11 Rotational symmetry
At this point, we’ve seen enough of one-dimensional quantum mechanics, and we’re ready to turn to our real three-dimensional world. As mentioned briefly in the notes and homework, the relationship between position \hat{x} and momentum \hat{p} generalizes simply to three dimensions; but let’s be concrete and establish the notation and conventions here.
11.0.1 Conventions and notation in higher dimensions
We start with position: you’ll often see \vec{r} = (x,y,z) as the three-dimensional position vector, but I’ll follow Sakurai and use \vec{x}. The key to generalizing from one to three dimensions is the fact that the canonical commutation relation only holds for position and momentum in the same direction; positions and/or momenta pointing in different Cartesian directions always commute. We can expand the set of commutation relations to read \begin{align*} [x_i, x_j] &= 0 \\ [p_i, p_j] &= 0 \\ [x_i, p_j] &= i \hbar \delta_{ij}. \end{align*} Most of our notation and work above passes through unchanged. Position eigenstates are labelled with \ket{\vec{x}} = \ket{x,y,z}, and similarly for momentum. They are normalized accounting properly for all three dimensions, i.e. \left\langle \vec{x} | \vec{x}' \right\rangle = \delta^3(\vec{x} - \vec{x}') = \delta(x-x') \delta(y-y') \delta(z-z') and similarly for momentum; here \delta^3(...) is just short-hand for the product of three delta functions, one for each dimension. Integrals for inner products etc. are now over all three dimensions. The representation of the momentum operator in position space becomes the gradient, \hat{\vec{p}} = \frac{\hbar}{i} \frac{\partial}{\partial \vec{x}} = \frac{\hbar}{i} \vec{\nabla}_x and the converse \hat{\vec{x}} = i\hbar \frac{\partial}{\partial \vec{p}} = i\hbar {\vec{\nabla}}_p. The position-space gradient is far more common, and I’ll typically write it without the subscript, especially if it’s obvious from context (i.e. if it’s acting on a function of position/momentum.)
Finally, the Fourier transform relation between momentum and position space looks almost the same, but with a different normalization: \psi(\vec{x}) = \frac{1}{(2\pi \hbar)^{3/2}} \int d^3p\ \exp \left(\frac{i\vec{p} \cdot \vec{x}}{\hbar} \right) \tilde{\psi}(\vec{p}) \\ \psi(\vec{p}) = \frac{1}{(2\pi \hbar)^{3/2}} \int d^3x\ \exp \left(\frac{-i\vec{p} \cdot \vec{x}}{\hbar} \right) {\psi}(\vec{x})
11.1 Rotational symmetry
When we turn to consider the full three-dimensional world, a new and extremely important symmetry operation appears: rotation. Rotational symmetry is everywhere, and has widespread implications in all areas of quantum mechanics. Also, the algebraic structure of spatial rotations re-appears in several places in the form of more abstract “rotations” appearing in nuclear and particle physics.
Let’s first remind ourselves of how finite rotations are described in classical physics. An arbitrary finite rotation can be described by a 3x3 matrix R which encodes the effects of the rotation on any vector as \vec{v}' = R \vec{v}. By definition, a rotation can’t change the length of a vector, so we must have |\vec{v}'| = |\vec{v}| for any \vec{v} that we choose; this implies that R must be an orthogonal matrix, R^T R = R R^T = 1. This is the analogue of a unitary matrix for a system where we only have real numbers. Notice also that this equation implies 1 = \det(R R^T) = \det(R) \det(R^T), so \det(R) = \pm 1. Technically, a rotation must satisfy the further constraint \det(R) = 1; the negative-determinant maps include parity reflections.
Since we’ve remarked on symmetry groups and representations before, I’ll point out that the structure for rotation matrices R noted above defines a Lie group called SO(3) - SO for “special orthogonal”, 3 for the size of the matrices (with “special” being basically equivalent to the condition \det(R) = 1.)
We won’t study the representation theory of SO(3) in detail, but I will state a result from mathematics here: all of the irreducible representations of SO(3) can be labelled by their dimensions, \mathbf{d}, where {\rm dim}(\mathbf{d}) = 2j+1 for integer j \geq 0 (so \mathbf{1}, \mathbf{3}, \mathbf{5}... they all have odd dimension.)
