# Klein Gordon equation

• A
Summary:
Second order time derivative
What problem actually arises when we take the second order time derivative in KG equation

vanhees71
Gold Member
We'd need a bit more detail, to which problems you are referring to. Sometimes it's claimed that the "problem" with "negative energy solutions" of the time-independent Klein-Gordon equations arises because of the 2nd-order time derivative, and then old-fashioned books make an argument that for this reason one wants the 1st-order Dirac equation. The modern answer to all these problems is, of course, quantum field theory and the reinterpretation of negative-frequency modes as antiparticle states with positive energy (just writing a creation instead of an annihilation operator in the mode-decomposition of the field operator, aka Feynman-Stückelberg trick).

We'd need a bit more detail, to which problems you are referring to. Sometimes it's claimed that the "problem" with "negative energy solutions" of the time-independent Klein-Gordon equations arises because of the 2nd-order time derivative, and then old-fashioned books make an argument that for this reason one wants the 1st-order Dirac equation. The modern answer to all these problems is, of course, quantum field theory and the reinterpretation of negative-frequency modes as antiparticle states with positive energy (just writing a creation instead of an annihilation operator in the mode-decomposition of the field operator, aka Feynman-Stückelberg trick).
The textbook simply mentions that the probability density is not positive definate because of the second order time derivative.so how actually the second derivative creates a problem for probabilistic interpretation

Gaussian97
Homework Helper
The problem is to try to use KG equation as an immediate relativistic substitute for the Schrödinger equation. In classical QM you know that the wave function that describes a particle must be a solution of the equation
$$\left(i\hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi=0$$
and then, the probability of the particle to be in some region is given by
$$\int |\psi|^2 dx$$
If you want to describe a relativistic particle just by using KG equation, i.e. that now the wave function must be a solution of
$$\left(\hbar^2\frac{\partial^2}{\partial t^2} - c^2\hbar^2\frac{\partial^2}{\partial x^2} + m^2 c^4\right)\psi=0$$
then you have problems because the probabilities become negative.

So constructing a relativistic QM is not as easy as that, you need to change some fundaments. And that's why one needs Quantum Field Theory.

• PeroK and vanhees71
vanhees71
Gold Member
The textbook simply mentions that the probability density is not positive definate because of the second order time derivative.so how actually the second derivative creates a problem for probabilistic interpretation
Which textbook?

The problem is that there are many textbooks around, staring with the old-fashioned idea that you could formulate relativistic quantum theory in a similar way in terms of a wave function as Schrödinger did for the non-relativistic case. The problem with that is that it simply doesn't work out, and that in fact one needs quantum field theory to formulate relativistic quantum theory. That is, because at relativistic energies in scattering processes you can always destroy and create particles, and that's why one most conveniently uses quantum-field theory which is the most simple way to describe such annihilation and creation processes. So to get positive probabilities you need relativistic QFT, and then you can describe also "Klein-Gordon particles" (e.g, scalar or pseudoscalar particles like the pions) without problems with "negative probabilities".

martinbn
The problem is to try to use KG equation as an immediate relativistic substitute for the Schrödinger equation. In classical QM you know that the wave function that describes a particle must be a solution of the equation
$$\left(i\hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi=0$$
and then, the probability of the particle to be in some region is given by
$$\int |\psi|^2 dx$$
If you want to describe a relativistic particle just by using KG equation, i.e. that now the wave function must be a solution of
$$\left(\hbar^2\frac{\partial^2}{\partial t^2} - c^2\hbar^2\frac{\partial^2}{\partial x^2} + m^2 c^4\right)\psi=0$$
then you have problems because the probabilities become negative.

So constructing a relativistic QM is not as easy as that, you need to change some fundaments. And that's why one needs Quantum Field Theory.
How are the probabilities computted? If it is ##\int |\psi|^2 dx## again, then it is always non-negative.

stevendaryl
Staff Emeritus
How are the probabilities computted? If it is ##\int |\psi|^2 dx## again, then it is always non-negative.

I don't know the proof that that interpretation doesn't work, but there is a consistency check for the non-relativistic Schrodinger equation (and the Dirac equation) that fails for the Klein Gordon equation.

In the case of the Schrodinger equation, we can derive it from a Lagrangian density

##\mathcal{L} = i \hbar \psi^* \dfrac{d\psi}{dt} - \frac{\hbar^2}{2m} (\nabla \psi^* \cdot \nabla \psi)##

The Lagrangian equations of motion give rise to Schrodinger's equation. For this Lagrangian, we can also see that it is invariant under a change of phase:
##\psi \rightarrow e^{i \phi} \psi##
##\psi^* \rightarrow e^{-i\phi} \psi^*##.
Noether's theorem about such symmetries implies the existence of a conserved current:

##\dfrac{d\rho}{dt} + \nabla \cdot j = 0##

where ##\rho = \psi^* \psi## and ##j = \frac{i \hbar}{2m} (\psi^* \nabla \psi - \psi \nabla \psi^*)##

So ##\rho## can consistently be interpreted as a probability density (If it is normalized to integrate to 1) and ##j## can be consistently interpreted as a probability current.

