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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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- Thread starter koustav
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- #1

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- Summary:
- Second order time derivative

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

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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 interpretation

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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?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.

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

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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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.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.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...

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.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 currentAnother 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.

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