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The quantity $\hbar$ appears in quantum mechanics by the canonical commutation relation $$[x, p] = i \hbar.$$ Would it be sensible to quantize different conjugate variable pairs with different values of $\hbar$, or is such a theory nonsensical? For example, we might postulate two types of particles, $A$ and $B$, with $$[x_A, p_A] = i \hbar_A, \quad [x_B, p_B] = i \hbar_B$$ where both $(x_A, p_A)$ and $(x_B, p_B)$ are classically canonically conjugate. This question also applies to quantum fields, where we can use different values of $\hbar$ for the field and field momentum commutation relations. Is such a theory automatically not mathematically self-consistent, or is it simply not physically useful? Is there any physical situation to which this would apply?


This really is a nontrivial change that cannot be "scaled away". For example:

  • Excitations will have "incompatible" de Broglie relations, i.e. $E_A = \hbar_A \omega_A$ and $E_B = \hbar_B \omega_B$. A system with definite energy would not necessarily have definite frequency!
  • Another example is that semiclassical states take up phase space area $h$, so I'm considering a setup where states of different systems are allowed to take up different areas.
  • Alternatively, think about quantizing the system by canonical quantization where $(x_A, p_A)$ and $(x_B, p_B)$ are canonically conjugate pairs. Obviously, in the standard canonical quantization, the value of $\hbar$ matters, because it sets the scale of quantum effects. So if I choose different values of $\hbar$ to quantize $(x_A, p_A)$ and $(x_B, p_B)$, the result will not be standard quantum mechanics.

I'm already aware that this is not what happens in ordinary quantum mechanics; I'm wondering if this modification of quantum mechanics has any use.

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    $\begingroup$ A "variable" $\hbar$ is commonly used in the semiclassical approximation. Anyways, if you consider the exponentiated commutation relations (as you should), in quantum mechanics there is only one way to irrepresent $[x,p]=i$ (in exponentiated form). This means that essentially $x_A/p_A$ and $x_B/p_B$ are multiples of each other by a scalar factor. $\endgroup$ – yuggib Mar 2 '17 at 7:51
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    $\begingroup$ Why would you ever want to keep this annoying scaling instead of just redefining $\tilde{x}_B = \sqrt{\hbar_B/\hbar_A} x_B$ and likewise $\tilde{p}_B$? Note that you could already rescale all $x_i,p_i$ in the usual theories by different factor to get "different $\hbar$" on the r.h.s. of the CCR. I don't see why you think this $\hbar_A,\hbar_B$ could have any physical impact, could you explain that? $\endgroup$ – ACuriousMind Mar 2 '17 at 10:22
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    $\begingroup$ This seems relevant: arxiv.org/abs/quant-ph/9503023 $\endgroup$ – Qmechanic Oct 29 '17 at 23:09
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    $\begingroup$ @ACuriousMind $x_B$ is an observable, the position of the particle - your $\tilde{x}_B$ might fulfill the canonical commutation relation but it does not have the meaning of particle position. If you fix the meaning of observables such as position, velocity, and energy, you cannot do away with the different Planck constant by any redefinition of parameters or operators. $\endgroup$ – Void Oct 31 '17 at 22:55
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    $\begingroup$ Simply stated, an electron with the same mass but a different Planck constant will have a different de Broglie wavelength in all the possible (very real!) experiments. And all people who try to tell otherwise are, honestly, the wrong type of theorists who have gotten themselves lost too deep in mathematical formalism. $\endgroup$ – Void Nov 2 '17 at 23:25
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Having different Planck constants for different particles violates energy-momentum conservation unless the different types of particles do not interact with each other. This can be seen by following the arguments in "New Test of Quantum Mechanics: Is Planck's Constant Unique?", E. Fischbach, G.L. Greene, R.J. Hughes, Physical Review Letters 66 (1991) 256-259.

Consider a simple non-relativistic one-dimensional system of two spinless particles with the same mass $m$ but different Planck constants $h_A$ and $h_B$, interacting through a potential $V$. Their Hamiltonian is $$H=\frac{p^2_A}{2m}+\frac{p^2_B}{2m}+V(x_A-x_B) =\frac{P^2}{2M}+\frac{k^2}{m}+V(r)$$ where $r=x_A-x_B$, $k=(p_A-p_B)/2$, $M=2m$, and $P=p_A+p_B$. The Planck constants for each particle relate their momentum and position through the commutation relations $$[x_A,p_A]=i\hbar_A,\quad[x_B,p_B]=i\hbar_B, \qquad\textrm{with }[x_A,x_B]=[p_A,p_B]=0$$ A quantity is conserved if it commutes with the Hamiltonian, but we find that $$[H,P]=[V(r),P]=i(h_A-h_B)\frac{\partial V}{\partial r}$$ which is not zero unless either $h_A=h_B$ or $V(r)$ is independent of $r$ (i.e. there is no force between the particles). So for momentum to be conserved, the two particles must have the same Planck's constant or they must not interact.

