Explanation of a failed approach to relativize Schrodinger equation

Click For Summary
SUMMARY

The discussion focuses on the challenges of relativizing the Schrödinger equation, particularly regarding the probability density and current. It highlights the transition from the non-relativistic Schrödinger expression to a relativistic framework, introducing the 4-current density \( J^\mu \) and the modified probability density \( \rho \). The key issue identified is that while the continuity equation remains valid, the new expression for density is not positive definite, which contradicts the requirements for a legitimate probability density. This failure indicates that a naive generalization of the Schrödinger equation does not suffice for relativistic contexts.

PREREQUISITES
  • Understanding of the Schrödinger equation and its implications in quantum mechanics.
  • Familiarity with the Dirac equation and its role in relativistic quantum mechanics.
  • Knowledge of probability density and current in quantum field theory.
  • Basic concepts of covariant expressions and continuity equations in physics.
NEXT STEPS
  • Explore the derivation of the Dirac equation and its implications for relativistic wave functions.
  • Study the properties of complex scalar fields and their probability densities.
  • Investigate the mathematical formulation of continuity equations in quantum mechanics.
  • Learn about the implications of negative probability densities in quantum field theories.
USEFUL FOR

Physicists, particularly those specializing in quantum mechanics and quantum field theory, as well as students seeking to understand the complexities of relativistic formulations of quantum equations.

nacadaryo
Messages
2
Reaction score
0
I'm reading the Wikipedia page for the Dirac equation
[itex]\rho=\phi^*\phi\,[/itex]

...

[itex]J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*)[/itex]

with the conservation of probability current and density following from the Schrödinger equation:

[itex]\nabla\cdot J + \frac{\partial\rho}{\partial t} = 0.[/itex]

The fact that the density is positive definite and convected according to this continuity equation, implies that we may integrate the density over a certain domain and set the total to 1, and this condition will be maintained by the conservation law. A proper relativistic theory with a probability density current must also share this feature. Now, if we wish to maintain the notion of a convected density, then we must generalize the Schrödinger expression of the density and current so that the space and time derivatives again enter symmetrically in relation to the scalar wave function. We are allowed to keep the Schrödinger expression for the current, but must replace by probability density by the symmetrically formed expression

[itex]\rho = \frac{i\hbar}{2m}(\psi^*\partial_t\psi - \psi\partial_t\psi^*).[/itex]

which now becomes the 4th component of a space-time vector, and the entire 4-current density has the relativistically covariant expression

[itex]J^\mu = \frac{i\hbar}{2m}(\psi^*\partial^\mu\psi - \psi\partial^\mu\psi^*)[/itex]

The continuity equation is as before. Everything is compatible with relativity now, but we see immediately that the expression for the density is no longer positive definite - the initial values of both ψ and [itex]∂_t ψ[/itex] may be freely chosen, and the density may thus become negative, something that is impossible for a legitimate probability density. Thus we cannot get a simple generalization of the Schrödinger equation under the naive assumption that the wave function is a relativistic scalar, and the equation it satisfies, second order in time.

I am not sure how one gets a new [itex]\rho[/itex] and [itex]J^\mu[/itex]. How does one do to derive these two? And can anyone show me why the expression for density not positive definite?
 
Physics news on Phys.org
I think those formulas are guessed, and it can be shown that they satisfy the relevant equations afterwards. As you can see from the nonrelativistic probability flow, it is nothing completely new...

And can anyone show me why the expression for density not positive definite?
Consider a function ##\psi## where ##\rho>0## and look at ##\psi^*##.
##\rho(\psi)=-\rho(\psi^*)##
 
Take ψ to be a plane wave, ψ(x,t) = ei(k·x - ωt), which will be a solution provided ω2 = k2 + m2. For this solution, ρ = ħω/m, obviously positive/negative whenever ω is positve/negative. For the general solution which is a superposition of plane waves, ρ is an integral over the positive frequency solutions minus an integral over the negative frequency ones.

According to the continuity equation, Q = ∫ρ d3x is a conserved quantity. Although the 'derivation' of this usually consists of simply writing it down, its existence is no accident. For a complex scalar field, Q represents the total charge. 'Charge' can mean either the ordinary electric charge or some other charge such as strangeness.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 19 ·
Replies
19
Views
2K