I'm not sure if I'm asking this right. I'm reading ''Introduction to Quantum Mechanics'' by Griffiths and in the chapter 2 exercises he asks to prove that the separation constant, $E$, must be real. The hint he gives is to substitute $E_0 +i \Gamma$ for $E$ and use the equation for the probability density of the particle's position.

The given wave function is $\Psi (x,t) = \psi (x) e^{\frac{-iEt}{\hbar}}$

When I make the substitution and square it, the $\Gamma$ cancels out, so I don't see how being a real number affects whether or not it well cancel. Even if you substitute in $i$ for $\Gamma$ it still cancels. What am I missing here?

  • 2
    $\begingroup$ The $\Gamma$ doesn't cancel out. You should get $|\Psi(x,t)|^2 = |\psi(x)|^2e^{2\Gamma t/\hbar}$. $\endgroup$ – march Jun 27 '16 at 16:02

To make a comment a bit more formal, a state $\Psi(x,t)=\psi(x)e^{-iEt}$ with a nonzero imaginary component of energy $E=E_0+i\Gamma$ will tend to either decay or grow exponentially in norm. This can be seen from the explicit probability density, $$|\Psi(x,t)|^2=|\psi(x)|^2e^{2\Gamma t},$$ or from its evolution equation, \begin{align} i\frac{\partial}{\partial t}|\Psi(x,t)|^2 & = \left(-i\frac{\partial\Psi(x,t)}{\partial t}\right)^*\Psi(x,t) + i\Psi(x,t)^*\frac{\partial\Psi(x,t)}{\partial t} \\ & = -E^*\Psi(x,t)^*\Psi(x,t) + \Psi(x,t)^*E\Psi(x,t) \\ & = (E-E^*)|\Psi(x,t)|^2 \\ & = (E_0+i\Gamma-E_0+i\Gamma)|\Psi(x,t)|^2 \\ & = 2i\Gamma|\Psi(x,t)|^2, \end{align} where the conjugate imaginary parts no longer cancel and instead combine to give a global exponential-type evolution equation $\frac{\partial}{\partial t}|\Psi(x,t)|^2=2i\Gamma|\Psi(x,t)|^2$.


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