7
$\begingroup$

Consider the equality \begin{equation}\exp\left(-\frac{i}{\hbar}\boldsymbol{\phi J}\right)\left|x\right>=\left|R(\phi)x\right>,\end{equation} where $\left|x\right>$ denotes a position eigenstate, $J$ the total angular momentum operator on the ket space, and $R(\boldsymbol{\phi})$ the $\mathbb{R}^3$ rotation matrix for rotations around $\boldsymbol{\phi}=\phi\cdot\mathbf{n}$ with $||\mathbf{n}||=1$.

Some consider this equation to be the definition of total angular momentum. How can this equation be proven by using the arguably more popular and classically motivated definition? \begin{equation} \mathbf{J}:=\mathbf{L}+\mathbf{S}=\mathbf{X}\times\mathbf{P}+\mathbf{S} \end{equation}

Note: Thanks to the work of @Valter Moretti and @Adam Latosiński an equality between the two defintions has been established for a spinless particle.

$\endgroup$
  • 1
    $\begingroup$ is all of $h\boldsymbol{\phi J}$ in the denominator? $\endgroup$ – Luyw Jul 14 at 7:49
  • $\begingroup$ Shouldn't it be $\hbar$ instead of $h$ ? $\endgroup$ – Thomas Fritsch Jul 14 at 8:16
  • $\begingroup$ I tried to address the misunderstandings. $\endgroup$ – TheoreticalMinimum Jul 14 at 9:09
  • 3
    $\begingroup$ I believe that the most broad definition of the angular momentum operator ${\bf J}$ is the operator that generates the rotations, i.e. the equality you want to be proven is the definition of what angular momentum operator is. If you have a different definition of angular momentum operator, please provide it. $\endgroup$ – Adam Latosiński Jul 14 at 9:42
  • 1
    $\begingroup$ @AdamLatosinski If one wants to follow your route, he/she must prove that the identity we are discussing implies that $J$ has the standard form in terms of $X_k$ and $P_j$. I mean, that is not a proof, but just another way to state the initial question. $\endgroup$ – Valter Moretti Jul 14 at 10:05
4
$\begingroup$

Consider a smooth function $\psi$ in $L^2(\mathbb{R}^3, d^3x)$ (more precisely $\psi$ is assumed to belong to Schwartz' space) and define for a fixed $n \in \mathbb{S}^2$ and $\phi \in \mathbb{R}$ $$\psi_\phi(x)= \psi(R^{-1}_n(\phi)x)$$ Since $R^{-1}_n(\phi) = e^{-\phi n \cdot S}$, where $S= (S_x,S_y,S_z)$ are the three generators of $SO(3)$: $$(S_j)_{rs}= \epsilon_{jrs}$$ we have $$\frac{d}{d\phi}\psi_\phi(x) = -\sum_{j,r,s=1}^3 n_j \epsilon_{jrs}x_r\frac{\partial }{\partial x_s}\psi_\phi(x)\:.\tag{1}$$ Now consider $$\psi'_\phi(x)= \left(e^{\frac{i}{\hbar}\phi n \cdot J}\psi\right)(x)\:,$$ where, omitting the spin part of $J$ since as far as I unserstand, you are interested in the spatial part of the state only, $$J_j = \sum_{r,s=1}^3\epsilon_{jrs} X_r P_s = -i\hbar \sum_{r,s=1}^3\epsilon_{jrs} X_r \frac{\partial}{\partial x_s} $$ Computing the $\phi$-derivative (using Stone's theorem and some careful analysis about the use of different topologies. I do not want to enter into the details here, I just say that here smoothness of $\psi$ matters), we have $$\frac{d}{d\phi}\psi'_\phi(x) = \frac{i}{\hbar}\sum_{j=1}^3 n \cdot \left(J_j\psi'_\phi\right)(x)= -\sum_{j,r,s=1}^3 n_j \epsilon_{jrs}x_r\frac{\partial }{\partial x_s}\psi'_\phi(x)\:.\tag{2}$$ In summary, for $x$ fixed, $\psi_\phi(x)$ and $\psi'_\phi(x)$ satisfy the same first order differential equation (in normal form with smooth known term) and furthermore they satisfy the same initial condition $$\psi_0(x)=\psi'_0 (x)\:.$$ The uniqueness theorem for the solutions of first-order differential equations implies that $$\psi_\phi(x)=\psi'_\phi (x)\:.$$ In other words $$\left(e^{-\frac{i}{\hbar} \phi n \cdot J}\psi\right)(x) = \psi\left(R_n^{-1}(\phi)x\right)\:.$$ (The result extends to the whole $L^2$ space made als of non-smooth functions exploiting the fact that Schwartz' space is dense therein.) The final step is quite formal, but it can be made rigorous adopting the theory of rigged Hilbert spaces to make rigorous the bra-ket notation for improper eigenvectors of the position operators. In position representation $|y\rangle = \delta(x-y)= \delta_y(x)$, so that, using rotational invariance of the Dirac delta function, $$\left(e^{-\frac{i}{\hbar} \phi n \cdot J}\delta_y\right)(x) = \delta_y\left(R_n^{-1}(\phi)x\right)= \delta\left(R_n^{-1}(\phi)x -y\right)= \delta\left(x -R_n(\phi)y\right)\:.$$ Coming back to the abstract notation, the found identity reads $$e^{-\frac{i}{\hbar} \phi n \cdot J}|y\rangle = |R_n(\phi)y\rangle\:.$$

(A completely rigorous proof can be found in this book of mine.)

