Admitting the ansatz

$$ψ=e^{i(kx-ω t)} \tag{1}$$ then $$k^2=-ψ^{-1} \frac {∂^2ψ}{∂x^2} \tag{2}$$ and

$$ω=iψ^{-1} \frac {∂ψ}{∂t} \tag{3}$$

If one admits that the total energy ($E$) is related to momentum ($p$) as $E=\frac{p^2}{2m}+U$, admiting also the De Broglie relations $E=ħω$; $p=ħk$ it follows that

$$\frac {-ħ^2}{2m} \frac {∂^2ψ}{∂x^2}+Uψ= iħ \frac {∂ψ}{∂t} \tag{4}$$

This is Schrödinger's equation. This equation is said to be non relativistic because of its use of $E= \frac{p^2}{2m}+U$ (rigorously speaking though, it is non relativistic because it is not Lorentz invariant).

However, starting from the relativistic total energy equation

$$E=\frac{1}{ \sqrt{1- \frac{v^2}{c^2}}}mc^2=T+mc^2 \tag{5}$$

Where, $T$ is the kinetic energy and $mc^2$ the particle’s self energy. Now, using the expansion of $\frac{1}{ \sqrt{1- \frac{v^2}{c^2}}}mc^2$

$$E=mc^2 + \frac{mv^2}{2} + \frac{3mv^4}{8c^2} + \frac{5mv^6}{16c^4}+...$$

and ignoring members dividing by $c$ (because we are considering $v\ll c$). It becomes

$$E=\frac{1}{2} mv^2+mc^2=T+mc^2 = \frac{p^2}{2m}+mc^2 \tag{6}$$


$$\frac{p^2}{2m}+mc^2=ħω=\frac{ħ^2k^2}{2m}+mc^2 \tag{7}$$

So, $mc^2$ do not vanishes even under classical approximation.

Admitting that Planck's and De Broglie's equations holds in every situation and that $E$ in Planck equation is the total energy, substituting equation (2) and (3) into (7) the Schrödinger equation “would” have the form

$$\frac {-ħ^2}{2m} \frac {∂^2ψ}{∂x^2}+mc^2 ψ = iħ \frac {∂ψ}{∂t} \tag{9}$$

Now we could postulate this equation, making the steps of getting it less fundamental then the end result.

I tried to consider that $T \ll mc^2$ in Schrödinger equation, but I realize that an electron in hydrogen atom moving with half the speed of light (using classical equations as we are analyzing Schrödinger’s equation) it would have less than $\rm 100keV$ ($\rm ≈64keV$ if my math is not wrong) of kinetic energy, but $\rm 511keV$ of self energy.

So, my question is: why Schrödinger equation do not have an $mc^2$ term, if $ħω$ is supposed to be the total energy and not just the kinetic energy.

  • $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$ – Qmechanic Nov 16 '16 at 18:04
  • 4
    $\begingroup$ Note that in classical mechanics you can also add any constant to potential without affecting equations of motion. Thus you can also go from relativistic Hamiltonian to non-relativistic one and drop $mc^2$ in the process. $\endgroup$ – Ruslan Nov 17 '16 at 13:57
  • 2
    $\begingroup$ You have two perfectly good answers already. In what sense had this question "not received enough attention"? $\endgroup$ – Emilio Pisanty Nov 6 '17 at 8:26
  • 2
    $\begingroup$ Simple: if you change $V(r,t)\to V(r,t)+W$, you'll get an extra $\exp(iWt/\hbar)$ factor in the eigenstates, but it isn't physically relevant, since all differences in eigen-energies will be the same. Don't forget that global phase is not an observable. $\endgroup$ – Ruslan Nov 6 '17 at 9:33
  • 2
    $\begingroup$ @J.Manuel If the existing answers are unsatisfactory, I would advise you to actually engage with the answerers and ask for clarification instead of just chucking rep at the problem and hoping that other people will magically know what it is about the existing answers that you found confusing. There is nothing really to be said that's not already in Valter and knzhou's answers - and that includes the two new answers. $\endgroup$ – Emilio Pisanty Nov 6 '17 at 11:24

Just because to introduce the constant added term $mc^2I$ to the Hamiltonian operator would be equivalent to redefine $\psi \to \psi'= e^{imc^2 t/\hbar}\psi$. This sort of phases do not matter in QM. You cannot see them by measuring any observable. Pure states are actually operators of the form $|\psi \rangle \langle \psi|$ and you see that these phases cancel each other.

Instead, if the mass were replaced by a mass operator with discrete spectrum the picture would change. In the classical limit the rapid temporal oscillations of the phases (I am assuming that the mass is big if compared with the typical energies of the system), would destroy the coherence of superpositions of different masses giving dynamically rise to superselection rule of the mass Bargmann's superselection rule (see here or here).

