• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  The non relativistic limit of the real Klein-Gordon equation

+ 2 like - 0 dislike

The non relativistic limit of the Klein-Gordon equation (K-G) is the Schrodinger equation (S). A complex field remains a complex field as expected for non-relativistic problems (where we are all familiar with wave functions and their interpretation). A non charged particle, though, is described in Relativistic Quantum Mechanics by a real solution of the K-G equation. Apparently the non-relativistic limit for such a particle would give us a real valued wave-function for any state. How is this possible and what is the relation between charge and the presence of complex numbers in quantum mechanics? What is the role of the factor $e^{imt}$ used in the derivation of  S equation from K-G in this case?

asked Jan 8, 2017 in Theoretical Physics by Lucas [ revision history ]
edited Jan 8, 2017

1 Answer

+ 2 like - 0 dislike

1. Relativistic QM for spinless particles is not a physically consistent theory. Let the scalar wave function \(\phi\) satisfy the KG eq. \(\partial_a \partial^a\phi+m^2\phi=0\). One proves directly from the KG eq. that the currents \(j_a\propto \phi^*\partial_a\phi-\phi\partial_a\phi^*\) are conserved, i.e., its covariant divergence vanishes: \(\partial^aj_a=0\). In a relativistic version of single-particle QM (read: first quantized theory), one would expect the time-like component of the KG current, \(j_0\propto\phi^*\partial_t\phi-\propto\phi\partial_t\phi^*\), to be interpreted as a probability current density. Differently from the non-relativistic QM, where the wave-function \(\psi\) obeying a Schrodinger eq. \(i\partial_t\psi=H\psi\) have as probability density the positive quantity \(\rho\propto\phi^*\phi\), the KG current \(j_0\) allows both positive as well as negative values, spoiling its probabilistic interpretation. This was one of the reasons that lead Dirac to find a new relativistic wave equation, which indeed cures the non-positiveness of the time-like component of the current.

Remark. Even if one insists with carrying on a relativistic QM theory for a spinless particle described by KG,  ignoring therefore the problem mentioned above regarding its probabilistic interpretation as well the one which I will mention below (about the negative-energy solutions), the expression for \(j_0\) given above shows that it vanishes for real-valued \(\phi\). The situation now is then even worse, that now its not only that negative values are allowed for \(j_0\), but what should be regarded as a "probability" current vanishes identically in the whole space-time, and then there is no particle left at all.

2. The KG eq. have as plane wave solutions both positive as well as negative frequency solutions, namely, \(\phi_\boldsymbol{k}^{(+)}(t,\boldsymbol{r})=e^{i(\omega_k t - k\cdot r)}/N\) and \(\phi_\boldsymbol{k}^{(-)}(t,\boldsymbol{r})=e^{-i(\omega_k t + k\cdot r)}/N\) (\(N\) is the normalization constant), where \(\omega_k=+\sqrt{\boldsymbol{k}^2+m^2}\) for all possible momenta \(\boldsymbol{k} \in \mathbb{R}^3\). In a single particle theory, both positive and negative frequency yields both positive as well as negative energy states, since relativity only imposes that \(E_k^2=\omega_k^2=\boldsymbol{k}^2+m^2\), giving a physically unstable vacuum for the theory (the particle could radiate its energy away down to \(-\infty\)). This situation reappears in the solutions to Dirac equation, and the latter tried to solve this issue by introducing the hypothesis that all negative-energy states are already filled, so that, his particle having spin one-half and therefore obeying Pauli exclusion principle, no transition to negative-energy states are allowed. This theory, historically known as the Dirac hole theory, is a manifestation that relativistic QM is only consistent with many-particle states.

3. This still left open the problem of defining a Lorentz invariant quantum theory for spinless particles. The hole theory interpretation is not possible since these particles do not obey the exclusion principle. Pauli (in what he dubbed his "anti-Dirac" paper) was one of the first to indicate that the right solution is to introduce quantum fields (which has been known from the 1930s as second quantization, since one treats the wave-function for a single-particle state as a classical field and then quantize again, but now the field; this is outdated terminology, however.) As opposed to the single-particle theory discussed above, we begin by interpreting \(\phi\) as a classical field. Its currents \((j_a)\) can be seen as the Nother charges derived from the KG Lagrangian \(\mathcal{L}=(1/2)\partial_a\phi\partial^a\phi+(1/2)m^2\phi^2\), so that real-valued \(\phi\) for which \(j_a=0\) should describe a classical non-charged field, while complex-valued \(\phi\) are to be identified with classical fields carrying charges. Quantum theory is obtained either by introducing the quantum commutation relations on the field variables, for example, the equal-time commutation relations on (the now field-operators) \(\phi\) and \(\pi=\partial\mathcal {L}/\partial{(\partial_t\phi)}\), or going to the path-integral formalism by working out the partition function \(Z_0[J]=\int\mathcal{D}\phi e^{i\int \mathcal{L}+J\phi}\), whose functional derivatives generates the theory Green's functions. In any case, one then obtains a consistent, Lorentz invariant, theory for spinless particles, as a quantum theory of the field \(\phi\). (Note. For Fermi-Dirac fields, one replaces commutation relations by anticommutation ones, or in the path-integral approach, one introduce Grassman variables).

