }_s {\big\vert 1\:1\big\rangle }_l$$

Here the subscripts $j$, $s$, and $l$ indicate net, spin, and orbital angular momentum, respectively.

There is a quicker way to get this result than going through the above algebraic mess. You can simply read off the coefficients in the appro­priate column of the bottom-right tabulation in figure 12.6. (In this figure take $a$ to stand for spin, $b$ for orbital, and $ab$ for net angular momentum.) Figure 12.6 also has the coefficients for many other net spin states that you might need. A derivation of the figure must wait until chapter 12.

The parity of vector wave functions is also important. Parity is what happens to a wave function if you invert the positive direction of all three Cartesian axes. What happens to a vector wave function under such an inversion can vary. A normal, or “polar,” vector changes sign when you invert the axes. For example, a position vector ${\skew0\vec r}$ in classical physics is a polar vector. Each position coordinate $x$, $y$, and $z$ changes sign, and therefore so does the entire vector. Similarly, a velocity vector $\vec{v}$ is a polar vector; it is just the time derivative of position. A particle with a vector wave function that behaves like a normal vector has negative intrinsic parity. The sign of the wave function flips over under axes inversion. Particles of this type turn out to include the photon.

But now consider an example like a classical angular momentum vector, ${\skew0\vec r}$ $\times$ ${m}\vec{v}$. Since both the position and the velocity change sign under spatial inversion, a classical angular momentum vector stays the same. A vector that does not change under axes inversion is called a “pseudovector” or “axial” vector. A particle whose wave function behaves like a pseudovector has positive intrinsic parity.

Note however that the orbital angular momentum of the particle also has an effect on the net parity. In particular, if the quantum number of orbital angular momentum $l$ is odd, then the net parity is the opposite of the intrinsic one. If the quantum number $l$ is even, then the net parity is the intrinsic one. The reason is that spherical harmonics change sign under spatial inversion if $l$ is odd, but not when $l$ is even, {D.14}.

Particles of all types often have definite parity. Such a particle may still have uncertainty in $l$. But if parity is definite, the measurable values of $l$ will need to be all even or all odd.