I am new to the area of quantum computing, but as I study it I can't help wonder about the role of the Planck's constant. When one studies quantum mechanics one often finds statements that quantum mechanics reduces to classical physics if Planck's constant was taken to be zero. While this statement is a bit vague, there are clear aspects of quantum mechanics that would not hold without a finite non-zero value of the Planck's constant. Uncertainty in "simultaneous" measurements of conjugate observables is one, for example. Since in many quantum algorithms measurements are carried out at the end, it appears that the Planck's constant can be chosen to be anything, including, zero, throughout all the unitary transformations of the state up until the end. In that case, what exactly is uniquely quantum about all the unitary transformations of the state? After all, states of classical systems can be defined as elements of Hilbert space. Entanglement and superposition of such classical states would be the result of them being elements of Hilbert space. Bottom line is that I have hard time seeing the reason for "quantumness" of quantum computing without the need for setting some non-zero value of the Planck's constant. Thank you.



3 Answers 3


I like this question although the question and answers may be a bit vague/a moving target.

Initially I'll quote Gil Kalai who asked a very similar question on MathOverflow:

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.

We can take Marin Vesely's observation and work in the natural units system, as that appears to hide the (dimensioned) $\hbar$ by setting it to $1$, and enabling exponentiation of (dimensionless) time as in the time-dependent Schrodinger equation.

I think this may depend upon how qubits are embodied, as we must trace this out to find a hidden $\hbar$.

For example, Carlo Beenakker (whose profile pic is of $\hbar$!) answers Kalai's question. Although I don't understand all of Beennakker's answer, if qubits are embodied as spin states that quantum computing is silent on Planck because:

That is simply a choice (made in the early days of [quantum computing]) 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.

We can also ponder/embody qubits as polarization states of a photon. There, it's not so clear to me what the action corresponds to; however, that's not to say that Planck's constant is or could be set to $0$.

For example, what is still fundamentally "quantum" are the photons of light itself. That is, if we work with linear optical quantum computing, we can measure non-commuting properties of photon polarization, but we are acting on the individual quanta of light.


Great question, and David Mermin dedicates an entire section of his great paper "From Cbits to Qbits" to it:

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,gedankenapplied) 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 infinitessimal 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 aren't setting $\hbar = 0$, instead, they're implicitly setting it equal to 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.

To put it another way, Shankar lists the four fundamental postulates of nonrelativistic quantum mechanics as (greatly abbreviating):

  1. Physical states are represented by vectors in a Hilbert space
  2. Measurements yield eigenvalues of the corresponding operator with probability given by the Born rule and change the state to the measured eigenstate.
  3. The canonical commutation relation $[x_i, p_j] = i \hbar \delta_{ij}$
  4. The Schrodinger equation for time evolution.

As Mermin points out, by abstracting away the microscopic continuum details of how the qubits and gates actually work at the physical level and representing the gates as unitary operators that act over a finite time span, algorithm developers can usually get away with ignoring $\hbar$. (Although the device engineers certainly need to understand the postulates 3 and especially 4 very well. And $\hbar$ is often buried very shallowly beneath the surface - for example, if your qubits are physically instantiated by the spin states of a spin-1/2 particle or a photon, then the $|0\rangle$ and $|1\rangle$ computational basis states are angular momentum states that differ by $\hbar$ or $2\hbar$.)

There is clearly some sense in which postulates 3 and 4 "go away" in the formal limit $\hbar \to 0$ (although we need to very careful what that limit actually means), since they explicitly contain $\hbar$, but postulates 1 and 2 clearly "survive", at least when the limit is taken in the appropriate way. And indeed, any quantum algorithms researcher needs to be extremely familiar with those first two postulates, even if the second two can be abstracted away.

Finally, I'll mention that $\hbar$ rarely appears in the gate-based formalism of quantum algorithms, but it comes up all the time in the (even purely theoretical) study of adiabatic quantum computing algorithms, which by their nature are a little more "analog" and more directly tied the underlying physics of the hardware.

  • $\begingroup$ Thank you for a very thoughtful and detailed answer. I have tried to think carefully through it and went back to David Mermin's paper. The problem I continue to have with this argument is that unitary transformations and states defined over HIlbert space can be implemented without quantum devices. This would imply that quantum algorithms can be implemented without quantum computer technology (devices). In other word, if one does not care for the value of Planck constant in algorithm design, one may set it equal to zero and quantum mechanics becomes sufficient, but not necessary for algorithms. $\endgroup$
    – QC-Novice
    Apr 14, 2020 at 10:58

Time evolution of quantum systems is described by Schrodinger equation $$ i \frac{h}{2\pi}\frac{\partial}{\partial t}|\psi(t)\rangle = H |\psi(t)\rangle. $$

So any change on quantum computer can be described by the equation. Or in other words any quantum gate is described by its Hamiltonian and the equation describe how to the gate acts. When we discretize time, we can describe acting of quantum gate with unitary matrix.

As you can see in the equation, Planck constant is inseparable part of it and it cannot be neglected. However, you can switch to so-called natural units system where reduced Planck constant (i.e $\frac{h}{2\pi}$), speed of light, gravitational constant, Boltzman constant and others are set to be equal 1. As a result, Planck constant seems not to be presented in equations describing quantum computing.

  • $\begingroup$ I think there's a difference between setting $\hbar=0$ and $\hbar=1$, or even $\hbar=\epsilon\gt 0$. In the "natural units system" $\hbar=1$. But the OP seems to ask if we can replace $\hbar=0$. $\endgroup$ Apr 10, 2020 at 22:28
  • $\begingroup$ @MarkS: I see. I think that it is imposible to set reduced Planck constant to zero as then there is no quantum effect and a quantum computer would not work. Right? $\endgroup$ Apr 10, 2020 at 22:32
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    $\begingroup$ That's my interpretation, especially after reading the question and answers to Gil Kalai's post on MO! But then, where is Planck's constant in qubits that are represented as the polarity of light? Where is it set to $1$ (i.e. in the natural units system?) $\endgroup$ Apr 10, 2020 at 22:36
  • $\begingroup$ Thank you. Of course, if you assume to begin with that quantum computing can be implemented only using quantum devices, you will certainly need non-zero Planck's constant in some unit system. The fact that Planck's constant is not used in formulating the unitary transformations means to me that it could be possible to implement equivalent computation without devices operating on the basis of quantum evolution (Schrodinger equation). Unitary gates, after all, are just norm preserving transformations. So, what am I missing? $\endgroup$
    – QC-Novice
    Apr 11, 2020 at 0:28
  • $\begingroup$ @QC-Novice I don't know, it's a great question. How do you define "on the basis of quantum evolution?" If you let your qubits be polarization states of photons, then there's no apparent hidden Planck's constant that I can see. You can entangle the qubits together, act on them unitarily, apply Shor's algorithm AFAIK, etc. But Planck isn't set to $0$ (or even $1$) in that case, it's just set to a "don't-care", right? $\endgroup$ Apr 11, 2020 at 1:18

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