Remark. The Fourier decomposition of the field operator \(\phi\) in terms of the (positive and negative frequency) plane-wave solutions \(\phi_\boldsymbol{k}^{(+)}\) and \(\phi_\boldsymbol{k}^{(-)}\), namely, \(\phi \propto\int (d^3\boldsymbol{k}/\sqrt{\omega_\boldsymbol{k}}) \big( \phi_\boldsymbol{k}^{(+)} a_\boldsymbol{k}+\phi_\boldsymbol{k}^{(-)} a_\boldsymbol{k}^{+}\big)\), shows at once that: (i) only positive-energy solutions exists, but (ii) the theory must be contain many-particle states, where the coefficients \(a_k\) and \(a_k^+\) acts as annihilation and creation operators.

4. Finally, two further remarks. (i) About the the role of the term \(e^{imt}\) in the derivation of the Sch. eq. from the KG, the author of the question should give more details about what reference he is using to study this. Nevertheless, I guess it must came from the expression of the time-evolution operator using the relativistic Hamiltonian, \(U(t)=e^{i\sqrt{p^2+m^2}t}\). In this case, the non-relativistic limit is \(\propto e^{imt+i\boldsymbol{p}^2t/2m}\), and so the first term can be ignored since it only gives an overall phase to the quantum state. (ii) As a further inquire, suppose now you begin with the Dirac equation, and then decide to take its non-relativistic limit. How one obtains the spinless Schrodinger eq. from it? (Tip: the state space of nonrelativistic QM will be recovered from the nonrelativistic limit of the Dirac eq. as an eigenspace of the spin operator, namely, the nonrelativistic wave-functions are going to arise as spin eigenfunctions.)

5. I think that the best way to relate the KG current for a complex scalar field to electric charge is by discussing the gauge-invariance of the theory. Begin with the KG Lagrangian for a complex field:


Under the transformation \((\phi,\phi^*)\mapsto(\phi e^{-i\Lambda},\phi^* e^{i\Lambda})\) for a constant \(\Lambda\), the complex KG Lagrangian is invariant. Using Noether theorem, you can show that under \(\delta \phi=-i\Lambda\phi\), \(\delta \phi^*=i\Lambda\phi^*\), the Noether current

\[\mathcal{J_a}=\frac{\delta \mathcal{L}}{\delta (\partial^a \phi)}\delta\phi+\frac{\delta \mathcal{L}}{\delta (\partial^a \phi^*)}\delta\phi^* =-i(\phi \partial_a \phi^*-\phi^* \partial_a \phi) \]

is proportional to the KG current which I introduced above (namely, \(\propto j_a\)). One might worry that a proportionality constant, in particular the electric charge \(e\), is missing. This happens because we have been considering a global gauge transformation, where the "generator" of the gauge \(\Lambda\) was assumed to be constant. If we allow it to vary in space-time, so that we begin to consider local transformations, the infinitesimal variations becomes \(\delta \phi=-i\Lambda(x_\mu)\phi\), \(\delta \phi^*=i\Lambda(x_\mu)\phi^*\). But now, the field derivatives transforms as

\[\delta (\partial_a \phi)=-i\Lambda \partial_a\phi-i\phi\partial_a\Lambda, \\ \delta (\partial_a \phi^*)=i\Lambda \partial_a^*\phi+i\phi^*\partial_a\Lambda.\]

KG Lagrangian is not invariant under these variations. One is then lead to introduce the following term to the Lagrangian density,


which couples with coupling strength \(e\) the KG current \(\mathcal{J}_a\) to a new vector field \(A^a\), which is supposed to transform as covariant gradient \(e^{-1}\partial_a\Lambda\). This is nothing but the gauge transformation that we have meet in classical EM. In order to give dynamical content to the field \(A^a\) (in the sense of making it contribute itself to the Lagrangian), we add the Maxwell term to the total Lagrangian, \(\mathcal{L}''=-(1/4)F_{ab}F^{ab}\) for the covariant curl \(F^{ab}=\partial^{[a}A^{b]}\), and we end up with the theory of electrodynamics coupled to a charged complex scalar field.

