Mathematica What is the Mathematical Signature of the deBroglie Function for an Electron?

[meopemuk]

Originally Posted by jostpuur:

This series does not converge for | P| > 1, and I would be suprised if it converged when an operator is subtituted in place of P.

So, it seems it would works as long as cp doesn't exceed the rest mass
energy. That would be up to v=\sqrt{1/2}\ c.

The series could be non-convergent, but the square root itself is a well-defined function for all real P or v. I think that your condition v=\sqrt{1/2}\ c should not be interpreted as any physical limit.

Eugene.

 

[jostpuur]

Unfortunately I don't know much of square roots of operators. If I was now given a task of defining the operator \sqrt{-\nabla^2 + m^2}, I would define it like this

<br /> \sqrt{-\nabla^2+m^2}\psi(x) := \int\frac{d^3x&#039;\; d^3p}{(2\pi)^3} \psi(x&#039;) \sqrt{|p|^2 + m^2} e^{ip\cdot(x-x&#039;)}<br />

Would other definitions agree with this one?

But this is not fully making sense. Isn't it well established fact, that Klein-Gordon equation does not conserve the probability \int d^3x\;|\psi|^2? Or could it be, that it actually conserves it, but it is just not manifestly apparent? Now when you define the time evolution with the Shrodinger's equation (with this Hamiltonian with square root), the time evolution also satisfies the Klein-Gordon equation. So if Shrodinger's equation conserves the probability, and Klein-Gordon equation does not, this is quite paradoxical.

I would sure like some well done mathematical analysis conserning this operator. Merely noting that the Taylor series don't work for it doesn't seem very satisfactory.

 

[meopemuk]

Unfortunately I don't know much of square roots of operators. If I was now given a task of defining the operator \sqrt{-\nabla^2 + m^2}, I would define it like this

<br /> \sqrt{-\nabla^2+m^2}\psi(x) := \int\frac{d^3x&#039;\; d^3p}{(2\pi)^3} \psi(x&#039;) \sqrt{|p|^2 + m^2} e^{ip\cdot(x-x&#039;)}<br />

Would other definitions agree with this one?

In my opinion, this definition is perfectly OK.

But this is not fully making sense. Isn't it well established fact, that Klein-Gordon equation does not conserve the probability \int d^3x\;|\psi|^2? Or could it be, that it actually conserves it, but it is just not manifestly apparent? Now when you define the time evolution with the Shrodinger's equation (with this Hamiltonian with square root), the time evolution also satisfies the Klein-Gordon equation. So if Shrodinger's equation conserves the probability, and Klein-Gordon equation does not, this is quite paradoxical.

I think that in order to answer this it is useful to take a bit more abstract and general approach. In quantum mechanics, probabilities are preserved if the state vector gets transformed by a unitary operator U (Wigner theorem)

|\Psi \rangle \to U |\Psi \rangle

So, the operator of time evolution U(t) must be unitary

|\Psi (t)\rangle = U(t) |\Psi (0)\rangle

Time evolution is a one-parametric subgroup of transformations from the Poincare group. Then by Stone theorem, representatives of time shifts in the Hilbert space must have the form

U(t) = \exp(\frac{i}{\hbar} Ht)

where H is a Hermitian operator (called Hamiltonian). So, we obtain

|\Psi (t)\rangle = \exp(\frac{i}{\hbar} Ht)|\Psi (0)\rangle

which can be differentiated by t to obtain

-i \hbar \frac{\partial}{\partial t}|\Psi (t)\rangle = H|\Psi (t)\rangle

So, it follows from very general principles (conservation of probability, group structure of inertial transformations, the principle of relativity) that the time dependence of state vectors (and/or wave functions) should be described by the above equation, where H is an Hermitian operator.

The Klein-Gordon equation does not have this form. This is the reason why KG does not conserve probability and why it is wrong to consider KG as a relativistic generalization of the Schroedinger equation.

 

[Hans de Vries]

Unfortunately I don't know much of square roots of operators. If I was now given a task of defining the operator \sqrt{-\nabla^2 + m^2}, I would define it like this

<br /> \sqrt{-\nabla^2+m^2}\psi(x) := \int\frac{d^3x&#039;\; d^3p}{(2\pi)^3} \psi(x&#039;) \sqrt{|p|^2 + m^2} e^{ip\cdot(x-x&#039;)}<br />

Would other definitions agree with this one?

That's OK. The operator working on the wave function in configuration space
is indeed a convolution with the Fourier transform of the propagator in
momentum space.

Regards, Hans.

 

[Hans de Vries]

But this is not fully making sense. Isn't it well established fact, that Klein-Gordon equation does not conserve the probability \int d^3x\;|\psi|^2? Or could it be, that it actually conserves it, but it is just not manifestly apparent? Now when you define the time evolution with the Shrodinger's equation (with this Hamiltonian with square root), the time evolution also satisfies the Klein-Gordon equation. So if Shrodinger's equation conserves the probability, and Klein-Gordon equation does not, this is quite paradoxical.

The whole point is, I think, that the Schrödinger wave function is defined as:

\psi \ =\ \Psi\ e^{-imc^2t/\hbar}

Where only \Psi can vary and \exp(-imc^2t/\hbar) is "hardcoded" in the solution.
The same is true in QFT working with plane waves in momentum space.

Now while the Schrödinger and QFT solutions are also the solutions of
the Klein Gordon equation, it does not generate them in a lattice
simulation if not given the proper initial conditions.

\partial^2_t \psi\ =\ \nabla\psi - m^2\psi

is a real equation without imaginary parts. If you give it a real wave function
as input, together with 0 or real initial values for \partial_t \psi then it will stay
forever real in a lattice simulation, and the integral over space can grow
indefinitely in some cases.

Given the proper initialization for \partial_t \psi, which is entirely imaginary if the
wave function is real, then it should continue to simulate exactly as with
Schrödinger's equation. (ignoring the small terms which are thrown away if
one goes from Klein Gordon to Schrödinger)Regards, Hans

 

[jostpuur]

meopemuk, have you been able to derive this relativistic Shrodinger's equation out of relativistic Lagrangian L=-mc^2\sqrt{1-|v|^2/c^2} with Feynman's path integral (or so called slice of path integral)? I've noticed it is notably more difficult than in non-relativistic case. At least I haven't been able to do it.

I was quite suprised when I noticed last night, that the Klein-Gordon equation actually does conserve probability, if only positive frequency solutions are considered. Are you aware of this?

btw this discussion has got quite far from the original post, but I guess it doesn't disturb anyone...

 

[meopemuk]

meopemuk, have you been able to derive this relativistic Shrodinger's equation out of relativistic Lagrangian L=-mc^2\sqrt{1-|v|^2/c^2} with Feynman's path integral (or so called slice of path integral)? I've noticed it is notably more difficult than in non-relativistic case. At least I haven't been able to do it.

No I haven't tried to do that. Personally I find least action, Lagrangian, and path integral methods more confusing than illuminating. Perhaps, it's just me.

I was quite suprised when I noticed last night, that the Klein-Gordon equation actually does conserve probability, if only positive frequency solutions are considered. Are you aware of this?

Even if this is so, it wouldn't convince me that the Klein-Gordon equation can be treated as an analog of the Schroedinger equation for relativistic particles. I've just written a long post about this in https://www.physicsforums.com/showpost.php?p=1379494&postcount=171

btw this discussion has got quite far from the original post, but I guess it doesn't disturb anyone...

It doesn't disturb be. We can open a new thread if somebody is disturbed.

Eugene.