Experimental constraints on differences in Planck's constant are set by how well theories such as quantum electrodynamics work. If different types of charged particles had different $h$, each would also have their own value for the fine structure constant $\alpha = \frac{e^2}{4\pi\epsilon_0\hbar c}$. The extremely good agreement between measurements of $\alpha$ in systems involving different types of particles means that any differences in Planck's constant between those particles must be tiny. Fischbach, Greene, and Hughes set limits on fractional differences in the Planck's constants of electrons, photons, and neutrons at $< 10^{-7}$ in 1991, and newer measurements set even stronger constraints.

You may also want to look at the answer to the similar questions Why is Planck's constant the same for all particles? and Universality of Planck's Constant.

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    $\begingroup$ But note that a modified momentum $\tilde{P} = p_A + \frac{h_A}{h_B} p_B$ will be conserved. $\endgroup$ – Void Nov 4 '17 at 18:55
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Experimentally, the uniqueness of the Planck's constant is established through measurements based on the photoelectric effect, Hall effect, black body radiation etc. I don't know of any serious work contradicting this assertion.

However, the operator algebra given in the question with two different "Planck's" constants describes a valid quantum system, although it cannot be obtained as a quantization of the usual commutative algebra of translation operators in the canonical phase space $(\mathbb{R}^{2n}, dp_i\wedge dx^i )$ (at least not in any conventional theory of quantization).

Put in another way, not both $\hbar$s come into existence as a consequence of the quantization process, and the classical theory obtained by the conventional classical limit will depend on their ratio. Thus an independent limiting procedure of the two $\hbar$s like the one adopted in 9503023 mentioned in the comments above, will lead not only to different quantum systems but also to different classical systems.

A constructive way to obtain this operator algebra is to start from a phase space with a noncanonical symplectic structure: $$ \omega = \frac{\hbar}{\hbar_1}dp_1 \wedge dx^1 + \frac{\hbar}{\hbar_2}dp_2 \wedge dx^2 + ...$$ (This is a special case of a symplectic vector space $\omega$) In this case, the Poisson brackets will take the form: $$\left \{x_1, p_1 \right \} = \frac{\hbar_1}{\hbar}$$ $$\left \{x_2, p_2 \right \} = \frac{\hbar_2}{\hbar}$$ When this algebra is quantized according to the rule: $$\left \{A, B \right \} = C \rightarrow \left [\hat{A} ,\hat{B} \right ] =i \hbar \hat{C}$$

A remark

It is possible to scale the positions and momenta to make the algebra canonical. However mathematically the scaling transformation modifies the symplectic structure, therefore it is not a symplectomorphism; and strictly speaking it describes a different mechanical system.

The uniqueness of the Planck’s constant can be heuristically understood from the path integral quantization point of view. There, the paths are weighted by the complex factor: $$ e^{ i \frac{S}{\hbar}}$$ Where $S$ is the action. If we believe that there is an action describing all phenomena in nature (theory of everything) and a procedure making the Feynman path integral rigorous, then all systems of nature will be subject to the same Planck’s constant, the one in the denominator of the complex factor.

However, people use different notions of $\hbar$ in different branches of research, please see the articles in nlab about Planck’s constant and deformation quantization.

Let me please go into details of two cases: In the case of the classical phase space $(\mathbb{R}^{2n}, dp_i \wedge dx^i)$, the algebra of the Hamiltonian vector fields on $ \mathbb{R}^{2n}$ representing the translations on the phase space is commutative. This algebra acts on the functions on the phase space which consist of the classical observables. After quantization, the translation symmetries of the classical system are lifted to act on sections of a line bundle (consisting of the quantum Hilbert space). The lifted algebra is no longer commutative. It receives a central extension. In a given representation of this algebra, the value of the center must be a scalar, since it commutes will all other observables. The value of this scalar is $\hbar$, please see Tuynmann and Wiegerinck). This is also the reason why the two $\hbar$s in the question cannot be of quantum origin, as the geometric quantization process produces a single central extension.

In the problem of quantization of spin can be obtained from the geometric quantization of the two dimensional sphere $S^2$ (please see section 3.5 in the following lecture notes for a concise introduction). In this case, the classical algebra of Hamiltonian vector field is $SU(2)$ and it does not receive a central extension upon quantization. The same algebra acts on the quantum Hilber space. In this case, the Planck’s constant can only introduced as a scaling factor of the symplectic structure: $$ \omega = \frac{i}{\hbar}\frac{dzd\bar{z}}{(1+\bar{z} z)^2}$$ Where $z$ is the stereographic projection coordinate of the sphere. The problem of geometric quantization of the sphere has a solution only if the reciprocal of the Planck’s constant ${1}{\hbar}$ is quantized to $2S$ where $S$ is the spin representation of the quantum Hilbert space. The classical limit corresponding to $\hbar \rightarrow 0$ corresponds to $S \rightarrow \infty$. Thus very large spin representations correspond to the classical limit. There are many works discussing the quantization of Planck’s constant, see for example the following work by Eli Hawkins.

These two examples illustrate the difference in the notions of Planck’s constant in different quantum systems and why the conceptual problem of the uniqueness of the Planck’s constant is still open.

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