| cite | improve this answer | |
  • $\begingroup$ It's a nice trick to explain cancellation of phase by switching to density operator formalism. But wouldn't it also cancel any changes in sign under particle exchange operator, thus rendering fermions and bosons sort of equivalent? $\endgroup$ – Ruslan Nov 17 '16 at 13:55
  • 1
    $\begingroup$ Sorry, I removed my previous comment because I though you was referring to my last remark. No, regarding the cancellation of phases no problem arises. The density operator formalism is completely enough to deal with QM. Every operation can be performed using that formalism. Pure states are density matrices of the form $P=|\psi\rangle \langle \psi|$. The only thing you can compute are expectation values and it holds $\langle\psi |A \psi\rangle = tr(PA)$. $\endgroup$ – Valter Moretti Nov 17 '16 at 16:22
  • $\begingroup$ You can however distinguish between density matrices of the form $P=|\psi\rangle \langle \psi|$ with symmetric or antisymmetric identical-particles vector $|\psi\rangle$ under the action of permutation group... Though it does not arise by a sign in front of $P$ when acting with the unitary representation of permutation group this way $UPU^*$ because signs cancel. $\endgroup$ – Valter Moretti Nov 17 '16 at 16:29
  • $\begingroup$ @ValterMoretti. If you see equation (9), it equals to the Schrödinger equation for a particle at constant potential, even though (in present case) it represents the free particle. In Schrödinger’s equation free particles have oscillatory solutions as long as their energy is positive. However, Equation (9) has oscillatory solutions only if $E>mc^2$, this could have a measureable effect on things like the tunnel effect, by increasing the height of the barrier, or on interference patterns of low energy particles, couldn’t it? $\endgroup$ – J. Manuel Nov 6 '17 at 8:41
  • $\begingroup$ @J.Manuel indeed the spectrum of $-\hbar^2(2m)^{-1}d^2/dx^2 + mc^2$ is $[mc^2, +\infty)$ and there are no elements under $mc^2$. The eigenfunctions are oscillatory as is due... $\endgroup$ – Valter Moretti Nov 6 '17 at 10:28

In nonrelativistic quantum mechanics, particles cannot be created or destroyed, and each particle has constant mass $m$. That means the extra $E = mc^2$ energy is just a constant, so it can be subtracted out by adding a constant to the Hamiltonian; only energy differences matter.

The $E = mc^2$ can play a role in quantum field theory, since particles can be created or destroyed there; for example, it is released in pair annihilation, giving the products extra energy.

| cite | improve this answer | |
  • 2
    $\begingroup$ Where do QM and especially Schroedinger equation enter your answer? $\endgroup$ – Valter Moretti Nov 16 '16 at 18:43

Let's say we start with this equation of yours :

$$-\frac {\hbar^2}{2m} \frac {\partial^2\psi}{\partial x^2}+V\psi +mc^2\psi=i\hbar \frac {\partial \psi}{\partial t}$$

Now a simple transform reduces it to the original form :

$$\psi = e^{Wt}\phi$$

$$-\frac {\hbar^2}{2m} e^{Wt} \frac {\partial^2\phi}{\partial x^2}+Ve^{Wt}\phi+mc^2e^{Wt}\phi=i\hbar e^{Wt} \frac {\partial \phi}{\partial t} + i\hbar W e^{Wt}\phi$$

Which reduces to :

$$-\frac {\hbar^2}{2m} \frac {\partial^2\phi}{\partial x^2}+V\phi+mc^2\phi=i\hbar \frac {\partial \phi}{\partial t} + i\hbar W \phi $$

And it is easy to see that $i\hbar W := mc^2$ recovers the original form of the equation.

So as we did in adding the rest mass term was add a rather pointless phase term that does nothing for us :

$$\psi = \phi \,\, exp\left( i\frac{mc^2}{\hbar} t\right)$$

This is the phase change that has no net effect that is discussed in the answers by Valter Moretti and Ruslan

| cite | improve this answer | |
  • $\begingroup$ The genius remark was not intended as an insult. I'll reword that. $\endgroup$ – StephenG Nov 6 '17 at 11:12

Consider a Schrödinger equation:

$$-\frac {\hbar^2}{2m} \frac {\partial^2\psi}{\partial x^2}+V(x,t)\psi=i\hbar \frac {\partial \psi}{\partial t}.$$

Let some wave function $\psi(x,t)$ be its solution. Let's now replace $V(x,t)\to V(x,t)+\hbar W$, where $W=\mathrm{const}$. Corresponding solution of the new equation will change: $\psi(x,t)\to\psi(x,t)\exp(-iWt).$

Consider now an observable $K$, with corresponding operator $\hat K$. Its expected value, calculated for the solution of the original equation, will be

$$\overline K(t)=\int\limits_{-\infty}^{\infty}dx\,\psi(x,t)^*\hat K\psi(x,t).$$

Now let's replace $\psi$ in the above integral with the solution of the modified equation where we shifted the potential energy:

$$\int\limits_{-\infty}^{\infty}dx\,(\psi(x,t)\exp(-iWt))^*\hat K(t)(\psi(x,t)\exp(-iWt))=\\ =\int\limits_{-\infty}^{\infty}dx\,\psi(x,t)^*\exp(iWt)\exp(-iWt)\hat K(t)\psi(x,t)=\\ =\int\limits_{-\infty}^{\infty}dx\,\psi(x,t)^*\hat K(t)\psi(x,t)=\overline K(t).$$

You can see that regardless of the global phase, the observable $K$ appears the same. Similarly can check that matrix elements of an operator will also be independent of the global phase of the basis functions you calculate the matrix elements in.

Any physically-relevant calculation ultimately is about observables, not about particular values of abstract functions like a wavefunction. Thus you shouldn't worry too much about gaining an extra phase factor when adding or removing a constant term to the potential.

| cite | improve this answer | |

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.