feynmann said:
Does anyone know a deeper reason why the quantum mechanical wavefunction has to be complex?
In the non-relativistic QM, the complex nature of the wave function is as fundamental as the Plank constant itself.
1) If the dynamical equation EXPLICITLY contains the imaginary number i, as does Schroginger's equation, then a PAIR of REAL functions ( or their equivalent in the form of a SINGLE COMPLEX function) are needed for complete physical description.
If you insert \psi = f + ig into Schrodinger's equation (2m=\hbar =1):
i \partial_{t}\psi = - \nabla^{2}\psi
you end up with 2 COUPLED equations
<br />
\partial_{t}f + \nabla^{2}g = 0, \ \ \ \partial_{t}g - \nabla^{2}f =0<br />
Thus we see that f and g are not INDEPENDENT of each other,i.e., neither of them alone is a solution of Schrodinger's equation.
Therefore, a complete solution requires the REAL pair (f,g) or their equivalent COMPLEX \psi.
Notice that the probability current
j_{i} = 2(f \partial_{i}g - g\partial_{i}f)
is a mutual property of f and g, which vanishes when either is identically zero.
2) If the imaginary (i) does not show up explicitly in your equation, then the USE of COMPLEX function is just an auxiliary device.
Take the wave equation
\frac{\partial^{2}\phi}{\partial t^{2}} = \nabla^{2}\phi
Now put \phi = f +ig. This leads to the following two independent equations
(\frac{\partial^{2}}{\partial t^{2}} - \nabla^{2})f = 0
(\frac{\partial^{2}}{\partial t^{2}} - \nabla^{2})g = 0
Thus, f and g remain independent of each other,i.e., a complete solution can be given in terms of f alone or g alone. Therefore, in this case, the wave function \phi need not necessarily be complex function.
3)You might now say this: But the first-order Schrodinger's equation can be replaced by the "equivalent" second-order equation
\frac{\partial^{2}\psi}{\partial t^{2}} + \nabla^{4}\psi = 0 \ \ \ (X)
Therefore, according to (2), the wave function \psi does not need to be a complex function! Does this mean that the complex nature of Schrodinger's wave function is not FUNDAMENTAL after all?
This quetions is, of course, meaningless, because it is based on a wrong conclusion. Let me explain this.
It is true (as explained in (2)) that \psi in the second-order equation (X) does not need to be complex function. It is also true that the equation (X) is "equivalent" to the first-order Schrodinger's equation. IT IS, however, a WRONG equation to use in QM! One way to see this is to prove that NO CONSERVED PROBABILITY can be set up which depends on \psi only and not on \partial_{t}\psi.
To do this, write
P = P(\psi)
Then
\partial_{t} \int dx \ P(\psi) = \int dx \frac{\partial P}{\partial \psi} \frac{\partial \psi}{\partial t}
If this expression is to vanish for arbitrary \psi, it is necessary that \partial_{t}\psi be given in terms of \psi. But this implies a FIRST-ORDER wave equation. In a SECOND-ORDER differential (like eq(X)), \partial_{t}\psi can be given an arbitrary initial value. Therefore, the above expression cannot vanish for all \psi. For example, if we put
\frac{\partial \psi}{\partial t} = \frac{\partial P}{\partial \psi}
then
\frac{\partial}{\partial t} \int dx \ P(\psi) = \int dx \ (\partial P / \partial \psi )^{2} > 0
So, we see that EQ(X) can not be PHYSICALLY equivalent to Schrodinger's equation.
regards
sam