There are also a set of representations of even dimension \mathbf{2}, \mathbf{4}, ..., corresponding to half-integer j = 1/2, 3/2, ...; as we will see later, there is something funny about these representations (if you’re familiar with the math already, they are “projective representations”), which means that they can never appear in rotations of classical objects - they are exclusively used to describe spin.
It’s useful to recall the explicit form of the rotation matrix for rotation about a specific axis. I’ll reproduce them all here for completeness: R_x(\phi) = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \phi & -\sin \phi \\ 0 & \sin \phi & \cos \phi \end{array} \right), \\ R_y(\phi) = \left( \begin{array}{ccc} \cos \phi & 0 & \sin \phi \\ 0 & 1 & 0 \\ -\sin \phi & 0 & \cos \phi \end{array} \right), \\ R_z(\phi) = \left( \begin{array}{ccc} \cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1 \end{array} \right). The minus sign in the top right/bottom left for each axis is a convention (corresponding to the usual right-hand rule.) Any arbitrary R can be constructed as a product of sequential rotations, at most three, about different axes (you may be having flashbacks to Euler angles right about now.) As I have pointed out before, rotations are non-commutative; the order in which we apply rotations matters, which is very easy for you to verify experimentally with your textbook or cell phone.
In classical mechanics, one typically starts by writing down the expression for an arbitrary finite rotation, as we just have. However, in quantum mechanics we like to build symmetries up from infinitesmal operations and identify the generators. We can easily work backwards to see what a rotating by an infinitesmal angle \epsilon looks like: R_z(\epsilon) = \left( \begin{array}{ccc} 1 - \epsilon^2/2 & -\epsilon & 0 \\ \epsilon & 1 - \epsilon^2 / 2 & 0 \\ 0 & 0 & 1 \end{array} \right), and similarly for R_x(\epsilon) and R_y(\epsilon). Even for these infinitesmal rotations, we can see the non-commutativity of rotations: it’s easy to work out that R_x(\epsilon) R_y(\epsilon) - R_y(\epsilon) R_x(\epsilon) = \left( \begin{array}{ccc} 0 & -\epsilon^2 & 0 \\ \epsilon^2 & 0 & 0 \\ 0 & 0 & 0 \end{array} \right). Notice that if we ignore terms at order \epsilon^2, the two matrices actually do commute. However, it’s also interesting to notice that this looks similar to an infinitesmal rotation about the z axis. In fact, we can write R_x(\epsilon) R_y(\epsilon) - R_y(\epsilon) R_x(\epsilon) = R_z(\epsilon^2) - 1 + \mathcal{O}(\epsilon^4). So for infinitesmal rotations, the commutator of rotations about two perpendicular axes depends explicitly on a rotation about the third axis.
Now to quantum mechanics. For other operators we’ve considered so far, the algebra in the classical limit was simple, with everything commuting. However, in this case our quantum algebra must reproduce the classical non-commutativity of rotation, so we’ll have to construct it from the beginning with the above structure in mind. We start by supposing that for any classical rotation R(\vec{n}, \phi) there exists a unitary operator \hat{U}(\vec{n}, \phi) which transforms a ket from the un-rotated coordinate system to the rotated one, \ket{\psi}_R = \hat{U}(\vec{n}, \phi) \ket{\psi}.
