We assume the answer is no because a minus sign appears to the left:
-h^2/2m (d^2Ψ(x,t)/dx^2]partial)+V(x,t)Ψ(x,t)=ihdΨ(x,t)/dt]partial
And,
-h^2/2m (d^2Ψ*(x,t)/dx^2]partial)+V(x,t)Ψ*(x,t)=-ihdΨ*(x,t)/dt]partial
The question is what is the mathematical/physical meaning of the minus sign?
No, if It's any complex transformation-yes, if it's the particular complex transformation known as complex conjugation.
This is a standard exercise in any course on quantum mechanics:
IF Ψ(x,t) is a solution of idΨ/dt = HΨ, THEN Ψ* is a solution of the equation
-idΨ*/dt = HΨ* idΨ*/d(-t) = HΨ*
which is the same Schrödinger equation, but for Ψ(x,-t)* (and vice versa), ASSUMING that the Hamiltonian is Hermitian; if it's not Hermitian, then complex conjugation maps the solution Ψ(x,t) of the Schrödinger equation with Hamiltonian H to a solution Ψ(x,-t)* of the Schrödinger equation with Hamiltonian H*.
The minus sign comes about as the result of complex conjugation and this is equivalent to changing t to -t (time reversal).
So the interesting statement is that IF Ψ(x,t) is a solution to the Schrödinger equation, THEN Ψ(x,-t)* is a solution of the SAME equation, assuming the Hamiltonian is a Hermitian operator, i.e. H = H*, which is what this-now standard-exercise in any quantum mechanics course is really about.
This isn't a research topic anymore, it's an exercise for undergraduates...
Incidentally, it's wrong to state ``we assume...'' about a statement that isn't an assumption! Ìt's wrong to mix up what's assumed and what's to be proved...
This is just an introductory answer to shed some light on the question and its proposed answer.
Some people may think that this subtle question is no longer a subject of research but it is a student exercise which is not true.
There is no definitive answer yet, either in undergraduate or postgraduate studies.
Here are some comments from today's top mathematicians and physicists:
*Just because ψ is a solution of Schrodinger's equation SE does not mean that its complex conjugate is also a solution.
**while it may be tempting to simply replace ψ∗ for ψ
in Schrödinger's equation, it's not something you're allowed to do.
*** I personally (perhaps incorrectly) see this feature as an early sign of CT
symmetry where C is the conjugate symmetry operation of charge and T
is the symmetry operation by time inversion.
&Coming back to the question, I personally assume the answer comes from the Q transition matrix which can replace the SE itself in many situations:
For a 1-D SE with 6 nodes, it is necessary to start from the transition matrix B. For RO=0 the B-matrix is expressed as follows:
0 1/2 0 0 0 0
1/2 0 1/2 0 0 0
0 1/2 0 1/2 0 0
0 0 1/2 0 1/2 0
0 0 0 1/2 0 1/2
0 0 0 0 1/2 0
And the Q-transition matrix, Q=B^1/2 is given by,
0.317+i*0.317 0.321-i*0.321 0.089+i*0.089 -0.065+i*0.065 -0.023-i*0.023 0.041-i*0.041
0.321-i*0.321 0.406+i*0.406 0.256-i*0.256 0.066+i*0.066 -0.023+i*0.023 -0.023-i*0.023
0.089+i*0.089 0.256-i*0.256 0.383+i*0.383 0.298-i*0.298 0.066+i*0.066 -0.065+i*0.065
-0.065+i*0.065 0.066+i*0.066 0.298-i*0.298 0.383+i*0.383 0.256-i*0.256 0.089+i*0.089
-0.023-i*0.023 -0.023+i*0.023 0.066+i*0.066 0.256-i*0.256 0.406+i*0.406 0.321-i*0.321
0.041-i*0.041 -0.023-i*0.023 -0.065+i*0.065 0.089+i*0.089 0.321-i*0.321 0.317+i*0.317
Note that all entries of the matrix Q have the form,
a+/-ia
which means that when multiplied by any complex conjugate number, the minus sign appears automatically.
To be continued.
Once more: IF Ψ is a solution to idΨ/dt = HΨ AND H is a Hermitian operator, THEN Ψ^* is a solution to -idΨ^*/dt = H Ψ^*. That's why this statement is now a trivial exercise.
The non-trivial part is ensuring that H is a Hermitian operator, if the potential doesn't have the properties presented in all textbooks.
The stochastic approach is equivalent to the study of the properties of the Schrödinger equation, so it's not possible to get something from one that's not present in the other. Which one uses is just a matter of taste.
It is perfectly legitimate to compute (idΨ/dt)^* = -idΨ^*/dt, by trivial manipulations, replace in the LHS of the above expression idΨ/dt by HΨ, deduce that (HΨ)^*= HΨ^*, if H is Hermitian and deduce that -idΨ^*/dt = HΨ^*.
The reason is that idΨ/dt is a complex function, so its conjugate is well defined. Nothing beyond high school algebra...
On principe c'est la solution du complexe conjugue de la fonction d'onde
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