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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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    Though this is not an answer to your question, it might interest you to know that in one-dimension, the wavefunction can always be taken to be real. (it's quite simple to show that if $\psi$ is a solution of the Schrodinger Equation, so is $\psi + \psi^*$.) – Philip Oct 22 '20 at 18:03
  • @Philip Can you give a reference for this? – my2cts Oct 26 '20 at 11:16
  • Possible duplicate: https://physics.stackexchange.com/q/77894/2451 and links therein. – Qmechanic Oct 26 '20 at 12:29
  • @my2cts I don't remember off the top of my head, by Qmechanic's link works just as well. For 1D, it's really trivial to do: just add the Schrodinger Equation to its complex conjugate, and you'll see that $\psi + \psi^*$ satisfies the same equation. – Philip Oct 26 '20 at 13:04
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    @Philip It all comes back ... – my2cts Oct 26 '20 at 14:33

1 Answers1

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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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  • -1 I do not think this is correct at all: it seems like you are claiming that a complex function cannot be normalised to 1, which is certainly not true. It is true that in the case that you've provided, $\psi$ cannot be normalised, but you have not added the boundary conditions! Your wavefunction blows up both at $\pm \infty$, irrespective of whether $A$ and $B$ are real or complex. Furthermore, there are many complex wavefunctions that are allowed in QM: why, the very evolution of a wavefunction in time is complex! And that comes directly from the Schrodinger equation! – Philip Oct 26 '20 at 14:27
  • It doesn't blow up if you choose $A=0$ and $B= 1$ and $a = 2m/\hbar^2 (V-E)$. Maybe my formulation was a bit misleading. For the specific case mentioned in the question there is only a real solution possible. Of course there are many complex wavefunctions out there, which can be normalised and which are solutions (just look at H-atom). Also you have to consider that the potential $V$ has a starting point $x$. Otherwise no wavefunction at all can exist or propagate, but then the question makes no sense at all. – Leviathan Oct 27 '20 at 00:32
  • The case $A=0$ $B=1$ is only valid if the potential is in the positive $x$ regime. If you consider the potential in the negative $x$ regime then its vice versa and you have $A=1$ and $B=0$. – Leviathan Oct 27 '20 at 00:35
  • You're right, for these two specific choices, the wavefunction is allowed (you have essentially added the boundary conditions). However, in this case, I hope you will agree that you can choose $A$ (or $B$) to be complex, and can still normalise the wavefunction? Furthermore, if you agree that the wavefunction can be complex, I don't understand what you mean by the second paragraph of your answer: what do you mean by "only real valued functions are allowed"? Clearly, complex function are allowed, even in this case (just multiply any solution you get by $e^{i\phi}$ and it's still a solution) – Philip Oct 27 '20 at 05:12
  • Ah I see your point. Yes you can also choose $A$ and $B$ to be complex, but since $\psi + \psi^*$ is a real valued function and is also a solution to the differential equation you can simply assume $\psi$ to be real and skip the complex phase I would argue. But you're right that you can solve the differential equation only up to a complex phase. – Leviathan Oct 27 '20 at 10:14