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Basically if we have a wavefunction $$\psi(x)=\alpha e^{ik_nx}+\beta e^{-ik_nx}\tag{1}$$ and I wish to factor out the positive and negative arguments in the exponentials, then is it plausible (mathematically correct) to simply write $$\psi(x) = Ae^{\pm ik_n x}?$$ Does this mean that $\alpha + \beta = A$?

For context this is about finding the appropriate wavefunction for a particle constrained to move in a circle of circumference $L$ as a 1-dimensional problem in $x$.

For research I looked at this youtube video which works through almost exactly the same problem I have here. The only problem is at 15min 20s the person in the video simply normalizes the first term in $(1)$, which I think is incorrect.

So can the wavefunction for a particle constrained to move in a circle with a potential $V(x)=0$ be written as $$\psi(x) = Ae^{\pm ik_n x}?$$

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When you write:

$$ \psi(x) = Ae^{\pm ik_n x} $$

This is normally shorthand for a pair of equations:

$$\begin{align} \psi_+(x) &= Ae^{+ik_n x} \\ \psi_-(x) &= Ae^{-ik_n x} \end{align}$$

So no it isn't a suitable notation in this case. I don't think there is a more compact notation than the one you started with. Possibly you could express $\psi(x)$ as a sum, but is this really any better?

Response to comments:

The eigenfunctions of the Hamiltonian can be written as:

$$ \psi(x) = Ae^{\pm ik_n x} $$

where $A$ is:

$$ A = \frac{1}{\sqrt{2\pi R}} $$

and is the same for all the eigenfunctions. But the wavefunction you started with is not an eigenfunction. It is a superposition of eigenfunctions.

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  • $\begingroup$ Thanks for your answer. Is the reason why $A$ is the same for both cases due to the fact that the particle can only move in a clockwise or counter clockwise direction? $\endgroup$
    – user138066
    Commented Oct 24, 2017 at 7:00
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    $\begingroup$ $A$ is the same because that's what you've written. You could write something like $\psi(x) = A_\pm e^{\pm ik_n x}$ to indicate that $A$ could be different for the two different directions. I have to confess it isn't clear to me what you are trying to achieve. $\endgroup$
    – John Rennie
    Commented Oct 24, 2017 at 7:11
  • $\begingroup$ What i'm trying to achieve is justification that the wavefunction for a particle confined to a move along a circle of zero potential is given by the form of $\psi(x) = Ae^{\pm ik_n x}$. The form $Ae^{\pm ik_n x}$ really is the correct wavefunction to model the situation. I simply need to justify that it is of that form. My thoughts were since the circle is of fixed radius only it's direction around the circle can change and the constant out front should be the same; is it any clearer now? $\endgroup$
    – user138066
    Commented Oct 24, 2017 at 7:18
  • $\begingroup$ The eigenfunctions of the Hamiltonian are $\psi_n = Ae^{\pm ik_n x}$ where $A$ is the same for all eigenfunctions, $A = 1/\sqrt{2\pi R}$. Is that what you're asking? $\endgroup$
    – John Rennie
    Commented Oct 24, 2017 at 8:00
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No, that’s not correct

What you started up with is the most general solution to this problem.

The general solution to a differential equation is the sum of the entire set of its linearly independent solutions. In this case you have just 2: $$\psi(x)^+=\alpha e^{ik_nx}\tag{1}$$ Which represents a particle moving clockwise, and $$\psi(x)^-=\beta e^{-ik_nx}\tag{2}$$ Which represents a particle moving counter-clockwise. These are linearly independent functions and there is no way to write then as a product of one-another. The full solution is $$\psi(x)= \psi(x)^{+} + \psi(x)^{-}=\alpha e^{ik_nx}+\beta e^{-ik_nx}\tag{3}$$ The thing is that after you find the solution for the motion of a particle, that solution must be normalized, i.e., one should "force" the magnitude of the wave equation to be in between 0 to 1. The reason for that comes from the interpretation that the magnitude of the wave equation at a certain point in space in a given moment in time represents the probability of finding the particle at that point in the given moment. Mathematically, $$ \int_{- \infty}^{+ \infty} | \psi(x)|^2 dx= \int_{- \infty}^{+ \infty} \psi(x) \psi^*(x)dx=1 \tag{4}$$ In this case considering that the particle is bound to move from $0$ to $L$ $$ \int_{0}^{L} \psi(x) \psi^*(x)dx=1 \tag{5}$$ Please notice the $ \psi^*(x)$ in this equation, because it seems to be where your misunderstanding of the video is. $ \psi^*(x)$ is the complex conjugate of $ \psi(x)$ and for an equation of the type $e^{ik_nx}$ it is just $e^{-ik_nx}$. We do that because $ \psi(x)$ is a complex function, but the probability is a real function.

If we admit that the particle must move just in one way or the other, then we must normalize individually equation (1) and (2) rather than (3), in other words, if it’s moving clockwise, it’s not moving counter-clockwise and vice-versa, meaning $$ \int_{0}^{L} \alpha e^{ik_nx}( \alpha^* e^{-ik_nx})dx= \int_{0}^{L} \left|\alpha\right|^2 e^{ik_nx}(e^{-ik_nx})dx=1 \tag{6}$$ And $$ \int_{0}^{L} \beta e^{-ik_nx}( \beta^* e^{ik_nx})dx= \int_{0}^{L} \left|\beta\right|^2 e^{-ik_nx}(e^{ik_nx})dx=1 \tag{7}$$

Considering that by the boundary condition $k_n=2n \pi$, equations (6) and (7) mandate $ \alpha= \beta = \sqrt{ \frac{1}{L}}= \sqrt{ \frac{1}{2 \pi r}}$, $r$ being the radius of the cirle.

As you see the $-ik_nx$ in brackets on (6) doesn’t come from (2). It is the complex conjugate of (1). The same is valid to (7).

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