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I am trying to analyze in general simple one dimensional QM problems. To be more specific let's consider this kind of Hamiltonian: $$H=\frac{\hat{p}^2}{2m}+V(\hat{x})$$ From this one we can derive the following: $$\frac{\partial^2\psi(x)}{\partial x^2}=\frac{2m}{\hbar ^2}(V(x)-E)\psi(x) \ \ \ \ \ \ (1)$$ Where of course $\psi(x):\mathbb{R}\to\mathbb{C}$ is the wave function.

Now for example let's consider the case of a symmetric potential well, and also let's consider the case in which $V(x)>E$. In this case my lecture notes state that we can see from (1) that $\psi ''$ has the same sign as $\psi$. Now here I have a problem: this statement would of course be true for a real value function $\psi(x):\mathbb{R}\to \mathbb{R}$, but the wave function assumes complex values, why are we authorized to treat the wave function as a real value function?

This is not the only example of this way of handling the wave function, often in my lecture notes the wave function is depicted in a $2D$ graph as if it was a real value function and not a complex one. Why can we do this? If we want to talk about real functions in this context shouldn't we consider the probability density $|\psi(x)|^2$ instead of the probability amplitude $\psi(x)$?

Qmechanic
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Noumeno
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1 Answers1

-1

Maybe I'm thinking to simple, but for $V(x)>E$ the equation has the form $\psi'' = a \cdot \psi$, where $a>0$. For this cases $\psi$ can only be of the form $\psi = A\cdot e^{\sqrt{a} x} + B \cdot e^{-\sqrt{a}x}$, which is entirely real. Maybe you could argue that $A$ and $B$ can be complex but this will not happen if you use the condition $\int \mathrm{d}^{3} r|\psi|^{2}=1$ and $\int \mathrm{d}^{3} r|\psi|^{2}<\infty$.

So I think the idea here goes the over way around like you stated it: We are not authorized to treat the wavefunction as a real valued function, but we can see from the differential equation that only real valued functions are allowed.

The fact that a quantum mechanical particle can have a finite spatial probability even in such a region leads to very characteristic phenomena (e.g. tunnel effect). If you examine this example in detail you will find that for general $V(x)$ the wavefunction is exponentially decaying (see also Nolting, Quantum Mechanics 6, Chapter 4).

Leviathan
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