Derivation of L, X and P Operators from Commutation Relations

This document provides a bottom-up derivation starting with the commutation relations of the operators for position $X_i$, momentum $P_i$, and angular momentum $L_i$. We compute their form, demonstrate their infinite-dimensional nature, and prove their uniqueness in the Hilbert space $L^2(\mathbb{R}^3)$.

Commutation Relations¶

The four fundamental commutation relations are:

1. $[L_i, L_j] = i \hbar \epsilon_{ijk} L_k$

2. $[L_i, X_j] = i \hbar \epsilon_{ijk} X_k$

3. $[L_i, P_j] = i \hbar \epsilon_{ijk} P_k$

4. $[X_i, P_j] = i \hbar \delta_{ij}$

All generators are Hermitian. The relation 1. describes the angular momentum Lie algebra, relations 2. and 3. indicate that $X_i$ and $P_i$ are vector operators with repsect to $L_i$. The relation 4. is the canonical commutation relation for position and momentum.

Here, $\epsilon_{ijk}$ is the Levi-Civita symbol, $\delta_{ij}$ is the Kronecker delta, and $\hbar$ is the reduced Planck constant.

Infinite-Dimensional Nature of $X_i$ and $P_i$¶

The operators $X_i$ and $P_i$ must be infinite-dimensional due to the commutation relation $[X_i, P_j] = i \hbar \delta_{ij}$. In a finite-dimensional Hilbert space of dimension $n$, the trace of a commutator $[A, B]$ is zero:

Thus:

• Left side: $\text{Tr}([X_i, P_j]) = 0$.

• Right side: $\text{Tr}(i \hbar \delta_{ij} I) = i \hbar \delta_{ij} n$, which is non-zero for $i = j$ and $n \neq 0$.

This contradiction implies that $n$ cannot be finite; hence, the Hilbert space must be infinite-dimensional, such as $L^2(\mathbb{R}^3)$.

Both $X_i$ and $P_i$ must be infinite-dimensional, and consequently also $L_i$, otherwise the commutators like $[L_i, X_j]$ would not make sense. All the generators must be from the same space. When infinite-dimensional space is needed, the $L^2(\mathbb{R}^3)$ is usually used, so we will use it as well.

Form of $X_i$ and $P_i$¶

The Stone-von Neumann theorem says that $X_i$ and $P_i$ are uniquely determined as below, except a unitary transformation.

In the position representation on $L^2(\mathbb{R}^3)$:

• Position Operator: $X_i \psi(x) = x_i \psi(x)$, where $x_i$ is the $i$-th coordinate.

• Momentum Operator: $P_i = -i \hbar \frac{\partial}{\partial x_i}$.

To verify:

Using the product rule:

Thus, $[X_i, P_j] = i \hbar \delta_{ij}$, confirming the forms.

Unitary Rotation Operator $U$¶

In a previous chapter we derived for any vector operator (we will use the vector operator $X_i$):

Now we can write, using the known properties of $X_i$ derived above:

Comparing with $X_i |x'\rangle = x_i' |x'\rangle$ we get $U |x\rangle=|x'\rangle$ and $x_i' = (R^{-1})_j{}^i x_i$, so:

Then we can compute:

So we derived:

For a rotation by angle $\theta$ around axis $\mathbf{n}$, the unitary operator $U$ is:

Acting on a wave function:

where $R$ is the rotation matrix, and $R^{-1}$ is its inverse.

Derivation of $L_i$¶

Consider an infinitesimal rotation by $\delta \theta$ around $\mathbf{n}$. The rotation matrix approximates as $R_{ij} \approx \delta_{ij} + \delta \theta \epsilon_{ijk} n_k$, and:

Expand $\psi(R^{-1} x)$:

For $U$:

Equate:

Subtract $\psi(x)$ and simplify:

Since this holds for all $\mathbf{n}$, equate coefficients:

Substitute $P_j = -i \hbar \partial_j$, so $\partial_j \psi = \frac{1}{-i \hbar} P_j \psi = \frac{i}{\hbar} P_j \psi$:

Multiply by $\frac{\hbar}{i}$:

Thus, $L_i = \epsilon_{ijk} X_j P_k$.

Uniqueness of $L_i$¶

Suppose $L_i'$ also satisfies the commutation relations. Define $D_i = L_i - L_i'$:

• $[D_i, X_j] = [L_i, X_j] - [L_i', X_j] = 0$

• $[D_i, P_j] = [L_i, P_j] - [L_i', P_j] = 0$

$D_i$ commutes with all $X_j$ and $P_j$. In the irreducible representation on $L^2(\mathbb{R}^3)$, Schur’s lemma implies $D_i = c_i I$. Check the algebra:

Thus, $c_k = 0$, so $D_i = 0$, and $L_i = L_i'$.

Summary¶

The four commutation relations uniquely determine the generators:

• $X_i = x_i$ (position operator)

• $P_i = -i \hbar \frac{\partial}{\partial x_i}$ (momentum operator)

• $L_i = \epsilon_{ijk} X_j P_k$ (angular momentum operator)

• Acting in the infinite-dimensional Hilbert space $L^2(\mathbb{R}^3)$

No other solution is possible. One can then extend this Hilbert space for example to also include spin, but in the smallest space $L^2(\mathbb{R}^3)$ the above solution is unique.

Group¶

To determine the group elements, we can define:

That’s a translation operator: $T(a)\psi(x) = \psi(x+a)$. We derived that ${\bf P}$ is a translation generator.

Next we compute from BCH:

This is a momentum translation element, generated by a momentum translation generator ${\bf X}$.

Finally:

is an operator of rotation, due to $U\psi(x) = \psi(R^{-1} x)$.

Explicit Example: Rotation Operator Around the Z-Axis¶

For a rotation by angle $\theta$ around the z-axis, with $\mathbf{n} = (0, 0, 1)$:

Angular Momentum Operator $L_z$¶

In cylindrical coordinates: $L_z = -i \hbar \frac{\partial}{\partial \phi}$

Rotation Operator $U$¶

Explicitly:

• Cartesian: $U = \exp\left(-\theta \left( x \frac{\partial}{\partial y} - y \frac{\partial}{\partial x} \right)\right)$

• Cylindrical: $U = \exp\left(-\theta \frac{\partial}{\partial \phi}\right)$

Action on Wave Function¶

Matches $\psi(R^{-1} \mathbf{x})$, where: