65
$\begingroup$

The questions

Q1. What are simple ways to think mathematically about the physical meanings of the Planck constant?

Q2. How does the Planck constant appear in mathematics of quantum mechanics? In particular, quantization is an important notion in mathematical physics and there are various forms of quantization for classical Hamiltonian systems. What is the role of the Planck constant in mathematical quantization?

Q3. How does the Planck constant relate to the uncertainty principle and to mathematical formulations of the uncertainty principle?

Q4. What is the mathematical and physical meaning of letting the Planck constant tend to zero? (Or to infinity, if this ever happens.)

Motivation:

One purpose of this question is for me to try to get better early intuition towards a seminar we are running in the fall. Another purpose is that the Planck constant plays almost no role (and, in fact, is hardly mentioned) in the literature on quantum computation and quantum information, and I am curious about it.

Related MO questions: Does quantum mechanics ever really quantize classical mechanics? ;

$\endgroup$
  • 2
    $\begingroup$ In my humble opinion (I'm not a professor and don't know physics) one of the best talks (from an informative point of view) of quantum gravity is from the official channel of YouTube Instituto de Física Teórica IFT with title Gravedad y Mecánica Cuántica (César Gómez), posted at December 3rd, 2014. I believe that it doesn't answer all your questions, and you will need a friend knowing Spanish. The ponent César Gómez has other talks in this channel about what is the energy, or also the talk ¿Por qué los agujeros negros NO son negros? January 29th, 2019. Are the best talks that I know. $\endgroup$ – user142929 Sep 11 '19 at 11:02
  • 3
    $\begingroup$ $h$, like the Boltzmann constant $k_B$ and the speed of light $c$, is just a human construct that doesn’t appear in the laws of nature in their simplest form. It is an artifact of the fact that, historically, humans didn’t know that energy and frequency are “really the same thing”, much like energy and temperature ($E = k_B T$) and length and duration ($x = ct$) are “really the same thing” (the latter in particular mixing under Lorentz boosts). It’s as if we’d decided to use different units for different forms of energy, and had to introduce arbitrary conversion factors in every equation. $\endgroup$ – user76284 Sep 11 '19 at 20:56
  • 53
    $\begingroup$ If I had asked this question, it would have been closed. $\endgroup$ – bobuhito Sep 12 '19 at 3:01
  • 18
    $\begingroup$ @bobuhito We low-points users ("lusers", for short) just have to hope that the high-point users ask the questions we really want to know. Just like how poor people just have to hope that rich peoples' interests align with their own. It's just how The System works. $\endgroup$ – Somatic Custard Sep 12 '19 at 20:13
  • 3
    $\begingroup$ @Nat both constants are famously equal to 1, of course $\endgroup$ – AlexArvanitakis Feb 7 at 20:41

10 Answers 10

74
$\begingroup$

Let's give it a try. Of course, the precise mathematical meaning is perhaps absent, so the answers are sort of heuristic. But if I understand correctly, you want to gain intuition ;)

The first observation is that Planck's constant has units, it is not a numerical constant but carries physical dimension. In some sense, this makes all the difference: in your favorite unit system, $\hbar$ has the numerical value $1$. Period. Nothing more to say...

Now what is the impact? If you think of e.g. Fourier transform, the phase in your integral is $e^{ikx}$ and for a mathematician, everything is fine with this. Now for a physicist (depending on the application), $x$ has a unit of length. So there is no way to exponentiate a length per se, you can only plug in dimension-less quantities in a transcendental function. This means that the $k$ has a physical dimension of $1/\text{length}$. The physical interpretation is that $k$ is an inverse wave length. Now in quantum physics, there comes the time that you want your wave functions in the momentum representation. So you have to replace $k$ by the physical momentum $p$ which has a different unit, namely that of momentum. This requires dividing the product $px$ in your phase by a quantity of dimension momentum times length which is action. So the observation in physics is that there is a universal constant providing a scale for doing exactly that, $\hbar$. The familiar uncertainty relation you know from Fourier transform becomes thereby scaled by $\hbar$ as well.

Another, perhaps more important observation is that in quantization in the Schrödinger approach you consider wave functions on configuration space, depending on the position $x$ of dimension length. Now the relevant operators are the momentum operators encoded by a derivative. But derivatives have dimension $1/\text{length}$ instead of momentum. Thus you have to rescale the derivative by a physical constant of dimension action to get a quantity of dimension momentum. Again, $\hbar$ is the one doing the job.

From what I said (and much more can be said) it should be clear that $\hbar$ does not tend to zero at all (it's $1$, right?). So what should these statements $\hbar \to 0$ then really mean from a physical point of view? This is in fact a quite subtle and, I guess, ultimately not well understood point. The observation in daily life is that quantum effects do not play any role, the world behaves classically. So classical physics is at least a perfect approximation in many situations. Quantum physics only enters the picture if we perform very precise measurements etc. Now the idea is that the more fundamental theory (quantum) has a certain less fundamental theory (classical) as limit. The interpretation of this limit is subtle. But in many situations, the limit relates to parameters of the system which carry dimensions, typically the dimension of action. Now it is the ratio of this system-dependent parameter (say a combination of masses, lengths etc) and $\hbar$ which tells us whether the classical theory is a good approximation or not. The main point is that we need parameters $\alpha$ of the same dimension as $\hbar$ to have a dimension-less ratio $\hbar/\alpha$. Only such a dimension-less parameter can be considered to be "small" or "large". The classical limit is thus better understood as $\hbar/\alpha \to 0$, meaning that we look at many different systems with different values of $\alpha$ (whatever its concrete physical interpretation may be) and get the classical limit as limiting scenario if these parameters assume values much larger than $\hbar$. We can compare them since they have both the same physical dimension.

Now why don't we see these arguments in math? The (perhaps pretty obscure) observation is that in the explicit situations we can handle, the mathematical limit $\hbar/\alpha \to 0$ can also be understood as $\hbar \to 0$ while $\alpha$ is fixed. Mathematically this is not a big deal, but physically an absurd interpretation: $\hbar$ is a fundamental constant of nature and we have no $\hbar$-wheel where we can adjust its value.

Hmm, lot of blabla, but I hope that this clarifies the physical side of the story a bit. Mathematically, one has several ways to incorporate "dimensional" constants. One nice way is to look at graded algebras where the grading refers to the power of the dimension you are looking at. Then the grading helps to keep track of the correct dimensions. In particular in quantization theory this turned out to be a very useful tool.

| cite | improve this answer | |
$\endgroup$
  • 4
    $\begingroup$ I really like this answer, $\endgroup$ – Nik Weaver Sep 11 '19 at 13:03
  • $\begingroup$ Dear Stefan, many thanks for the answer! $\endgroup$ – Gil Kalai Sep 11 '19 at 16:32
  • 1
    $\begingroup$ Seconded: this is an awesome answer. $\endgroup$ – Alon Amit Sep 11 '19 at 20:14
  • 2
    $\begingroup$ @JohnJiang --- yes, it's because you cannot "add apples and pears" (you can only add quantities that have the same dimension) $\endgroup$ – Carlo Beenakker Sep 13 '19 at 4:22
  • 1
    $\begingroup$ “The familiar uncertainty relation you know from Fourier transform”: I wondered for how long this was known, and found that according to Primas (2017, p. 37), it was first proved by Karl Küpfmüller (Einschwingvorgänge in Wellenfiltern, Elektrische Nachrichtentechnik 1 (1924) 141–152) and simultaneously Norbert Wiener (Göttingen seminar, for which he takes credit in (1956, p. 107)). $\endgroup$ – Francois Ziegler Sep 17 '19 at 5:52
33
$\begingroup$

To build intuition for the Planck constant $\hbar$, which I understand is the purpose of the OP, I would start by noting that $\hbar$ is not a dimensionless number: it has dimensions of energy $\times$ time, or of momentum $\times$ position, meaning that it represents an action. (Equivalently, it could represent an angular momentum, I will get back to that at the end.)

Q1 & Q2: Mathematically, the action $S=\int_{t_1}^{t_2} Ldt$ is the integral of the Lagrangian $L[q(t),\dot{q}(t)]$, a functional of position $q(t)$ and velocity $\dot{q}(t)$ over time, along a path that is bounded in space and time. The evolution in time of a physical system is given by an integral over all paths of the exponent of $iS$. Since the exponent must be dimensionless, one needs to divide $S$ by some quantity with the dimension of action, and that quantity is the Planck constant. To properly define this path integral is the central problem of the mathematics of quantization.

Q4: Since $h$ is not dimensionless, what is meant by "taking the limit $h\rightarrow 0$ is that one first identifies a typical value $L_0$ of the Lagrangian $L$. There is no unique prescription for this. One then takes the limit $\epsilon=h/S_0\rightarrow 0$, where $S_0=(t_2-t_1)L_0$. In this limit one will find that the path integral of $e^{iS/\hbar}$ is dominated by the paths at which $S$ is extremal, with corrections that vanish as powers of $\epsilon$.

The OP also asks about the opposite limit, $h/S_0\rightarrow\infty$, which appears when one considers vanishingly small time intervals $t_2-t_1\rightarrow 0$. In that limit one will find that paths with very large velocity dominate. This is one manifestation of the uncertainty principle.

Q3: Even when $\epsilon\approx 0$, so a single path predominantly contributes to the path integral, there remain contributions from close-by paths, so one should actually think of this path as a thin tube. The resulting uncertainty in position and velocity corresponds to an uncertainty in the action $S$ of order $\hbar$.

Finally, the OP remarks on the absence of Planck's constant in the literature of quantum computation and quantum information. That is simply a choice (made in the early days of the field) to normalize the action $S$ by twice the angular momentum of the electron spin, which is just $\hbar$. So with that choice $S$ is dimensionless and Planck's constant $\hbar=1$, which is why it vanishes from sight.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Dear Carlo, many thanks for the answer! $\endgroup$ – Gil Kalai Sep 11 '19 at 16:33
14
$\begingroup$

To supplement the excellent answers previously given, some more remarks situated in the canonical quantization formalism, which is equivalent (at physicist level) to the path integral formalism referred to by Carlo Beenakker:

When we open our eyes, we see that we're given a playground in which we can organize what we see according to position $x$, time $t$, angle $\phi $, etc. Soon afterwards, as inquisitive beings, we try to figure out how we can move around in our playground. We define conjugate objects, which, quantum mechanically, are the generators of these movements. The momentum operator $p$ generates a spatial translation by $x_0 $ of the playground via the operation $\exp (-ix_0 p/\hbar )$ on a state vector describing the playground. The Hamiltonian $H$ generates a temporal translation by $t_0 $ via $\exp (-it_0 H/\hbar )$. The angular momentum operator $L$ generates a rotation by angle $\phi_{0} $ via $\exp (-i\phi_{0} L/\hbar )$. These operations form groups - the described structure is very natural from that point of view.

So, our inquisitive nature leads us to organize our physical quantities in conjugate pairs, and in each instance, the product of the pair has dimensions of an action - hence, action takes on such a central role. In the case of the angle/angular momentum pair, since angles are dimensionless, angular momentum itself has same the units as an action, and one can swap measuring in units of $\hbar $ with measuring in units of an angular momentum characteristic of the system, as Carlo notes.

For any dimensionful quantity, one has to provide a scale. It doesn't really make sense to declare a dimensionful quantity "large" or "small" without stating what one is comparing to. A fly will disagree with me about the size of a penny. $\hbar $ is an important scale for action because it controls whether our system will display quantum behavior (typical action comparable to $\hbar $) or classical behavior (typical action large compared to $\hbar $ - or, $\hbar $ small compared to a typical action, which is where the notion of taking $\hbar \rightarrow 0$ comes from, as described by Stefan Waldmann).

One place where this becomes very apparent is, indeed, in uncertainty relations. The standard example is the position/momentum pair. If we define a position operator $x$ to partner the momentum operator $p$, the operator algebra encoding the transformation behavior described above is given by $[x,p]=i\hbar $. The corresponding uncertainty relation, $\Delta x \cdot \Delta p \geq \hbar /2$, follows directly from this. If the typical positions and momenta in the system are very large, as in a classical system, the uncertainties forced by this inequality become negligible by comparison, and quantum effects become invisible.

(Note that we don't argue in quite the same simple fashion for the other conjugate pairs mentioned above. In the case of time/energy, this is because we don't treat time as a dynamical variable, and hence do not define a time operator - time is an external parameter, rigidly given by a ticking clock. Until we consider relativity. The case of angles is tricky because of the compact nature of an angle - so, also there, we set up the algebra differently).

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Dear Michael, many thanks for the answer! $\endgroup$ – Gil Kalai Sep 12 '19 at 6:21
12
$\begingroup$

About your Q1: I think that the simplest—and most obvious—way to think mathematically about the physical meaning of Planck's constant $h$ is that it is a kind of quantitative measure of the departure from commutativity:
Although the word "quantization" does not have a uniquely defined mathematical meaning, it is almost always connected with the passage of our mathematical description from models built on commutative algebras of functions of (commutative) space-time coordinates to models involving non-commutative algebras of operators acting on Hilbert spaces.
In this sense, the most familiar noncommutative object in physics is the algebra of canonical commutation relations CCRs: $[x,p]=i\hbar\mathbf{1}$, which comes in place of the Poisson bracket of canonical coordinates, as a non-commutative analogue of phase-space coordinates.

Regarding Q3, Heisenberg's uncertainty relation $\Delta x\Delta p\geq \frac{\hbar}{2}$ is a direct consequence of the CCRs (as has already been mentioned in previous answers). So in some sense it might be reasonable to say that the quantitative measure of the departure from commutativity—that is, Planck's constant—provides at the same time a definite lower bound indicating the limits (imposed by physical reality itself and not technological limitations) of our probabilistic knowledge, as this is described in the frame of QM.

About your Q2: On the side of physics, the value of $h$ comes from Planck's interpretation of black-body radiation. Planck considered a multiple-oscillator model with discrete energy spectrum (borrowing an older "trick" of Ludwig Boltzmann), he then obtained an expression for the entropy per oscillator and demanded that this expression should be consistent with Wien's law of radiation. This led him to $E=h\nu$.
On the mathematics side, things are not so straightforward, mainly because, as mentioned by the OP,

there are various forms of quantization for classical Hamiltonian systems.

I think the simplest viewpoint is the one provided by (formal) deformation quantization: QM is considered as a deformation of classical mechanics, with the Planck constant being the deformation parameter. The algebra of smooth functions on the Poisson manifold is replaced with the same vector space but equipped with a new noncommutative associative unital product $\star$ (the Moyal product), whose commutator agrees, up to order $\hbar$, with the underlying Poisson bracket. This, combined with the Wigner–Weyl transform, provides the opportunity to consider QM as a "smooth" deformation of the algebra of classical observables rather than a "sharp" or "discontinuous" change of our view of the (mathematical) nature of the observables (functions on a manifold which suddenly become operators on a Hilbert space).

Regarding Q4, i understand that the question is intimately related to the Correspondence principle or the so-called classical limit of QM introduced by Niels Bohr. This roughly states that our formulation of QM should be able to recover classical mechanics via some continuity argument while $\frac{\hbar}{S}\to 0$. In my understanding this is not a rigorous mathematical postulate but rather a heuristic argument (quite common in the development of physical theories). However, its mathematical formulation is quite subtle: in some cases of simple systems and under quite binding assumptions (for example if we restrict ourselves to coherent states etc) this can be interpreted as $\hbar\to 0$ but in general, simply setting $\hbar$ equal to zero produces indefinite limits or physically meaningless (see arXiv:1201.0150 [quant-ph]) results: it is the values of various parameters -usually depending on the particular system- that should be taken into account, as has already been outlined in Stefan Waldmann's answer.
Attempting to discuss these limiting procedures from a more rigorous mathematical viewpoint, i think it is group contractions with respect to some continuous subgroup -determined by the parameters to be considered together with $\hbar$- which provide reasonable methods and results. (For example, this is the case in frame of deformation quantization; Moyal bracket reduces to Poisson bracket up to $\hbar^2$ in this setting). Interesting details on the origin and the motivations of this description can be found in: The classical limit of quantum mechanical correlation functions, K. Hepp, Comm. Math. Phys., 35, 265-277, (1974).
It is always important to have a broader sense of what we are looking for in the classical limit: Given the probabilistic nature of QM predictions vs the deterministic predictions of classical physics, it is important to keep in mind that apart from special states, it is not reasonable to expect to get the classical trajectories in the classical limit of QM (see for example What is the limit $\hbar\to 0$ of quantum theory); instead it seems more reasonable to expect a more "modern" interpretation of the correspondence principle:

In the classical limit, QM probability distributions are expected to be identified with the probability distributions of suitable ensembles of classical trajectories.

See for example Classical Limit and the WKB Approximation, Am. J. of Phys., L.S. Brown, 40, 371 (1972), where this last point is analyzed in a quite rigorous and technical manner.
Maybe the article: The classical limit of quantum theory, R.F. Werner, arXiv:quant-ph/9504016 might be of some further interest.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Dear Konstantinos, many thanks for the answer! $\endgroup$ – Gil Kalai Sep 12 '19 at 6:21
8
$\begingroup$

The previous answers all give good discussions of the physical significance of $\hbar$ and the classical limit "$\hbar \to 0$" (in large quotation marks), but few of them discuss your "motivation" comment that $\hbar$ rarely appears in the study of quantum computing and information. David Mermin dedicates an entire section of his great paper "From Cbits to Qbits" to that question:

Like my disapproving colleague, some physicists may be appalled to have finished what purports to be an exposition of quantum mechanics — indeed, of applied (well, gedanken applied) quantum mechanics — without ever having run into Planck’s constant. How can this be?

The answer goes back to my first reason why enough quantum mechanics to understand quantum computation can be taught in a mere four hours. We are interested in discrete (2-state) systems and discrete (unitary) transformations. But Planck’s constant only appears in the context of continuously infinite systems (position eigenstates) and continuous families of transformations (time development) that act on them. Its role is to relate the conventional units in which we measure space and time, to the units in which it is quantum-mechanically natural to take the generators of the unitary transformations that produce translations in space or time.

If we are not interested in location in continuous space and are only interested in global rather than infinitesimal unitary transformations, then $\hbar$ need never enter the story. The engineer, who must figure out how to implement unitary transformations acting over time on Qbits located in different regions of physical space, must indeed deal with $\hbar$ and with Hamiltonians that generate the unitary transformations out of which the computation is built. But the designer of algorithms for the finished machine need only deal with the resulting unitary transformations, from which $\hbar$ has disappeared as a result, for example, of judicious choices by the engineers of the times over which the interactions that produce the unitary transformations act.

Deploring the absence of $\hbar$ from expositions of quantum computer science is rather like complaining that the $I$-$V$ curve for a $p$-$n$ junction never appears in expositions of classical computer science. It is to confuse computer science with computer engineering.

So basically, quantum algorithms researchers are implicitly setting $\hbar = 1$, or more precisely they're sweeping it up in the various fixed time constants and so on that determine the gates' operation at the microscopic level, and only considering the value of the dimensionless ratios $(\Delta E) T/\hbar$ that arise from the Schrodinger equation.

| cite | improve this answer | |
$\endgroup$
7
$\begingroup$

Universal physical constants like $c$ and $h$ have numerical values that are accidents of our choice of units, and often people choose units such that they equal 1. So the numerical value of Planck's constant has no significance.