If you do the same thing with the Klein-Gordon equation, you get a conserved current again, but a different one:

##\mathcal{L} = \dfrac{d\psi^*}{dt} \dfrac{d\psi*}{dt} - \nabla \psi^* \cdot \nabla \psi##

The corresponding density and current is given by:

##\rho = \psi^* \dfrac{d\psi}{dt} - \dfrac{d\psi^*}{dt} \psi##
##j = \psi^* \nabla \psi - (\nabla \psi^*)\psi##

So ##\psi^* \psi## does not play a role in the conserved current of the Klein Gordon theory.

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• PeroK and martinbn
Demystifier
Gold Member
How are the probabilities computted? If it is ##\int |\psi|^2 dx## again, then it is always non-negative.
But then it's not conserved in time, which is inconsistent because the sum of all probabilities ##\int_{-\infty}^{\infty} |\psi|^2 dx## should be equal to 1, at any time.

• dextercioby
martinbn
But then it's not conserved in time, which is inconsistent because the sum of all probabilities ##\int_{-\infty}^{\infty} |\psi|^2 dx## should be equal to 1, at any time.
Yes, my question was about that part, where he wrote that he probabilities are negative.

stevendaryl
Staff Emeritus
Another way to put my point about the Klein Gordon equation is this:
Yes, my question was about that part, where he wrote that he probabilities are negative.

You have two choices, and neither works: If you interpret ##\rho = \psi^* \psi## as a probability density, then that doesn't work because the total probability isn't conserved for Klein-Gordon. If you interpret ##\rho = \psi^* \frac{d\psi}{dt} - \frac{d\psi^*}{dt} \psi## as a probability density, then it is in fact conserved, but it can be negative.

I guess you can consistently interpret ##\rho = \psi^* \frac{d\psi}{dt} - \frac{d\psi^*}{dt} \psi## as a charge density, with both positive and negative charges...

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vanhees71
Gold Member
Sigh. Once more: There is NO consistent interpretation of the KG Gordon (nor any other relativistic) wave function in terms of a single-particle picture. That's why you don't get a consistent definition of a probability distribution as in the non-relativsitic (Schrödinger or Pauli) case. All these quibbles go away when quantizing the relativistic fields and doing quantum field theory. Skipt the textbooks called "relativistic QT" and start with QFT from the beginning. Though more formal, it's at the end much simpler than all the handwaving arguments making the single-particle picture at the end to a many-body picture as it was done by Dirac for his bispinor field, leading to a very murky representation of QED, called the "hole theory". It's indeed equivalent to the modern QFT version, but much more complicated to understand and handle in practice!

• physicsworks
Gaussian97
Homework Helper
How are the probabilities computted? If it is ##\int |\psi|^2 dx## again, then it is always non-negative.
Oh yes, completely true, my fault. Of course, the definition of probability density must change. Anyway, I think @stevendaryl has already answered your question.

vanhees71
Gold Member
Another way to put my point about the Klein Gordon equation is this:

You have two choices, and neither works: If you interpret ##\rho = \psi^* \psi## as a probability density, then that doesn't work because the total probability isn't conserved for Klein-Gordon. If you interpret ##\rho = \psi^* \frac{d\psi}{dt} - \frac{d\psi^*}{dt}## as a probability density, then it is in fact conserved, but it can be negative.

I guess you can consistently interpret ##\rho = \psi^* \frac{d\psi}{dt} - \frac{d\psi^*}{dt}## as a charge density, with both positive and negative charges...
The invariance of the Lagrangian of the Klein-Gordon equation for the complex scalar field leads, via Noether's, theorem to the conserved current
$$j_{\mu}=\mathrm{i} (\psi^* \partial_{\mu} \psi-\psi \partial_{\mu} \psi^*),$$
which implies that
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 x j^0=\text{const}.$$

Of course, you cannot interpret this as a probability density, because it's not positive definite, but you can "gauge" the symmetry under multiplication with a phase factor by making the phase space-time dependent and introduce a gauge field. In this way you get the coupling of the KG field to the electromagnetic field. Then indeed the above current is (up to a factor ##q##) interpreted as the electric charge and current densities. Adding a "kinetic term" for the gauge field and quantizing both the KG and the gauge field leads to scalar electrodynamics.