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 – **k**∙**x** = (E∙t – **p**∙**x**)/ħ formula for the argument of the wavefunction exactly *because *relativity theory had already established the invariance of the four-vector product p_{μ}x_{μ} = E∙t – **p**∙**x** = p_{μ}‘x_{μ}‘ = 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 – **p**∙**x** product is invariant, so is (E∙t – **p**∙**x**)/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:

- Four-vector dot products – like x
_{μ}x_{μ}= x_{μ}^{2}, p_{μ}p_{μ}= p_{μ}^{2}, the spacetime interval s^{2 }= (**Δr**)^{2 }– Δt^{2}, or our p_{μ}x_{μ}product here – are invariant under a Lorentz transformation (aka as a Lorentz*boost*). To be formally correct, I should write x_{μ}x^{μ}, p_{μ}p^{μ}, and p_{μ}x^{μ}, because the product multiplies a*row*vector with a*column*vector, which is what the sub- and superscript indicate. - Four-vector dot products are referred to as
*Lorentz scalars*. - 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:

**∇**_{μ}ψ= (∂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:

- ħ = 1 ⇒ θ = (E∙t – p∙x)/ħ = E∙t – p∙x = E
∙t − (m_{v}∙_{v}*v*)∙x - E
= E_{v }_{0}/√(1−*v*^{2}) and m= m_{v }_{0}/√(1−*v*^{2}) ⇒ θ = [E_{0}/√(1−*v*^{2})]∙t – [m_{0}∙*v*/√(1−*v*^{2})]∙x *c*= 1 ⇒ m_{0}= E_{0 }⇒ θ = [E_{0}/√(1−*v*^{2})]∙t – [E_{0}∙*v*/√(1−*v*^{2})]∙x = E_{0}∙(t −*v*∙x)/√(1−*v*^{2})

⇔ θ = E_{0}∙t’ = E’·t’ with t’ = (t − *v*∙x)/√(1−*v*^{2})

The t’ in the θ = E_{0}∙t’ expression is, obviously, the *proper time* as measured in the inertial reference frame. Needless to say, *v *is the relative velocity, which is usually denoted by β. Note that this derivation uses the *numerical* m_{0} = E_{0} 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 E
t and/or E_{v}_{0}∙t’ is (N∙m)∙s, i.e. the dimension of (physical)*action.* - The dimension of the (m
∙_{v}*v*)∙x term must be the same, but how is that possible? Despite us using natural units – so the*value*of*v*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 m∙_{v}*v*∙x term is kg∙m = (N∙s/m)∙(m/^{2}~~s~~)∙~~m~~= N∙m∙s. - Hence, the dimension of the [E
_{0}∙*v*/√(1−*v*^{2})]∙x term only makes sense if we remember the m^{2}/s^{2}dimension of the*c*^{2}factor in the E = m∙*c*^{2}equivalence relation. We write: [E_{0}∙*v*∙x] = [E_{0}]∙[*v*]∙[x] = [(N∙m)∙(s^{2}/m^{2})]∙(m/s)∙m = N∙m∙s. In short, when doing the m= E_{v}_{v}and/or m_{0}= E_{0}substitution, we should not get rid of the physical 1/*c*^{2}dimension.

That should be clear enough. Let’s now do the example. The rest energy of an electron, expressed in Planck units, E_{eP} = E_{e}/E_{P} = (0.511×10^{6 }eV)/(1.22×10^{28 }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) ≈ 1×10^{20} cycles per second (or *hertz*). The relevant calculations are given hereunder.

Electron | |

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 10^{20}), 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 1.235×10^{20} 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 = m* _{v}*·t − m

*·*

_{v}*v*·

*v*·t = m

*·(1 −*

_{v}*v*

^{2})·t

Now, how does that compare to our θ = θ = E_{0}∙t’ = E’·t’ expression? Well… The value of the two coefficients is calculated below. You can, effectively, see it hardly matters.

m·(1 − _{v}v^{2}) |
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! 🙂