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koustav

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- Second order time derivative

What problem actually arises when we take the second order time derivative in KG equation

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In summary, the textbook mentions that the probability density is not positive definate because of the second order time derivative. So, in order to use the Klein Gordon equation as an immediate relativistic substitute for the Schrödinger equation, one has to change some fundamenta

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koustav

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- TL;DR Summary
- Second order time derivative

What problem actually arises when we take the second order time derivative in KG equation

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koustav

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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 interpretationvanhees71 said:

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Gaussian97

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$$\left(i\hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi=0$$

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.

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Which textbook?koustav said: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

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".

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martinbn

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How are the probabilities computted? If it is ##\int |\psi|^2 dx## again, then it is always non-negative.Gaussian97 said:

$$\left(i\hbar \frac{\partial}{\partial t} + \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\right)\psi=0$$

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.

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

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martinbn

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Yes, my question was about that part, where he wrote that he probabilities are negative.Demystifier said: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.

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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} \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...

martinbn said: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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Gaussian97

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

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The invariance of the Lagrangian of the Klein-Gordon equation for the complex scalar field leads, via Noether's, theorem to the conserved currentstevendaryl said: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...

$$j_{\mu}=\mathrm{i} (\psi^* \partial_{\mu} \psi-\psi \partial_{\mu} \psi^*),$$

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.

The Klein-Gordon equation is a relativistic wave equation that describes the behavior of spinless particles, such as the Higgs boson, in quantum field theory. It is a second-order partial differential equation that combines elements of both the Schrödinger equation and the relativistic energy-momentum relation.

The 2nd order time derivative in the Klein-Gordon equation represents the time evolution of the wave function of a particle. Solving for this derivative allows us to understand how the particle's wave function changes over time, and therefore, how the particle itself behaves and moves through space.

The Klein-Gordon equation is solved using various mathematical techniques, such as separation of variables, Fourier transforms, and Green's functions. These techniques allow us to find the wave function of a particle at any given time and location.

The Klein-Gordon equation has been used in various fields of physics, including particle physics, quantum mechanics, and cosmology. It has also been applied in other areas such as quantum field theory, condensed matter physics, and nuclear physics. Additionally, the equation has been used to study the behavior of particles in accelerators and to predict the properties of new particles.

One limitation of the Klein-Gordon equation is that it only describes spinless particles. It also does not take into account the effects of external forces or interactions between particles. Additionally, the equation does not account for relativistic effects at very high energies, where a more complex equation, such as the Dirac equation, is needed.

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