Suppose that Alice prepared a hydrogen atom in her laboratory such that the electron is in the state $|\psi\rangle:=|l=1,m_z =0\rangle$. Here $m_z$ refers to the magnetic quantum number with respect to the $z$-direction.

Now the expectation value of the orbital angular momentum in the $x$-direction is

$$\left \langle L_x \right \rangle_\psi=0.$$

Suppose now a second observer Bob is moving relativistically with a velocity $\beta=v/c$ in the $z$-direction relative to Alice. Let's denote Bob's coordinates by $(t', x', y', z')$, which are related to Alice's coordinates $(t, x, y, z)$ by a Lorentz transformation.

What is the expectation value $\left \langle L_{x'} \right \rangle$, which Bob will obtain, according to some relativistic theory, e.g. QFT?

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    $\begingroup$ I've updated the question: Bob is moving in the z-direction and wants to measure L_x', which is the generator of rotations in the 2-3 plane of Bob. $\endgroup$
    – oliverkn
    Apr 12, 2021 at 13:45
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    $\begingroup$ This problem can be made equivalent to the question: what is the expectation value of the generator of boosts ($K_z$) between single particle states (in the sense of the Wigner classification). I think it would be extremely interesting to see how someone might go about doing that calculation. $\endgroup$ Apr 22, 2021 at 18:51
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    $\begingroup$ May we assume that Alice prepared the atom in a state with strictly zero momentum in the lab frame? That idealization makes the atom completely delocalized, because the zero-momentum eigenstate is invariant under spatial translations, but it also makes the question easier to answer. (Still not easy, but easier.) The atom still has $\ell=1$ and $m_z=0$, like you specified. Is that okay? $\endgroup$ Apr 23, 2021 at 3:42
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    $\begingroup$ @ChiralAnomaly I think it's fine to make this assumption and probably gives a good approximation to a calculation where one assumes a Gaussian wave packet. $\endgroup$
    – oliverkn
    Apr 23, 2021 at 6:38

1 Answer 1


Formulation of the question

The only input we need from relativistic QFT is that it carries a unitary representation of the Poincaré group. For the purposes of this question, we can treat the atom as an elementary particle with spin $1$. The intrinsic spins of the electron and nucleus are ignored in this question, so the "intrinsic spin" of $1$ represents the orbital angular momentum of the bound electron.

Let $|m,\mathbf{p}\rangle$ denote a state in which the atom has momentum $\mathbf{p}$ and in which the eigenvalue of $L_z$ is $m$: $$ \newcommand{\la}{\langle} \newcommand{\ra}{\rangle} \newcommand{\bfp}{\mathbf{p}} \newcommand{\bfx}{\mathbf{x}} \newcommand{\bfy}{\mathbf{y}} \newcommand{\pl}{\partial} \newcommand{\bfzero}{\mathbf{0}} L_z|m,\bfp\ra = m|m,\bfp\ra. \tag{1} $$ Since the atom has spin $1$, the allowed values of $m$ are $m\in\{1,0,-1\}$. The question says that Alice prepares a state with $m=0$. I'll assume the atom is "at rest" in Alice's lab, but that's ambiguous in quantum theory, because the atom can't be localized in any finite region of space unless the state is a superposition of different momentum eigenstates. Using such a superposition would complicate the analysis, so I'll assume that Alice prepares the atom in the state that is arbitrarily close to a momentum eigenstate with eigenvalue zero ($\bfp=\bfzero$). I'll write this state as $$ |\psi\ra = |0,\bfzero\ra, \tag{2} $$ with the understanding that this is really a superposition of different momenta (to make it normalizable) but concentrated in an arbitrarily small neighborhood of $\bfp=\bfzero$ (to simplify the analysis).

The goal is to determine the expectation value of a boosted version of $L_x$ in the state (2), which is the same as the expectation value of the original $L_x$ in a boosted version of the state.

The Lie algebra of the Lorentz group

The analysis becomes more intuitive after we modify the notation slightly. $L_z$ is the generator of rotations in the $x$-$y$ plane: it rotates the $x$ and $y$ directions and leaves the $z$ direction unchanged. For that reason, it is more natural to write $L_{xy}$ instead of $L_z$. Similarly, the generator of boosts in the $z$-direction is naturally denoted $L_{zt}$, because a boost is just a hyperbolic "rotation" in the $z$-$t$ plane, where $t$ is the time direction. The generators are antisymmetric with respect to exchanging the two indices, and since we want a unitary representation, they should also be hermitian as operators on the Hilbert space: $$ L_{ab}^\dagger=L_{ab} \hspace{2cm} L_{ba}=-L_{ab}. \tag{3} $$ Now the Lie algebra of the Lorentz group can be written in a simple way: \begin{align*} [L_{ab},L_{bc}] &\propto L_{ac} && \text{if }a\neq c \\ [L_{ab},L_{cd}] &= 0 && \text{if $a,b,c,d$ are all distinct}. \tag{4} \end{align*} In particular, $$ [L_{xy},L_{zt}] = 0, \tag{5} $$ which says that $L_z$ (in the original notation) commutes with a boost in the $z$-direction.

Analysis and result

Let $U$ denote a unitary operator that implements a boost in the $z$-direction. Equation (5) says that this operator commutes with $L_z$, so we have $$ L_z U|\psi\ra = UL_z|\psi\ra = 0, $$ which implies $$ U|\psi\ra \propto |0,\bfp\ra \tag{6} $$ for some momentum $\bfp$ in the $z$-direction. The goal is to calculate the expectation value of $L_x$ in this boosted state, which is equivalent to calculating the expectation value of a boosted version of $L_x$ in the original state.

Now, let $R_y$ denote a unitary operator that implements a reflection along the $y$-direction. To see what this reflection does to the state (6), remember that $L_z$ is the generator of rotations in the $x$-$y$ plane: it rotates the $x$ and $y$ directions and leaves the $z$ direction invariant. A reflection along the $y$-axis reverses the direction of such a rotation, so we have $$ L_z R_y =-R_y L_z. \tag{7} $$ Similarly, $$ L_x R_y =-R_y L_x. \tag{8} $$ For any momentum $\bfp$ with zero $y$-component, equations (1) and (7) imply $R_y|m,\bfp\ra\propto |{-m},\bfp\ra$. In particular, $$ R_y U|\psi\ra\propto U|\psi\ra. \tag{9} $$ Since $R_y$ is unitary, we have \begin{align*} \la\psi|U^\dagger L_x U|\psi\ra &= \la\psi|U^\dagger R_y^\dagger R_y L_x U|\psi\ra \\ &= -\la\psi|U^\dagger R_y^\dagger L_x R_y U|\psi\ra \\ &= -\la\psi|U^\dagger L_x U|\psi\ra, \end{align*} where equations (8) and (9) were used in the last two steps. This says that $\la\psi|U^\dagger L_x U|\psi\ra$ is equal to its own negative, so it must be zero.


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