$\endgroup$
3
$\begingroup$

The most broad definition of the angular momentum operator ${\bf J}$ is the operator that generates rotation, so the equality you want to prove is just the definiton (assumin that the particle we consider does not have any internal degrees of freedom, which could include spin) and as such does not require a proof.

But what would then needs a proof is the relation between the angular momentum operator and the position operator $$ {\bf J} = {\bf X} \times {\bf P}$$ Let's see that this is satisfied for ${\bf J}_3$, i.e. $$ {\bf J}_3 = {\bf X}_1{\bf P}_2 - {\bf X}_2{\bf P}_1$$ (other components will be analogous).

Let us consider an rotation acting on a state $|\psi\rangle = \int \psi(x) |x\rangle dx$. $$ \exp(- \frac{i}{\hbar}\phi\, {\bf J}_3) \int \psi(x)|x\rangle dx = \int \psi(x)|R(\phi)x\rangle dx = \int \psi(R^{-1}(\phi)x) |x\rangle$$ That means that $$ {\bf J}_3 |\psi\rangle = i\hbar \left.\frac{d}{d\phi}\right|_{\phi=0} \exp(- \frac{i}{\hbar}\phi\, {\bf J}_3) \int \psi(x)|x\rangle dx = i\hbar \int \left.\frac{d}{d\phi}\right|_{\phi=0} \psi(R^{-1}(\phi)x) |x\rangle$$

We have $$\psi(R^{-1}(\phi)x) = \psi(x_1 \cos\phi + x_2 \sin\phi,-x_1\sin\phi+x_2\cos\phi,x_3) $$ so $$ \left.\frac{d}{d\phi}\right|_{\phi=0} \psi(R^{-1}(\phi)x) = x_2 \frac{\partial \psi(x)}{\partial x_1} - x_1 \frac{\partial \psi(x)}{\partial x_2}$$ so \begin{align} {\bf J}_3|\psi\rangle &= i\hbar \int \Big(x_2 \frac{\partial \psi(x)}{\partial x_1} - x_1 \frac{\partial \psi(x)}{\partial x_2}\Big) |x\rangle = \\ &= -{\bf X}_2 \int \Big(-i\hbar\frac{\partial \psi(x)}{\partial x_1}\Big) |x\rangle dx + {\bf X}_1 \int \Big(-i\hbar\frac{\partial \psi(x)}{\partial x_2}\Big) |x\rangle = \\ &= -{\bf X}_2 {\bf P}_1 \int \psi(x) |x\rangle dx + {\bf X}_1 {\bf P_2} \int \psi(x) |x\rangle = \\ &= ({\bf X}_1{\bf P}_2 - {\bf X}_2{\bf P}_1)|\psi\rangle \end{align} so $$ {\bf J}_3 = {\bf X}_1{\bf P}_2 - {\bf X}_2{\bf P}_1$$

If a particle has internal degrees of freedom, then $$ {\bf J} = {\bf X} \times {\bf P} + {\bf S}$$ where ${\bf S}$ is the spin operator acting on these internal degrees of freedom.

$\endgroup$
  • $\begingroup$ Thank you very much. Together with @Valter Morettis proof this proves the equality. $\endgroup$ – TheoreticalMinimum Jul 14 at 11:22
0
$\begingroup$

Let's just consider $e^{iJ_z\phi}$ where $\hbar=1$. We have $$R_z=\begin{bmatrix}\cos(\phi) & \sin(\phi)&0\\-sin(\phi)&\cos(\theta) & 0 \\0&0&1\end{bmatrix}$$ $$J_z=\frac{1}{i}\frac{dR_z(\phi)}{d\phi}|_{\phi=0}=\begin{bmatrix}0 & -i&0\\i& 0 & 0 \\0&0&0\end{bmatrix}$$

We can exapnd exponential as: $$e^{iJ_z\phi}=1+iJ_z\phi-J_z^2\phi^2/2!-iJ_z^3\phi^3/3!+....$$ Or $$e^{iJ_z\phi}=\begin{bmatrix}1 & 0&0\\0 & 1 & 0 \\0&0&1\end{bmatrix}+ \phi \begin{bmatrix}0 & 1&0\\-1 & 0 & 0 \\0&0&0\end{bmatrix} +\phi^2/2! \begin{bmatrix}-1 & 0&0\\0& -1 & 0 \\0&0&0\end{bmatrix}+ ...=R_z(\phi)$$ We can do the same for other axes. Check out Quantum Field Theory by Lewis H. Ryder/ Single-particle relativistic wave equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.