In classical mechanics, the angular momentum $\textbf{L}$ of a particle is $$ \mathbf{L = r\times p} $$ where the $\mathbf r$ is the position of the particle measured from some origin. That means that $\mathbf L$ is dependent on the choice of the origin. Indeed the angulur momentum $\mathbf L'$ wrt the point $\mathbf{r}_{0'}$ is $$ \mathbf L' = \mathbf r'\times \mathbf p = \mathbf{r}\times \mathbf p - \mathbf{r}_{0'}\times \mathbf p = \mathbf L - \mathbf{r}_{0'}\times \mathbf p $$ where $$ \mathbf{r'} = \mathbf{r} - \mathbf{r}_{0'}\\ $$ Furthermore scaling the mass of the particle scales $\mathbf L$ in the same way.

In quantum mechanics, the story of the angular momentum is different. It is only possible for the (square of) magnitude and one component to exist simultaneously. The eigenvalues of these two operators are $\hbar^2\ell(\ell+1)$ for $\hat L^2$ and $\hbar m$ for $\hat L_z$, where $-\ell \le m \le \ell$. But the only constant that appears in them is $\hbar$ and nothing else. It is not clear how these only-possible values are dependent on the mass $m_0$, the energy $E$ and the choice of the point $\mathbf{r}_{0'}$ wrt to which the angular momentum is measured.

I hope anyone can explain something explicity.


2 Answers 2


Quantum mechanical systems are usually defined in an inertial frame of reference associated typically to an observer. Let us consider two observers $O$ and $O'$ describing the same system $I$. Each observer uses a complete set of commuting operators to describe the system (its possible states, possible measured observables, and results of measurements, probabilities, etc.) acting on each own's Hilbert space $\mathcal {H}_{O}$ and $\mathcal{H}_{O'}$.

Let us describe by $T$ the space-time transformation which leads the coordinates of $O$ to be identical to the coordinates of $O'$ (either a rotation, a translation in space and/or time, or a Galilean boost or any combination of them). To this $T$, the mathematical formalism assigns a unitary or an antiunitary operator $U_T :\mathcal{H}_{O}\rightarrow \mathcal {H}_{O'}$.

Let us consider that both observers wish to measure angular momentum ($O$ along the $z$ axis and $O'$ along its $z'$ axis).

Then we have:

$$ \displaystyle{L_z'= U_T L_z U_T ^{\dagger}}.$$

This is the way the observables angular momentum are connected in QM. You can check for youself that the spectrum of $L_z$ is unchanged by switching between frames with the help of $U_T$.

The observable energy is described by the operator called Hamiltonian. Its transformation law by moving from one space-time observer to another is given by an identical rule. Mass is a trivial observable, so it is simply defined by a positive scalar number $m$ times the unit operator on the Hilbert space of each observer. We say that mass is not frame-dependent. Systems with variable mass (such as the rockets in the extension of Newtonian mechanics due to Tsiolkowskii and Meshcherskii) are not in the standard formalism (see, for example, a paper by Fushchych https://www.imath.kiev.ua/~fushchych/papers/1968_2.pdf).

  • $\begingroup$ Thanks for your attempt to answering my question, but it is about "DEPENDENCE of the Eingenvalues of the Angular Momentum on the Reference Point and the mass", not about how mass or energy transform between reference frames $\endgroup$
    – Physor
    Feb 17, 2021 at 18:20
  • $\begingroup$ There is no dependence of the angular momentum eigenvalues on reference point and mass. $\endgroup$
    – DanielC
    Feb 19, 2021 at 23:40

I think I found where the dependence on the reference point in the quantum case should be. This was a deceptive question to myself. Well this dependence has moved to the expansion coefficients and to the expectation values.

The eigenstates of the quantum angular momenutm are just possible states for the particle, the particle doesn't need to be in any of them. These are states wrt a chosen origin (reference point). The corresponding eigenvalues are just themselves. But the expectation values of the angular mmomentum of the state of a given particle in space are dependent on which reference point the eigenstates are wrt (centered at). More precisely, for two different reference points, the two sequences of expansion coefficients of the same wave function (same particle) are different, because these are two different bases (two different sets of the eigenfunctions, each set centered at a distinct reference point.) Therefore the expectation values can be expected to depend on the reference point.

Example: the particle in an eigensates $|\Psi\rangle = |l,m,\alpha \rangle$, $\mathcal{O}$ is a coordinate system, and $\alpha$ is rest of the quantum numbers. So $\langle L_z\rangle = m$ and is sharp. In a translated coordinate system $\mathcal{O}'$ (not along $z$) the same state $|\Psi\rangle$ is expanded wrt a different basis $\{|l',m',\alpha \rangle'\}$ and so the value of $\langle L_{z'}\rangle$ is not sharp.


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