However, no change of units can make a nonzero number equal zero, so the fact that $h\ne0$ does have physical significance.

One way in which this comes up is that in a phase-space plane of $(x,p)$, where $x$ is position and $p$ is momentum, we can't localize a state to an area smaller than $h$. In terms of concrete wavefunctions $\Psi(x)$, you can think of this as a property of Fourier transforms. More generally, it comes about because the position and momentum operators don't commute. In a universe where $h=0$, we would be able to localize states to points in phase space.

This has implications for statistical mechanics. The third law of thermodynamics holds because of this discreteness of phase space.

There is this whole idea of recovering classical physics by letting $h\rightarrow0$, which gives a correspondence principle between classical and quantum physics. This is not the only way of making the correspondence principle pop out, nor does it always work. Another common way in which this limit occurs is the limit of large numbers of particles. For example, nuclei have ~100 particles in them, so they are in between the classical and quantum limits. Since $h$ can't actually change (remember, it equals 1), a better way of thinking about the $h\rightarrow0$ limit is to instead think of a limit where the other variables describing your system are such that something else becomes big compared to $h$. For example, a spinning basketball has $\sim10^{34}\hbar$ of angular momentum, which is why we don't notice that its process of slowing down consists of quantum jumps.

Another purpose is that the Planck constant plays almost no role (and, in fact, is hardly mentioned) in the literature of quantum computation and quantum information and I am curious about it.

This is probably symptomatic of the fact that $h$ plays no role at all in the foundations of quantum mechanics. If you look at axiomatic formulations of quantum mechanics, they basically are descriptions of a kind of information theory. They don't have anything to do with space or motion. Examples below.

Hardy, "Quantum Theory From Five Reasonable Axioms," https://arxiv.org/abs/quant-ph/0101012

Masanes and Mueller, "A derivation of quantum theory from physical requirements," https://arxiv.org/abs/1004.1483

Mackey, The Mathematical Foundations of Quantum Mechanics, 1963

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Dear Ben, many thanks for the answer! $\endgroup$ – Gil Kalai Sep 12 '19 at 18:21
5
$\begingroup$

Q1. What are simple ways to think mathematically about the physical meanings of the Planck constant?

See: "Determination of the Planck constant using a watt balance with a superconducting magnet system at the National Institute of Standards and Technology" (Apr 24 2014), by Stephan Schlamminger, Darine Haddad, Frank Seifert, Leon S Chao, David B Newell, Ruimin Liu, Richard L Steiner, and Jon R Pratt along with this APS Physics article: "Living with the New SI" (March 25, 2019, Physics 12, 33):

"Over the past several decades, researchers have developed two kilogram-to-Planck experiments. The first, called the Kibble balance, works by offsetting the downward force of gravity on a chunk of metal with an upward magnetic force on a coil held in a magnetic field. Researchers tune the magnetic force by running current through the coil, and that current is measured in terms of Planck’s constant. The second experiment, conceived by the International Avogadro Project, involves fabrication of a near-perfect sphere of silicon. Using a combination of x-ray crystallography and optical interferometry, researchers can count the number of atoms in the sphere and connect its mass to Planck’s constant

In 2017, both of these methods returned values of Planck’s constant — based on the standard kilogram — having a precision of 10 parts per billion (ppb). These highly precise demonstrations have now allowed the metrology community to “turn the tables” by making Planck’s constant the defined quantity rather than the kilogram. As a result, the kilogram inherits the uncertainty that previously appeared in the Planck measurement.".

The Planck becomes: $h = 6.626 \, 069 \, 79(30) × 10^{−34} \;\text{J} s.$ That increases the uncertainty of the mass of the kilogram by 10 micrograms, a change at the 10 ppb level.

A better physical definition of the Planck permits normalization. A simple explanation without too much overlap is on the Wikipedia webpage "Natural Units":

In physics, natural units are physical units of measurement based only on universal physical constants. For example, the elementary charge $e$ is a natural unit of electric charge, and the speed of light $c$ is a natural unit of speed. A purely natural system of units has all of its units defined in this way, and usually such that the numerical values of the selected physical constants in terms of these units are exactly $1$.

If you nondimensionalize your variables and "normalize" the numerical values of certain fundamental constants to $1$ you can simplify and speed up your calculations. You need to be careful not to mix dimensioned with dimensionless quantities when using Planck base units and Hartree atomic units. You also can not normalize all your constants and need to choose your set carefully. There can also be a loss of precision, Planck units use the gravitational constant $G$, which is measurable in a laboratory only to four significant digits.

These are the Planck units based on Lorentz–Heaviside units (instead of on the more conventional Gaussian units), the rationalized Planck units are defined so that $c$ $=4\pi$$G$ = $\hbar$ = $\epsilon_{0}$ = $k_{\text{B}}$=$\,1$.

