I'm looking at a section of Griffiths and Schroeter's Introduction to Quantum Mechanics, pp. 355. It states a straightforward set of equations that got me thinking about the exact way in which complex numbers manifest in classical mechanics.

Specifically, here's an excerpt:

The Schrödinger equation $$-\frac{\hbar^2}{2m}\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}+V(x)\psi=E\psi$$ can be written in the following way: $$\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2}=-\frac{p^2}{\hbar^2}\psi,$$ where $$p=\sqrt{2m[E-V(x)]}$$ is the classical formula for the (magnitude of the) momentum of a particle with total energy $E$ and potential energy $V(x)$.

I'm interested in the classical interpretation of this expression for momentum. My question is about classical mechanics, I mentioned the excerpt only to provide background about what motivated my confusion.

In classical mechanics, I'm tempted to say that the particle can never have position $x$ such that $V(x)>E$ (i.e. that is the classically forbidden region) because in those places, the equation above gives implies that $p$ is imaginary, which is forbidden. I'd appreciate it if someone could elucidate this claim ("the state is forbidden because it implies that an observable is imaginary").

Here are examples that make me confused about why the claim is rigorous:

  • We often model oscillators with complex exponentials and simply ignoring the imaginary component when reporting the actual expected momentum or position.
  • We often have complex wavenumbers and frequencies, which we interpret as damped motion.
  • Sometimes, instead of working with cartesian vectors, we use complex numbers for positions. Then, the complex portion indicates that the quantity is a component along an orthogonal axis.

Clearly, classical mechanics is built with provisions that make complex numbers acceptable under certain circumstances. How do we know that the imaginary momentum found here is truly meaningless? I cannot think of a possible significance. But is there an argument for why a physical significance shouldn't exist? Furthermore, in classical mechanics, is there a rigorous way to tell when the complex nature of calculated observables indicates that said calculations predict a physically un-realizable state?

My guess at an answer: in classical mechanics, complex observables are always a red flag (i.e. if you have to use complex numbers, then you are wrong, as implied by Can one do the maths of physics without using $\sqrt{-1}$?), and this proposition is consistent with the bullet-point examples in the question. For instance, the case of oscillators being written as phasors is confusing because it's shorthand notation. In reality, a harmonic oscillator is not $x(t)=x_0\mathrm{e}^{\mathrm{i}\omega t}$. It is actually $x(t)=\mathrm{Re}(x_0\mathrm{e}^{\mathrm{i}\omega t})$. On the other hand, complex wavenumbers and frequencies are not actually observables: the real parts can be measured as instantaneous real frequencies by evaluating the derivatives of the exponentials, and the complex parts can be directly observed as real numbers by fitting maximum and minimum points of observed real-space damped oscillation curves to real-valued exponential decrease envelopes.

Is this correct?


4 Answers 4


When you solve a differential equation you don't get a single solution, but rather a family of solutions. The solution that describes your system is selected by imposing the boundary conditions e.g. the initial position and velocity.

You give the example of a harmonic oscillator where a solution is:

$$ x(t)=x_0\mathrm{e}^{\mathrm{i}\omega t} $$


$$ x(t)=x_0\mathrm{e}^{-\mathrm{i}\omega t} $$

is also a solution, and since the equation is linear a sum (or difference) of the two is also a solution. Suppose we find the position at $t = 0$ is $x_0$ and the velocity at $t = 0$ is zero then we can use these initial conditions to find the equation for our system and we get:

$$ x(t) = \frac{x_0}{2}(\mathrm{e}^{\mathrm{i}\omega t} + \mathrm{e}^{-\mathrm{i}\omega t} ) $$

And of course this is just:

$$ x(t) = x_0 \cos(\omega t) $$

That is, our real initial conditions determined a real equation of motion despite the fact we used complex functions as our solutions. And I think this is how you should look at the situation. If we had complex values as our initial conditions we should expect the equation of motion to give complex observables. However experiment suggests that we never get initial conditions that are complex values, so we would never expect the equation of motion to predict complex observables.

If we look at it this way then your question reduces to whether complex initial conditions are unphysical. There is no answer to this except to say that never in the thousands of years that physicists have being doing measurements have we observed a complex valued initial condition. This doesn't prove that complex valued observables are unphysical, but most of us would take it as a reasonable working hypothesis.

  • $\begingroup$ -1: Given that complex numbers have only been around for the last few hundred years, its not surprising that physicists 'thousands' of years ago didn't use them. Of course we wouldn't have called them physicists but geometers, natural philosophers amd astronomers. Moreover, have you seen an Argand diagram? This gives a natural geometric interpretation of complex numbers and hence position in polar coordinates, which has been understood for thousands of years, and is a physical observable, has a natural interpretation in complex numbers. $\endgroup$ Feb 1, 2022 at 11:24

Introductory textbooks have the responsibility of conveying a lot of knowledge to the beginner in a reasonably short amount of time. In this endeavor they have to make use of all kinds of metaphors in order to make the facts memorable. Only a small fraction of students would stay with a book that builds everything on axioms and lists all the preassumptions in tedious detail like a math book (actually, the math guys often wish that to be true, because they have been socialized differently, and they are confused by all the fuzzy assumptions they never heard about).

The problem begins if one of the mnemonics is taken too seriously. Then the learner is going to get stuck because he thinks that there has to be some sense in what this or that authority has claimed to be true.

Two things in your question strike the eye with respect to what I have said above:

  1. the claim that imaginary quantities are "unphysical"
  2. the expression $p=\sqrt{\dots}$ which is claimed to be the "classical formula for momentum"

As to 1), the imaginary unit is nothing special when it comes to physics. The mathematicians can do all sorts of interesting and meaningful stuff with complex numbers. But for a physicist what counts is whether the mathematical quantities are able to represent measurements (at least indirectly, e.g. via the probability amplitude in QM) in a way that allow predictions or at least correlations. You can simply represent the imaginary unit as a real matrix $$i=\left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)$$ As you can easily check, $i^2=-1$ where $1$ denotes the unit matrix in 2D. Would you say that matrices indicate something "physically illegal"? Certainly no. And, as it turns out, the above matrix is the infinitesimal generator (the tangent, so to say) of rotations, and so you can most naturally describe two-dimensional rotations by unit complex numbers (you can even describe 3D rotations by a generalization of complex numbers, called quaternions!). The harmonic oscillator is such a kind of rotation: in a sense, energy "rotates" between kinetic and potential energy, or more precisely, the system rotates between maximum speed and maximum displacement. The fact that you have two degrees of freedom for such a rotation reflects the fact that the oscillator is second order, linear and has real coefficients (making the complex conjugate oscillator equivalent to the original one) and therefore, its solutions can be decomposed into sine and cosine solutions without losing information.

That is just an example of how the imaginary unit can represent a physical quantity. On the other hand, if you start out with the assumption of real coordinates and the solution of your differential equation or whatever yields complex or imaginary coordinates, you are in the trouble of explaining what that means in terms of measurements. Usually, one is not inclined to assume that complex position means the existence of some sort of extra-dimension. Ockham's razor demands that we first check the simplest explanation. And in the case of the harmonic oscillator it is simply that we are always able to choose real solutions and superpose them, so why arbitrarily choose complex solutions? In that sense, complex solutions of the harmonic oscillator are unphysical: we don't need them to explain reality, but we may use them if we explain what using them means (namely superposition of two orthogonal solutions in a very practical complex exponential expression). With quantum mechanics it is completely different: we can't explain "quantum reality" by a single-valued real wave function. We need to use $i$ (or an equivalent matrix) to make everything fit together. So, the imaginary unit is not unphysical in quantum mechanics, at least not in the same sense as in classical mechanics. It is, however, partially redundant because it is related to the gauge symmetry of the electromagnetic field. If we gauge the electromagnetic potentials differently, we can partly change the phase of the wave function (which determines its complex nature), but that is only up to $2\pi$ rotations.

As to 2), there are lots of relations in quantum mechanics which look suspiciously "classical", and there are also a lot of relations in quantum mechanics that are in striking contradiction to classical mechanics. Just because an arbitrary formula looks like a classical one if you are using the right letters of the alphabet doesn't mean that the quantities they represent are the same that you can measure in an experiment. After all, they are just letters. Replace $p$ by $R$ and it doesn't look so intriguing anymore. The reason why the authors have chosen to present the stuff that way was to wake up the reader and tell him: "hey man, there is pot of gold lying in that quantum mechanics, and if you dig this up, you are going to be rich".