As I’ve written explicitly, any single rotation can be specified by choosing a rotation axis \vec{n}, and an angle of rotation \phi. Taking the angle to be an infinitesmal d\phi, we know that we can write the unitary \hat{U} in terms of some new Hermitian operator, \hat{U}(\vec{n}, \phi) = 1 - \frac{i}{\hbar} \mathcal{O}_{\vec{n}} d\phi + ... Of course, we don’t want to have an infinite number of operators for each possible choice of rotation axis. If we suppose that like the classical theory, any quantum rotation can be decomposed into rotations about the coordinate axes, then we can hypothesize a vector operator \hat{\vec{J}} as the generator, and \hat{U}(\vec{n}, \phi) = 1 - \frac{i}{\hbar} (\hat{\vec{J}} \cdot \vec{n}) d\phi + ... As always, we can go backwards to rebuild a finite rotation from a large number of infinitesmal ones, finding e.g. \hat{U}(\hat{z}, \phi) = \exp \left( \frac{-i \hat{J}_z \phi}{\hbar} \right). Now we remember that we want to reproduce the classical commutation relations between finite rotation operators, one of which we derived above. Using the exponential form and expanding to order \epsilon^2, we find for the quantum theory \left[ 1 - \frac{i \hat{J}_x \epsilon}{\hbar} - \frac{\hat{J}_x{}^2 \epsilon^2}{2\hbar^2}, 1 - \frac{i \hat{J}_y \epsilon}{\hbar} - \frac{\hat{J}_y{}^2 \epsilon^2}{2\hbar^2} \right] = \left( 1 - \frac{i \hat{J}_z \epsilon^2}{\hbar} \right) - 1 The commutator is a mess at first glance, but since everything commutes with 1, all of the terms of order \epsilon^0 and \epsilon^1 vanish immediately, and the \epsilon^2 term becomes quite simple: [-i\hat{J}_x \epsilon / \hbar, -i \hat{J}_y \epsilon/\hbar] = i \hat{J}_z \epsilon^2 / \hbar \\ \Rightarrow [\hat{J}_x, \hat{J}_y] = i \hbar \hat{J}_z. There is nothing particularly special about the z axis, of course. We can repeat this argument for other combinations of rotations, and the only difference will be a possible minus sign. Gathering all of the results, we find the following compact formula:
The components of an angular momentum operator \hat{\vec{J}} satisfy the commutation relation [\hat{J}_i, \hat{J}_j] = i \hbar \epsilon_{ijk} \hat{J}_k.
As noted, these are the defining commutation relations for the operator \hat{\vec{J}}, which we identify as the angular momentum operator since it generates rotations in the same way that linear momentum generates translations.
In fact, we can make this more precise through our group-theory language. The three \hat{J}_i are the generators of continuous rotations; when we exponentiate them, we get the group of rotations, SO(3). But since the exponential map is unique, these specific commutation relations between \hat{J}_i and \hat{J}_j are required in order to end up with an SO(3) group structure. So, the algebra of the generators fixes what group we get! The mathematical terminology is that the commutation relations above define a Lie algebra called so(3); exponentiating the elements of this Lie algebra gives us the corresponding Lie group.
To clarify this point, it’s helpful to compare with the ordinary momentum operators that generate translations. They satisfy a much simpler set of commutation relations: [\hat{p}_i, \hat{p}_j] = 0. The corresponding Lie algebra (and resulting Lie group of translations) is \mathbb{R}^3, the three-dimensional space of real numbers. Due to the lack of any structure to the algebra, our translation operations all commute with each other (this is known as an “Abelian” group, where commutators are always zero so order doesn’t matter.) In contrast, the non-trivial commutation of the \hat{\vec{J}} components is exactly what gives the required non-commuting structure that we need for rotations.
11.2 Eigenvalues and eigenstates of angular momentum
Let’s see what the commutation relations imply about a system with angular momentum in full generality. There is a new operator we can construct by summing up all of the individual angular momentum operators, \hat{J}^2 = \hat{J}_x{}^2 + \hat{J}_y{}^2 + \hat{J}_z{}^2. This operator commutes with all of the individual angular-momentum operators, [\hat{J}{}^2, \hat{J}_k] = 0. This is easy to prove: first, notice that we can write the general commutator [\hat{J}_i{}^2, \hat{J}_j] = \hat{J}_i [\hat{J}_i, \hat{J}_j] + [\hat{J}_i, \hat{J}_j] \hat{J}_i \\ = \sum_k i \hbar \epsilon_{ijk} (\hat{J}_i \hat{J}_k + \hat{J}_k \hat{J}_i). For both i and j fixed, the epsilon symbol just gives us a single term k \neq i \neq j with either a plus or minus sign. However, if we sum over the first index i, then we can use the antisymmetry of epsilon: for every term \epsilon_{ijk}, there will be a term \epsilon_{kji} with opposite sign. So \sum_i [\hat{J}_i{}^2, \hat{J}_j] = [\hat{J}{}^2, \hat{J}_j] = 0. We can thus choose one of the three directions \hat{J}_i to diagonalize simultaneously with \hat{J}{}^2; by convention, we usually take \hat{J}_z. (We of course can’t choose more than one, since the other directions don’t commute with each other.) Before we start to look at the common eigenstates of these operators, it’s useful to define the combinations \hat{J}_{\pm} = \hat{J}_x \pm i \hat{J}_y, which (just like their SHO counterparts) are known as angular-momentum ladder operators or raising and lowering operators, a name which will make more sense shortly. Notice that these are not Hermitian operators; in fact, the complex transpose of one is the other, \hat{J}_+^\dagger = \hat{J}_-. Their commutation relations are easily derived: \begin{align*} [\hat{J}_+, \hat{J}_-] &= 2\hbar J_z \\ [\hat{J}_z, \hat{J}_{\pm}] &= \pm \hbar \hat{J}_{\pm} \\ [\hat{J}{}^2, \hat{J}_{\pm}] &= 0. \end{align*}
Now let’s tackle the eigenvalue problem. Usually, the action of these operators and the nature of their eigenkets are just handed down from on high; I find it much more enlightening to go through the derivation as Sakurai does. So let’s for the moment assume that the two eigenvalues are completely arbitrary, and label them “a” and “b”: \hat{J}{}^2 \ket{a,b} = a \ket{a,b} \\ \hat{J}_z \ket{a,b} = b \ket{a,b}. The ladder operators have a familiar-looking effect on these eigenstates: \hat{J}_z (\hat{J}_{\pm} \ket{a,b}) = ([\hat{J}_z, \hat{J}_{\pm}] + \hat{J_{\pm}} \hat{J}_z) \ket{a,b} \\ = (b \pm \hbar) (\hat{J}_{\pm} \ket{a,b}). So in terms of the \hat{J}_z eigenvalue, the ladder operators give another \hat{J}_z eigenstate, with the eigenvalue raised or lowered by \hbar. On the other hand, since the ladder operators commute with \hat{J}^2, they don’t affect that eigenvalue: \hat{J}{}^2 (\hat{J}_{\pm} \ket{a,b}) = a (\hat{J}_{\pm} \ket{a,b}). Unlike the simple harmonic oscillator, it turns out that we can’t continue raising the \hat{J}_z eigenvalue forever. To see why, notice that we can rewrite the difference \hat{J}{}^2 - \hat{J}_z{}^2 = \frac{1}{2} (\hat{J}_+ \hat{J}_- + \hat{J}_- \hat{J}_+) \\ = \frac{1}{2} ( \hat{J}_+ \hat{J}_+^\dagger + \hat{J}_+^\dagger \hat{J}_+). If we take the expectation value, we find that it must be manifestly positive: \left\langle \hat{J}_+^\dagger \hat{J}_+ \right\rangle = \bra{a,b} \hat{J}_+^\dagger \hat{J}_+ \ket{a,b} = \left\langle a,b+\hbar | a,b+\hbar \right\rangle \geq 0, (remember that all states in our Hilbert space have positive norm!) The same is true for the complex conjugate, so we have immediately that \bra{a,b} (\hat{J}{}^2 - \hat{J}_z{}^2) \ket{a,b} = a - b^2 \geq 0. So we find both an upper limit and a lower limit on the eigenvalue of \hat{J}_z; its square can’t be larger than the eigenvalue a of \hat{J}{}^2. This means that for any a, there must be a b_{\textrm{max}} such that \hat{J}_+ \ket{a,b_{\textrm{max}}} = 0. Now, we can apply the lowering operator to 0 and we still have 0: \hat{J}_- \hat{J}_+ \ket{a,b_{\textrm{max}}} = 0. But this is an interesting combination of operators: \hat{J}_- \hat{J}_+ = (\hat{J}_x - i\hat{J}_y) (\hat{J}_x + i\hat{J}_y) \\ = \hat{J}_x{}^2 + \hat{J}_y{}^2 -i [\hat{J}_y, \hat{J}_x] \\ = \hat{J}{}^2 - \hat{J}_z{}^2 - \hbar \hat{J}_z. Thus, \hat{J}^2 - \hat{J}_z^2 - \hbar \hat{J}_z \ket{a,b_{\textrm{max}}} = (a - b_{\textrm{max}}^2 - \hbar b_{\textrm{max}}) \ket{a,b_{\textrm{max}}} = 0. By assumption, the eigenket \ket{a, b_{\textrm{max}}} certainly exists and is non-null, so the only way this can be true is if a = b_{\textrm{max}} (b_{\textrm{max}} + \hbar). A similar argument at the other end yields a minimum possible b, and a = b_{\textrm{min}} (b_{\textrm{min}} - \hbar). So by comparison, b_{\textrm{min}} = -b_{\textrm{max}}. Moreover, for all of this to be self-consistent we must be able to get from the minimum state to the max, and vice-versa, by successive use of the ladder operators. Thus, we find that b_{\textrm{max}} = b_{\textrm{min}} + n\hbar for some integer n, and therefore b_{\textrm{max}} = \frac{n\hbar}{2}. The integer n is fixed by the eigenvalue of \hat{J}{}^2, which must satisfy a = \frac{1}{4} \hbar^2 n(n+2). Physically, you can think about what’s happening here in fairly simple terms. Since \hat{J}{}^2 is in some sense the “length” of the angular momentum vector operator \hat{\vec{J}}, the “length” \hat{J}_z^2 of one component of the vector can’t be greater than the total vector length, hence the minimum and maximum conditions on b. Combination with the angular-momentum commutation relations then leads to quantization, with only certain values of the angular momentum allowed.