From this perspective then, we obtain the EM field (and the charge \(e\) coupling the scalar field to the EM one) as a result of requiring gauge invariance. I believe this point of view is more modern (certainly more sophisticated) than the traditional approach: relativistic QM through wave equations \(\mapsto\) second quantization.

answered Jan 8, 2017 by Igor Mol (550 points) [ revision history ]
edited Jan 8, 2017 by Igor Mol

Thanks for your detailed answer! 

The reference I am using for the factor $e^{-imt}$ is Greiner, Relativistic Quantum Mechanics. See page 7 here:


The author simply derives Sch. Equation from K-G in that page and he claims that it is the valid equation for an spin 0 particle in the non-relativistic limit. 

Also, since the Dirac equation is the 'square root of K-G' , solutions of Dirac (each component of the spinor) satisfy K-G as well. Just apply $(i\gamma^{\mu}\partial_{\mu}+m)$ to both sides of the Dirac equation.

In a sense, it is as if we have a proper non-relativistic limit for charged particles derived from complex K-G solutions and also a non-relativistic limit for real K-G but at the price of defining a complex field (via the extraction of the factor $e^{-imt}$) where only a real one was needed.

This seems to imply that real solution of the K-G equation contain information about the particle location since a proper conserved probability current can be defined for Sch. equation.

As a follow up and very related question let me ask you: K-G is a second order differential equation. A well defined initial value problem require the knowledge of $\phi (\mathbf{x},t_0)$ and $\dot{\phi}(\mathbf{x},t_0)$ where, $t_0$, is an arbitrary initial time. Equivalently we could supply $\phi (\mathbf{x},t_0)$ and $\phi (\mathbf{x},t_1)$ and apply a least action principle to find the field configuration between $t_0$ and any other fixed time $t_1$. Since $\phi$ contains the information concerning the probability of finding this uncharged particle would this mean that we can solve for the evolution of this probability in time just by knowing how likely is it to find the particle at times $t_0$ and $t_1$ at all spatial points?

1. Just to complement my answer, I have just read that part of Greiner's derivation of Sch. eq. from KG. He introduces the ansatz \(\psi(t,\boldsymbol{r})=e^{-imt}\phi(t,\boldsymbol{r})\) into KG eq. and shows that \(\phi\) satisfy Sch. eq. in first order. To see how this relates to what I wrote above, I have worked from the propagator point of view. If \(t>t_0\) and \(G(\boldsymbol{r}_a,\boldsymbol{x}_b;t)=\langle\boldsymbol{x}_b|e^{i\sqrt{p^2+m^2}t}|\boldsymbol{x}_a\rangle\), then the wave-function at a later time evolves according to \(\psi(t,\boldsymbol{x}_b)=\int d^3\boldsymbol{x}_a G(\boldsymbol{x}_a,\boldsymbol{x}_b;t)\psi(\boldsymbol{x}_a,t_0)\). Taking the propagator's n-r limit,

\[G(\boldsymbol{x}_a,\boldsymbol{x}_b;t)\approx e^{imt} \langle\boldsymbol{x}_b|e^{i\boldsymbol{p}^2t/2m}|\boldsymbol{x}_a\rangle=e^{imt}G_0(\boldsymbol{x}_a,\boldsymbol{x}_b;t)\]

where \(G_0\) is the propagator for the n-r Sch. eq. Then, to first-order, the wave-function evolves according to

\[\psi(t,\boldsymbol{x}_b)\approx e^{imt}\int d^3\boldsymbol{x}_a G_0(\boldsymbol{x}_a,\boldsymbol{x}_b;t)\psi(\boldsymbol{x}_a,t_0),\]

which implies at once that it obeys the Sch. eq. in first order (since \(G_0\) is the kernel to the above integral equation).

2. Exactly, taking the squared adjoint of Dirac eq. we see that each component of the Dirac spinor satisfy separately the KG eq. So when the n-r limit is taken, it happens that you end up with a projection of the full state space into eigenspaces of the spin operator.

(I see you are studying Greiner's book. I have seen other of Greiner's book on field quantization and symmetries in QM. They seem quite nice because of the great number of worked examples. However, I think that the best approach to QFT today is by using symmetries, in particular, by working out the representation of the Lorentz groups to deduce particle's properties. I would recommend you to see Maggiore's Modern Quantum Field Theory, which I am enjoying a lot.)

As to your latter question, give me a little time to think and write better.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights