The wavefunction and relativity

When reading about quantum theory, and wave mechanics, you will often encounter the rather enigmatic statement that the Schrödinger equation is not relativistically correct. What does that mean?

In my previous post on the wavefunction and relativity, I boldly claimed that relativity theory had been around for quite a while when the young Comte Louis de Broglie wrote his short groundbreaking PhD thesis, back in 1924. Moreover, it is more than likely that he suggested the θ = ω∙t – kx = (E∙t – px)/ħ formula for the argument of the wavefunction exactly because relativity theory had already established the invariance of the four-vector product pμxμ = E∙t – px = pμ‘xμ‘ = E’∙t’ – p’x’. [Note that Planck’s constant, as a physical constant, should obviously not depend on the reference frame either. Hence, if the E∙t – px product is invariant, so is (E∙t – px)/ħ.] However, I didn’t prove that, and I didn’t relate it to Schrödinger’s equation. Hence, let’s explore the matter somewhat further here.

I don’t want to do the academic thing, of course – and that is to prove the invariance of the four-vector dot product. If you want such proof, let me just give you a link to some course material that does just that. Here, I will just summarize the conclusions of such course material:

  1. Four-vector dot products – like xμxμ = xμ2, pμpμ = pμ2, the spacetime interval s= (Δr)– Δt2, or our pμxμ product here – are invariant under a Lorentz transformation (aka as a Lorentz boost). To be formally correct, I should write xμxμ, pμpμ, and pμxμ, because the product multiplies a row vector with a column vector, which is what the sub- and superscript indicate.
  2. Four-vector dot products are referred to as Lorentz scalars.
  3. When derivatives are involved, we must use the so-called four-gradient, which is denoted by  or μ and defined as:

 = μ = (∂/∂t, –) = (∂/∂t, –∂/∂x, –∂/∂y, –∂/∂z)

Applying the four-gradient vector operator to the wavefunction, we get:

μψ= (∂ψ/∂t, –ψ) = (∂ψ/∂t, –∂ψ/∂x, –∂ψ/∂y, –∂ψ/∂z)

We wrote about that in the context of electromagnetic theory (see, for instance, my post on the relativistic transformation of fields), so I won’t dwell on it here. Note, however, that that’s the weak spot in Schrödinger’s equation: it’s good, but not good enough. However, in the context in which it’s being used – i.e. to calculate electron orbitals – the approximation works just fine, so you shouldn’t worry about it. The point to remember is that the wavefunction itself is relativistically correct. 🙂

Of course, it is always good to work through a simple example, so let’s do that here. Let me first remind you of that transformation we presented a couple of times already, and that’s how to calculate the argument of the wavefunction in the reference frame of the particle itself, i.e. the inertial frame. It goes like this: when measuring all variables in Planck units, the physical constants ħ and c are numerically equal to one, then we can then re-write the argument of the wavefunction as follows:

  1. ħ = 1 ⇒ θ = (E∙t – p∙x)/ħ = E∙t – p∙x = Ev∙t − (mvv)∙x
  2. E= E0/√(1−v2) and m= m0/√(1−v2)  ⇒ θ = [E0/√(1−v2)]∙t – [m0v/√(1−v2)]∙x
  3. c = 1 ⇒ m0 = E⇒ θ = [E0/√(1−v2)]∙t – [E0v/√(1−v2)]∙x = E0∙(t − v∙x)/√(1−v2)

⇔ θ = E0∙t’ = E’·t’ with t’ = (t − v∙x)/√(1−v2)

The t’ in the θ = E0∙t’ expression is, obviously, the proper time as measured in the inertial reference frame. Needless to say, is the relative velocity, which is usually denoted by β. Note that this derivation uses the numerical m0 = E0 identity, which emerges when using natural time and distance units (c = 1). However, while mass and energy are equivalent, they are different physical concepts and, hence, they still have different physical dimensions. It is interesting to spell out what happens with the dimensions here:

  • The dimension of Evt and/or E0∙t’ is (N∙m)∙s, i.e. the dimension of (physical) action.
  • The dimension of the (mvv)∙x term must be the same, but how is that possible? Despite us using natural units – so the value of is now some number between 0 and 1 – velocity is what it is: velocity. Hence, its dimension is m/s. Hence, the dimension of the mvv∙x term is kg∙m = (N∙s2/m)∙(m/s)∙m = N∙m∙s.
  • Hence, the dimension of the [E0v/√(1−v2)]∙x term only makes sense if we remember the m2/s2 dimension of the c2 factor in the E = m∙c2 equivalence relation. We write: [E0v∙x] = [E0]∙[v]∙[x] = [(N∙m)∙(s2/m2)]∙(m/s)∙m = N∙m∙s. In short, when doing the mv = Ev and/or m0 = E0 substitution, we should not get rid of the physical 1/c2 dimension.

That should be clear enough. Let’s now do the example. The rest energy of an electron, expressed in Planck units, EeP = Ee/EP = (0.511×10eV)/(1.22×1028 eV) = 4.181×10−23. That is a very tiny fraction. However, the numerical value of the Planck time unit is even smaller: about 5.4×10−44 seconds. Hence, as a frequency is expressed as the number of cycles (or, as an angular frequency, as the number of radians) per time unit, the natural frequency of the wavefunction of the electron is 4.181×10−23 rad per Planck time unit, so that’s a frequency in the order of [4.181×10−23/(2π)]/(5.4×10−44 s) ≈ 1×1020 cycles per second (or hertz). The relevant calculations are given hereunder.

Rest energy (in joule) 8.1871E-14
Planck energy (in joule) 1.9562E+09
Rest energy in Planck units 4.1853E-23
Frequency in cycles per second 1.2356E+20

Because of these rather incredible numbers (like 10–31 or 1020), the calculations are not always very obvious, but the logic is clear enough: a higher rest mass increases the (angular) frequency of the real and imaginary part of the wavefunction, and gives them a much higher density in spacetime. How does a frequency like 1.235×1020 Hz compare to, say, the frequency of gamma rays. The answer may surprise you: they are of the same order, as is their energy! 🙂 However, their nature, as a wave ,is obviously very different: gamma rays are an electromagnetic wave, so they involve an E and B vector, rather than the two components of the matter-wave. As an energy propagation mechanism, they are structurally similar, though, as I showed in my previous post.

Now, the typical speed of an electron is given by of the fine-structure constant (α), which is (also) equal to the  is the (relative) speed of an electron (for the many interpretations of the fine-structure constant, see my post on it). So we write:

α = β = v/c

More importantly, we can use this formula to calculate it, which is done hereunder. As you can see, while the typical electron speed is quite impressive (about 2,188 km per second), it is only a fraction of the speed of light and, therefore, the Lorentz factor is still equal to one for all practical purposes. Therefore, its speed adds hardly anything to its energy.


Fine-structure constant 0.007297353
Typical speed of the electron (m/s) 2.1877E+06
Typical speed of the electron (km/s) 2,188 km/s
Lorentz factor (γ) 1.0000266267

But I admit it does have momentum now and, hence, the p∙x term in the θ = E∙t – p∙x comes into play. What is its momentum? That’s calculated below. Remember we calculate all in Planck units here!

Electron energy moving at alpha (in Planck units) 4.1854E-23
Electron mass moving at alpha (in Planck units) 4.1854E-23
Planck momentum (p = m·v = m·α ) 3.0542E-25

The momentum is tiny, but it’s real. Also note the increase in its energy. Now, when substituting x for x = v·t, we get the following formula for the argument of our wavefunction:

θ = E·t – p·x = E·t − p·v·t = mv·t − mv·v·v·t = mv·(1 − v2)·t

Now, how does that compare to our θ = θ = E0∙t’ = E’·t’ expression? Well… The value of the two coefficients is calculated below. You can, effectively, see it hardly matters.

mv·(1 − v2) 4.1852E-23
Rest energy in Planck units 4.1853E-23

With that, we are finally ready to use the non-relativistic Schrödinger equation in a non-relativistic way, i.e. we can start calculating electron orbitals with it now, which is what we did in one of my previous posts, but I will re-visit that post soon – and provide some extra commentary! 🙂


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