Now that we’ve done the derivation, let’s switch to the standard notation. We label our states by two numbers j and m, which may be integers or half-integers; in particular, from our derivation above we identify j=n/2 as the \hat{J}{}^2 label, and m = b/\hbar as the \hat{J}_z label. The eigenvalue equations are then \hat{J}{}^2 \ket{j,m} = \hbar^2 j(j+1) \ket{j,m} \\ \hat{J}_z \ket{j,m} = \hbar m \ket{j,m}, and the allowed values of m are determined by j, m = \{-j, -j+1, ..., j-1, j\}. If j is an integer, than m always is too; if j is half-integer, then so is m. The dimensionality of the space, i.e. the number of independent kets is always 2j+1. So a particle carrying intrinsic angular momentum with two possible states is “spin-1/2”, because for two states j=1/2. Incidentally, j=1 is three dimensional, i.e. vector valued; the fact that light polarization can be represented as a vector is related to the fact that photons have spin 1.
Tying back to our more general discussion of groups and representations above, the value of j labels what representation of SO(3) we are describing, and (as expected) the dimension of the representation is exactly 2j+1.
The raising and lowering operators \hat{J}_{\pm} were instrumental in this derivation, but although we know that they act to give us \hat{J}_z eigenstates, we still need to fix the normalization, which I leave as an exercise.
Find the normalization of the ladder operators \hat{J}_{\pm}, that is, give the full formula for what \hat{J}_{\pm} \ket{j,m} produces, including the coefficient in front of the raised/lowered state.
Answer:
The best way to do this is to consider matrix elements, assuming the basic states \ket{j,m} to be properly normalized: \bra{j', m'} \hat{J}{}^2 \ket{j,m} = \hbar^2 j(j+1) \delta_{j,j'} \delta_{m,m'}, \\ \bra{j', m'} \hat{J}_z \ket{j,m} = \hbar m \delta_{j,j'} \delta_{m,m'}. We can now recycle our earlier rewriting of the product \hat{J}_+^\dagger \hat{J}_+, starting first with the diagonal matrix elements: \bra{j,m} \hat{J}_+^\dagger \hat{J}_+ \ket{j,m} = \bra{j,m} (\hat{J}{}^2 - \hat{J}_z{}^2 - \hbar \hat{J}_z) \ket{j,m} \\ = \hbar^2 [j(j+1) - m^2 - m]. But we can also rewrite this product by letting \hat{J}_+ act on the state directly: \hat{J}_+ \ket{j,m} = c_{jm}^+ \ket{j,m+1} and so \bra{j,m} \hat{J}_+^\dagger \hat{J}_+ \ket{j,m} = |c_{jm}^+|^2 \left\langle j,m+1 | j,m+1 \right\rangle = |c_{jm}^+|^2. This fixes c_{jm}^+, up to an arbitrary phase; by convention we take it to be real and positive. Thus, repeating the same derivation for \hat{J}_-, we arrive at the formulas \hat{J}_+ \ket{j,m} = \hbar \sqrt{(j-m)(j+m+1)} \ket{j,m+1} \\ \hat{J}_- \ket{j,m} = \hbar \sqrt{(j+m)(j-m+1)} \ket{j,m-1}.
The result is the following: \hat{J}_+ \ket{j,m} = \hbar \sqrt{(j-m)(j+m+1)} \ket{j,m+1} \\ \hat{J}_- \ket{j,m} = \hbar \sqrt{(j+m)(j-m+1)} \ket{j,m-1}.