In Wikipedia's explanation of derived units it mentions:

  • A speed of 1 Planck length per Planck time is the speed of light in a vacuum, the maximum possible physical speed in special relativity; 1 nano-(Planck length per Planck time) is about 1.079 km/h.
  • Our understanding of the Big Bang begins with the Planck epoch, when the universe was 1 Planck time old and 1 Planck length in diameter, and had a Planck temperature of 1. At that moment, quantum theory as presently understood becomes applicable. Understanding the universe when it was less than 1 Planck time old requires a theory of quantum gravity that would incorporate quantum effects into general relativity. Such a theory does not yet exist.

With Planck units, the units are defined by properties of quantum mechanics and gravity. Not coincidentally, the Planck unit of length is approximately the distance at which quantum gravity effects become important. Likewise, atomic units are based on the mass and charge of an electron, and not coincidentally the atomic unit of length is the Bohr radius describing the "orbit" of the electron in a hydrogen atom.

In the Hartree system the numerical values of the following four fundamental physical constants are all unity by definition:

In Hartree atomic units, the speed of light is approximately 137 atomic units of velocity. See the webpage: "8.1: Atomic and Molecular Calculations are Expressed in Atomic Units" for an example of how setting the numerical values of four fundamental physical constants to unity permits simplification of the Hamiltonian.

Cosmology
Main article: Chronology of the Universe

In Big Bang cosmology, the Planck epoch or Planck era is the earliest stage of the Big Bang, before the time passed was equal to the Planck time, $t_P$, or approximately $10^{-43}$ seconds. There is no currently available physical theory to describe such short times, and it is not clear in what sense the concept of time is meaningful for values smaller than the Planck time. It is generally assumed that quantum effects of gravity dominate physical interactions at this time scale. At this scale, the unified force of the Standard Model is assumed to be unified with gravitation. Inconceivably hot and dense, the state of the Planck epoch was succeeded by the Grand unification epoch, where gravitation is separated from the unified force of the Standard Model, in turn followed by the Inflationary epoch, which ended after about $10^{-32}$ seconds (or about $10^{10} \; t_P$).

Q2. How does the Planck constant appears in mathematics of quantum mechanics. In particular, quantization is an important notion in mathematical physics and there are various forms of quantization for classical Hamiltonian systems. What is the role of the Planck constant in mathematical quantization.

See above, the Planck is a physical unit of “action” which sets the scale at which effects of quantum physics are genuinely important and physics is no longer well approximated by classical mechanics/classical field theory. See: "Planck's constant in geometric quantization".

Q3. How does the Planck constant relate to the uncertainty principle and to mathematical formulations of the uncertainty principle.

Any calculation of the position and momentum of an object (at the quantum level) involves some uncertainty and Heisenberg's uncertainty principle states that complementary properties cannot be observed or measured simultaneously. Knowing the value of a Planck reduces the error in the equation: $$\sigma _{x}\sigma _{p}\geq {\frac {\hbar }{2}}~~\text{where ħ is the reduced Planck constant, h/(2π).}$$

Q4. What is the mathematical and physical meaning of letting the Planck constant tending to zero. (Or to infinity, if this ever happens.)

I'll leave you with this answer: When does ℏ→0 provide a valid transition from quantum to classcial mechanics? When and why does it fail?

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Dear Rob, many thanks for the answer and welcome to MO! $\endgroup$ – Gil Kalai Sep 14 '19 at 19:24
2
$\begingroup$

Waves (electromagnetic, acoustic, etc) carry energy. Classical physics calculates this energy, or rather the energy flux which, for example, is responsible for heating the surface on which sunlight falls. This energy is always proportional to the square of the wave amplitude.

In order to explain an important puzzle with the radiation emitted by a "black body" in 1900 Planck introduced the hypothesis that there are "portions", or quanta, of radiation energy: ℏ*omega, where omega is frequency, and ℏ is the Planck constant. Thus for a given frequency there is a minimal portion of energy that can be transmitted by a wave of any nature.

This idea has allowed to understand (among many other things) the stability of atoms. A Hydrogen atom consists of an electron rotating around the proton. Since a rotating charge should emit electromagnetic radiation, classically the rotating electron should loose energy and finally fall on the proton, and calculations show that this should happen within ~1 nanosecond. According to Planck and quantum mechanics, this does not happen because the quantum becomes larger than the actual energy of the electron: it rotates, but does not have enough energy for emitting a single quantum. Thus our existence, and the existence of the Universe that we know is due to the fact that Planck constant is non-zero!

The internal angular momentum (called "spin") of electron, proton, neutron is equal to ℏ/2, it is a quantum-mechanical phenomenon.

For any quantum formula, the limit ℏ->0 gives the corresponding classical result (which, in some cases, can be zero or infinity, the latter case meaning that quantum mechanics is indispensable).