Actually, there is no such thing as "classical momentum" in quantum mechanics, and particularly an imaginary momentum is meaningless in quantum mechanics. Maybe you can use that imaginary classical momentum to describe other things nicely (e.g. the tunnel effect if "classical momentum" becomes imaginary), but these descriptions are only useful in a limited context (otherwise we wouldn't need QM in the first place). Therefore, the formula you have given does not represent momentum. In QM, momentum of a given state of the system can only be defined as a set of probability amplitudes $\langle p|\psi\rangle$ (the Fourier components of the wave function, represented by the eigenstates $|p\rangle$ of the momentum operator) for the possible momentum measurements. If you measure the system, the probability of getting a certain momentum value is the probability amplitude squared. Ehrenfest's theorem connects that to classical mechanics by dealing with the expectation values, but that is only valid for a huge number of repeated measurements on the same system (or measuring an ensemble of equivalent systems). Momentum of a single measurement is undetermined (unless the system is in a momentum eigenstate, which you have ruled out by referring to the time-independent Schrödinger equation with a potential).

So, summarizing the above and short answer to your question: the imaginary unit has no predefined meaning in classical mechanics. It can be anything you want it to be and that it is able to mathematically represent (like rotations in phase space, rotations in physical space, etc.). Hence, it is not at all unphysical per-se, unless you have constructed the "unphysicalness" into it by using a bare mathematical formalism.


We often model oscillators with complex exponentials and simply ignoring the imaginary component

To say we model oscillators a certain way means we claim that's how they are. That's not what happens here at all. In classical mechanics, the oscillator's phase space coordinates are real-valued functions of time satisfying differential equations whose complex solutions are easily found, and whose real-valued functions are most easily found by knowing the complex ones and noting those equations are linear.

We often have complex wavenumbers and frequencies, which we interpret as damped motion.

Wavenumbers, frequencies etc. can be defined however we like. How do we like to define them? In a way that's useful. It's useful to define $e^{ikx}$ as having wavenumber $k$, even if $k\in\Bbb C\setminus\Bbb R$. But physically, the implications of $k\in\Bbb R,\,k\in i\Bbb R,\,k\in\Bbb C\setminus(\Bbb R\cup i\Bbb R)$ are crucially different.

How do we know that the imaginary momentum found here is truly meaningless?

Because when you measure momentum, it's real. That's nature's decision.

is there a rigorous way to tell when the complex nature of calculated observables indicates that said calculations predict a physically un-realizable state?

If by "rigorous" you mean "theory says so", that's open to the objection "why is that our theory?" The bottom line is always explaining what's observed, which implies compatibility with same.

if you have to use complex numbers, then you are wrong, as implied by Can one do the maths of physics without using $\sqrt{-1}$?

That's not what any answer there seems to say. Physics doesn't presuppose some mathematical objects are off-limits; it only cares about whether predictions are empirically right.


I do not think complex quantities are a red flag in physics. Mathematics is a set of tools and we need not assign ontological weight to them, the question should be do they make the physics simpler and/or clarify it.

Moreover, not all mathematics can be reduced to real numbers. For example, category theory or set theory does not make use of numbers in an essential way.

  • $\begingroup$ I think this is an evasive answer. I do not expect all math to be reduced to real numbers. I do not know if there is consensus yet, but ongoing physics research suggests that real-number formalisms of QM are experimentally found to be inaccurate (physics.stackexchange.com/q/691623). I tried to emphasize the fact that this question only considers classical cases. It becomes quite opinion-based and speculative if this constraint is not imposed. $\endgroup$
    – Stack
    Feb 1, 2022 at 7:31
  • $\begingroup$ I've edited the question to further emphasize that I am talking only about classical mechanics. Not all of physics, and certainly not all of mathematics. $\endgroup$
    – Stack
    Feb 1, 2022 at 7:32
  • $\begingroup$ @Stack: No. Complex numbers can be modelled by real numbers hence QM could be reformulated in real terms. People don't because the resulting formalism is messy. The paper referred to in the link is attempting a different real formalism not the one you would get simply by translating Schrodingers equation into real coupled PDEs. $\endgroup$ Feb 1, 2022 at 7:36

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