It can be easy to mix these similar-looking formulas up, but an easy way to keep them straight is to remember that the \hat{J}_+ formula should vanish when m=j, and the \hat{J}_- one must vanish when m=-j. We can use these results to write a compact formula for the matrix elements of the ladder operators: \bra{j',m'} \hat{J}_{\pm} \ket{j,m} = \sqrt{(j \mp m) (j \pm m + 1)} \hbar \delta_{j,j'} \delta_{m',m\pm 1}.
Let’s see a few examples of what these various operators we’ve defined look like for systems with small, fixed j (i.e. for fixed irreducible representations of rotation.)
- j=0
This gives us m = 0 as the only possibility, so this is completely trivial: the operators \hat{J}{}^2 and \hat{J}_z give zero when acting on any state. Physically, this makes sense - a j=0 object has no angular momentum at all (it transforms in the trivial representation of rotation.)
- j=1/2
This is a familiar example in some new notation. With j=1/2 we find two possible states \ket{\frac{1}{2}, \pm \frac{1}{2}}. We can write out the matrix expressions for the various operators we’ve defined: \hat{J}_x = \frac{\hbar}{2} \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)\ \ \hat{J}_y = \frac{\hbar}{2} \left( \begin{array}{cc} 0 & -i \\ i & 0 \end{array} \right)\ \ \hat{J}_z = \frac{\hbar}{2} \left( \begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array} \right) \\ \hat{J}_+ = \hbar \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right)\ \ \hat{J}_- = \hbar \left( \begin{array}{cc} 0 & 0 \\ 1 & 0 \end{array} \right)\ \ \hat{J}{}^2 = \frac{3\hbar^2}{4} \left( \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right).
This is the same algebra we saw for spin-1/2 systems; since our derivation was totally general, our results apply to both spin angular momentum and orbital angular momentum (which we’ll come to soon.)
- j=1
For the next highest j, we find three states: m=-1,0,1. To write things in matrix form we can order them, as is conventional, from highest to lowest: \ket{1,+1} = \left(\begin{array}{c}1\\ 0\\ 0\end{array} \right) \ \ \ket{1,0} = \left(\begin{array}{c}0\\ 1\\ 0\end{array} \right) \ \ \ket{1,-1} = \left(\begin{array}{c}0\\ 0\\ 1\end{array} \right) \ \ Then the \hat{J}_z operator and the raising and lowering operators are \hat{J}{}^2 = 2\hbar^2 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right) \\ \hat{J}_z = \hbar \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array} \right) \\ \hat{J}_+ = \sqrt{2} \hbar \left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right) \ \ \hat{J}_- = \sqrt{2} \hbar \left( \begin{array}{ccc} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right) Of course, since \hat{J}{}^2 doesn’t depend on m, it’s easy to see that when we write things out in this way, the matrix \hat{J}{}^2 will always be proportional to the identity. This is an artifact of our current focus on systems with fixed j; when we come to consider addition of angular momentum, this operator won’t always be so trivial.
You might start to see a pattern here in the form of the other operators as matrices, but it may not be quite as simple as you think; let’s do one more example.
- j=3/2
We’ve now added one more state, for four possibilities: m=-3/2, -1/2, +1/2, +3/2. Going in order from large to small m again, we have \hat{J}_z = \hbar \left( \begin{array}{cccc} \frac{3}{2} &&& \\ &\frac{1}{2}&& \\ &&-\frac{1}{2}& \\ &&&-\frac{3}{2} \end{array} \right) \\ \hat{J}_+ = \hbar \left( \begin{array}{cccc} 0& \sqrt{3} &0&0\\ 0&0&2&0\\ 0&0&0&\sqrt{3} \\ 0&0&0&0 \end{array} \right) \\ \hat{J}_- = \hbar \left( \begin{array}{cccc} 0&0&0&0\\ \sqrt{3}&0&0&0\\ 0&2&0&0 \\ 0&0&\sqrt{3}&0 \end{array} \right) \\ Although the raising and lowering matrices are now more interesting - we can’t just pull an overall factor out front anymore - they still have the same off-diagonal structure, with \hat{J}_+ occupying the upper diagonal (due to our choice of m=+3/2 as the “top” vector state.)