Evidently, there are purely classical phenomena in which the elementary quantum is of no importance (e.g. the frequency of rotation of the weals of my car is so low that the energy quantum is negligible, compared to the mechanical energy of the rotating weal).

Consequence: all quantum formulas contain the Planck constant, the limit h->0 giving the classical result. In particular, quantum computing cannot exist in this limit. This means that there should be a condition for the possibility of QC of the form A*ℏ >1. The expression for this A is presently unknown.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Dear Michel, welcome to MO! $\endgroup$ – Gil Kalai Sep 24 '19 at 17:34
  • $\begingroup$ The Planck constant ℏ is fundamental for quantum mechanics, like the speed of light c is fundamental for relativity. The obvious fact that one can use arbitrary units for physical quantities (and, in particular, choose units where ℏ=1, or c=1) is trivial and has no meaningful content, because what matters are the ratios between physical quantities, e.g. the Empire State building is 100 (?) times taller than me, and this fact does not depend on the chosen units. $\endgroup$ – Michel Dyakonov Oct 2 '19 at 10:03
  • $\begingroup$ (cont) The harmonic oscillator with proper frequency omega can be described classically (with energy E being an arbitrary positive value) under the condition that E>> ℏomega, otherwise it becomes important that the values of energy are discreet with energy difference ℏomega. QM also explains why the electron in the Hydrogen atom does not loose energy in radiating em waves eventually falling on the proton (classical electrodynamics calculates the time for this as 1 nanosecond). Because of QM this does not happen: there is a ground state with minimal energy -me^4/2ℏ^2. $\endgroup$ – Michel Dyakonov Oct 2 '19 at 10:20
  • $\begingroup$ (cont) Thus, in the limit ℏ->0 all quantum mechanics (quantum computing including) disappears. It is rather bizarre that the theory of quantum computing has no use for the standard attributes of quantum mechanics: Planck constant, Energy, Hamiltonian, and Schroedinger equation. The single notion of "entanglement" is by far not sufficient to understand the properties and the evolution of a quantum system, like e.g. a quantum computer. $\endgroup$ – Michel Dyakonov Oct 2 '19 at 10:31
2
$\begingroup$

For the Planck constant the rule is simple : any time you have a derivative term $\partial$ then there is also the Planck constant $\bar{h}\partial$. This is true for space $\bar{h}\nabla$ and for time $\bar{h}\partial_t$. For example we have the Schrödinger equation $$i\partial_t \phi = -\frac{\bar{h}^2}{2m}\Delta\phi +V\phi.$$ As mention in the others post, a physicist would say it is a dimension question. A mathematician would say the world is a 4 dimension manifold, some quantity are scalar but other like $p=\bar{h}\partial$ belong to the tangent space or the cotangent space. These depend on the choice of the charts on the manifold (or just the basis if the manifold is simply a vector space). So one can see $\bar{h}$ as the constant associated to usual choice of basis which in real life has been made once for all by the International System of Units (at least if you follow the metric system.)

Q2 : Quantification can be a important and difficult step. The only rule here is to have $$[X,P]=i\bar{h}.$$The usual choice is $X=x$ (the multiplicative operator) and $P=i\bar{h}\partial_x$. But of course $X$ and $P$ are symmetric and one can choose $X=i\nabla_k$ and $p=\bar{h}k$ (multiplicative operator). Both are equivalent by the Fourier transform. We usually expect to recover the classical mechanics in some point. One should first mention here is the Ehrenfest theorem (https://en.wikipedia.org/wiki/Ehrenfest_theorem). With $H(x,p)$ a hamiltonian depending on $x$ and $p$ (for example $H(x,p)=p^2/m+V(x)$), we have $$\partial_t\langle X \rangle = \langle\partial_p H (X,P) \rangle ,\quad \partial_t\langle P \rangle = -\langle \partial_x H (X,P) \rangle $$ which are the classical Hamiltonian equations. This follows directly from the Schrodinger equation and the commutating rules of $X$ and $P$. The constant $\bar{h}$ appears if one now consider evolution of non linear term as $\partial_t \langle X^2 \rangle$. From a physical point of view the particle is a wave function diffusing in time.

Q3 : For a physicist point of view the Heisenberg uncertainty principle played a major role in the quantum revolution. It gives a clear proof that there is no way to keep thinking particle in the classical way with position and motion and one have to see particles also as wave (Schrödinger) or to use matrix mechanics (Heisenberg). From a mathematical point of view this is only a elementary remark : if a function is located in space, then its Fourier transform is spread. A other nice interpretation (when we deals with fermions) is that $\bar{h}$ ($\bar{h}^3$ in three dimension) is the space needed to "park" a particle in the phase space. Here we can mention the Weyl law https://en.wikipedia.org/wiki/Weyl_law.

