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I am having trouble understanding the reason for choosing solutions as $\psi(x) = exp(ikx)$ in the following problem. I am following MIT OCW 8.04 lecture by Prof. Barton Zwiebach (Energy eigen states for a particle on a circle). Can somebody please help?

There is a free particle moving on a circle (described as a periodic function along $x$-axis). The time independent Schroedinger equation can be written as

$$\frac{-\hslash^2}{2m}\frac{d^2}{dx^2}\psi(x) = E\psi(x).$$ This is a second-order differential equation. So my understanding is that the solution needs to have two constants, i.e., let $$\psi(x) = A e^{ikx} + B e^{-ikx},$$ where $A$ and $B$ are the constants and $$k^2 = \frac{2mE}{\hslash^2}.$$ But the solution assumed is $e^{ikx}$, why is it so chosen? Don't we need two constants?

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When he writes $\psi = e^{ikx}$ he is implicitly including negative values of $k$ so it includes the solutions $\psi = e^{-ikx}$. Note that at 3:25 into the video he writes:

$$ kL = 2\pi n \\ n \in\mathbb{Z}$$

so $n$ can be negative and that makes $k$ negative.

The set of all functions $\psi = e^{ikx}$ (for positive and negative $k$) is a perfectly good basis for a Hilbert space. You can certainly include sums of functions with the same $|k|$ but this adds nothing new.

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  • $\begingroup$ thank you for the help. But suppose I assume $Aexp(ikx)+Bexp(-ikx)$ as the solution, then can you please tell how I can find these constants? $\endgroup$ Commented Feb 4 at 9:50
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    $\begingroup$ You use the initial conditions i.e. $\psi(x=0)$ and $\psi'(x=0)$. You substitute these values into your equations and you get two simultaneous equations in $A$ and $B$. $\endgroup$ Commented Feb 4 at 9:56

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