11.2.1 Orbital angular momentum
So far, we’ve introduced the idea of a generic Hermitian angular momentum operator \hat{J}_i as the infinitesmal generator of rotations about axis i. But there’s another way we could have defined angular momentum: by taking the classical angular-momentum operator \vec{L} = \vec{r} \times \vec{p} and promoting \vec{r} and \vec{p} to operators, as we’ve done before. What do we find if we take this approach?
It’s easy to see that \hat{\vec{L}} is indeed an angular momentum operator as we’ve defined it, in particular its components satisfy the angular momentum commutation relation [\hat{L}_i, \hat{L}_j] = i \hbar \epsilon_{ijk} \hat{L}_k. You can convince yourself that this is true by expanding out the definition in terms of \hat{\vec{x}} and \hat{\vec{p}}. Furthermore, we can see that if we let this operator act on a position wavefunction, we do get an infinitesmal rotation as expected. For example, trying to generate an infinitesmal rotation \delta \phi about the z-axis, \bra{x,y,z} \left[ 1 - i \frac{\delta \phi}{\hbar} \hat{L}_z \right] \ket{\psi} = \left[ 1 - i \frac{\delta \phi}{\hbar} \hat{p}_y \hat{x} + i \frac{\delta \phi}{\hbar} \hat{p}_x \hat{y} \right] \ket{\psi} \\ = \bra{x,y,z}\left[1 - (\delta \phi) x \frac{\partial}{\partial y} + (\delta \phi) y \frac{\partial}{\partial x} \right] \ket{\psi} \\ = \psi(x,y,z) - x (\delta \phi) \frac{\partial \psi}{\partial y} + y (\delta \phi) \frac{\partial \psi}{\partial x} \\ = \psi(x+y(\delta \phi), y - x (\delta \phi), z). This is exactly what we expect for spatial rotation of a scalar function (here it’s a backwards rotation, since it’s acting on the position bra to the left.) In fact, this result is nicer if we switch to spherical coordinates, x = r \sin \theta \cos \phi \\ y = r \sin \theta \sin \phi \\ z = r \cos \theta. Then we have from above \bra{x,y,z} \left[ 1 - i \frac{\delta \phi}{\hbar} \hat{L}_z \right] \ket{\psi} = \psi(r, \theta, \phi - \delta \phi) which lets us identify \bra{x,y,z} \hat{L}_z \ket{\psi} = -i \hbar \frac{\partial}{\partial \phi} \left\langle x,y,z | \psi \right\rangle. It’s important to emphasize that this version of angular momentum is independent of the internal properties of whatever particle we’re considering; we’ve constructed it simply from the position and momentum operators. As a result, \hat{\vec{L}} is known as the orbital angular momentum. All of the results we’ve derived for general \hat{\vec{J}} still apply, with the one exception that the eigenvalues of \hat{\vec{L}}{}^2 can only take on integer values, and never half-integer; we’ll prove this soon.
Now, going through the same exercise for the other directions is a little tedious; in fact, it turns out to be slightly cleaner to write the answer in terms of the ladder operators L_{\pm} = L_x \pm i L_y: \bra{x,y,z} \hat{L}_{\pm} \ket{\psi} = -i \hbar e^{\pm i \phi} \left( \pm i \frac{\partial}{\partial \theta} - \cot \theta \frac{\partial}{\partial \phi} \right) \left\langle x,y,z | \psi \right\rangle. The last operator to consider is the squared total angular momentum, \hat{\vec{L}}{}^2. We can construct that from what we’ve already done: \hat{\vec{L}}{}^2 = \hat{L}_z^2 + \frac{1}{2} (\hat{L}_+ \hat{L}_- + \hat{L}_- \hat{L}_+), which with a bit of algebra gives \bra{x,y,z} \hat{\vec{L}}{}^2 \ket{\psi} = -\hbar^2 \left[ \frac{1}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{1}{\sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial}{\partial \theta} \right) \right] \left\langle x,y,z | \psi \right\rangle. This is precisely the angular part of the Laplacian operator \nabla^2 in spherical coordinates! In fact, if we expand out the full three-dimensional kinetic energy term in spherical coordinates, we find that \frac{\hat{\vec{p}}{}^2}{2m} = \frac{-\hbar^2}{2m} \nabla^2 = \frac{-\hbar^2}{2mr^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial}{\partial r}\right) + \frac{\hat{\vec{L}}{}^2}{2mr^2}. We could have started with the classical Hamiltonian in spherical coordinates and identified the angular momentum from there as well, leading to the same expression.