Q4 : Studying quantum system in the limit when $\bar{h}\rightarrow 0$ is a whole mathematical area called semiclassical regime. Some of the most interesting results reveal some correspondence between classical trajectories and eigenvectors of the Hamiltonian (or usually just the Laplacian) of large eigenvalue (In some case $\lambda\rightarrow \infty$ is the equivalent to $\bar{h}\rightarrow 0$) I would also mention here the buzz word "Quantum Chaos"

Motivations: For quantum computing and quantum information physicists only works with system already quantized with a few quantum states. $\bar{h}$ doesn't play any role since they have already reduce their system. It only implies that in order to play with a few quantum states one should use only system made of single atoms or electrons.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Dear Raphael, many thanks for the answer! $\endgroup$ – Gil Kalai Sep 14 '19 at 19:24
  • $\begingroup$ "It only implies that in order to play with a few quantum states one should use only system made of single atoms or electrons" - ANY system is made of single atoms or electrons! And $\hbar$ DOES play an important role, because a superposition of states with different energies $E_1$ and $E_2$ will oscillate in time with a frequency $\frac {E_1 - E_2} \hbar$ (according to Schroedinger's equation). $\endgroup$ – Michel Dyakonov Oct 3 '19 at 9:58
  • $\begingroup$ @RaphaelB4: The correct LaTeX code is \hbar instead of \bar h. $\endgroup$ – Alex M. Oct 3 '19 at 10:35
2
$\begingroup$

There are already good answers for Q1-Q3, so I'll clarify Q4.

In physics, quantities typically have a dimension, such as length, mass, time etc., and a unit, such as metre, kilogram, second, etc. Pure mathematical numbers have neither a dimension nor a unit.

Let's consider time. A possible mathematical model for time is to use a one-dimensional oriented affine space $\mathbb{T}$ over the real numbers. By choosing an origin ("time zero") $O \in \mathbb{T}$ and a ("positive") reference time $t_1 \in \mathbb{T}\setminus O$, such that $t_1$ appears after $O$ with respect to the orientation, the pair $(O, t_1)$ provides an oriented affine coordinate base. This coordinate base induces an isomorphism of oriented real affine spaces $\mathbb{T} \to \mathbb{A}_\mathbb{R}^1$, where the real affine space $\mathbb{A}_\mathbb{R}^1$ carries the orientation induced by the real numbers. In particular, we map $O \mapsto 0$ and $t_1 \mapsto 1$.

This isomorphism of real affine spaces also induces an isomorphism of the corresponding real vector spaces of oriented time intervals (durations). In $\mathbb{T}$, the vector $T_{ref}$ with $O + T_{ref} = t_1$ gives a reference unit $T_{ref}$ for durations. Given a duration $T$, we obtain a real number $t$ by $t := T/T_{ref}$. In this sense, the reference unit $T_{ref}$ provides a time scale.

An equation of time intervals can be made dimensionless by dividing each occurring time by the reference time $T_{ref}$ and also each variable of a duration by $T_{ref}$.

Example. If we are given an equation for time events, which is $$2 s + T = 5 s,$$ and if my reference time length is $T_{ref} = 1 s$, then I will declare $t := X/T_{ref}$ and I will obtain in dimension-less form $$2 + t = 5.$$ This equation basically has the same structure and the same look, but it is dimensionless. It is an equation for real numbers only.

Now, it is often reasonable to have physical processes, where different choices of a good reference time are appropriate. Say, we are given an atomic process, where a good reference time length is $T_1 := 10^{-12} s$, and a geological process, where a good reference time length is $T_2 := 10^{12} s$. If we make times dimensionless by $t := T/T_1$, then the other option would be to consider $T/T_2 = T_1 / T_2 \cdot T/T_1 = \varepsilon \cdot t$ with a parameter $$\varepsilon := T_1/T_2 \ll 1$$ which is (positive and) very small. So, if we make atomic times dimensionless by considering $t$, and if me make geological times dimensionless by considering $\varepsilon t$, then the small parameter $\varepsilon$ will occur in our mathematical equations.

A possible treatment would perhaps be to apply a perturbation method, where we take formally $\varepsilon \to 0+$ in the equations to obtain a reduced problem, which might be better treatable, and then try an expansion ansatz in $\varepsilon$. The limit $\varepsilon \to 0+$ is only a mathematical limit. As for the physical problem, it means, that the quotient of the choosen reference times $T_1/T_2 \to 0+$, because $T_1 \ll T_2$, in order to obtain a proper reduced physical problem.

Basically, the same thing happens with "$\hbar \to 0+$." One chooses a reference length, reference time, and reference mass, and one rewrites the equations in a dimensionless form. If we have another good reference time (or reference length or reference mass), then a small parameter $\varepsilon$ will appear, maybe in exactly the same positions, where $\hbar$ has occurred in the original physical equations. So one simply writes $\hbar$ instead of $\epsilon$, to keep the notational form of the original physical equations, but in fact, dimensionless pure mathematical equation are meant, that now contain an additional small parameter.

| cite | improve this answer | |
$\endgroup$

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.