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I am trying to understand the solution to the infinite square well centered at zero in Principles of Quantum Mechanics by Shankar. Here is how it goes:

Inside the well (region II - Outside left is I and outside right is III) the solution is that of the free particle:

$$\psi_{II}(x)=Ae^{ikx}+Be^{-ikx}$$

$k$ has the obvious value. Applying the boundary conditions, we require that

$$\psi_{I}\left(\frac L2\right)=\psi_{II}\left(\frac L2\right)=0$$

$$\psi_{III}\left(-\frac L2\right)=\psi_{II}\left(-\frac L2\right)=0$$

Setting the determinant of the augmented matrix corresponding to this system to zero (otherwise only the trivial solution holds):

$$e^{-ikL}+e^{ikL}=0$$

So,

$$k=\frac{n\pi}{L}$$

So,

$$Ae^{-in\pi/2}+Be^{in\pi/2}=0$$

$$B=-e^{in\pi}A = -A(-1)^n$$

We find that there are two families of solutions; those for odd n and those for even n. For the odd case:

$$\psi(x)=Ae^{ikx}+Ae^{ikx}=2A\cos\left(\frac{n\pi x}{L}\right)$$

But for the even case:

$$\psi(x)=Ae^{ikx}-Ae^{ikx}=2Ai\sin\left(\frac{n\pi x}{L}\right)$$

These intermediary steps are not gone over in the book. I am wondering if it is valid to multiply the even solutions by $i$. It is probably of fundamental importance. I know I can multiply certain things without messing up the physics but will this? Anyway, if I have made an error somewhere please point it out. I am a beginner.

Addendum: I suppose all that physically matters is whether the functions are eigenfunctions of the hamiltonian representing the system and obey the boundary and normalization conditions.

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    $\begingroup$ What is $A$ here? $\endgroup$ – jacob1729 Aug 3 at 0:19
  • $\begingroup$ A and B are just arbitrary coefficients determined by boundary conditions and normalization.@jacob1729 $\endgroup$ – Sean O'Gary Aug 3 at 0:58
  • $\begingroup$ What happens if you apply the substitution $A \rightarrow iA$? $\endgroup$ – Sandejo Aug 3 at 1:53
  • $\begingroup$ Related: physics.stackexchange.com/q/77894/2451 and links therein. $\endgroup$ – Qmechanic Aug 3 at 3:59
  • $\begingroup$ Also worth keeping in mind that the probability density function ψψ* is what we interpret physically and that will *always be real. $\endgroup$ – Rahul Chowdhury Aug 3 at 22:52
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Yes, you can multiply the even solutions by $i$. In general, you can multiply a wavefunction by any constant number of unit modulus (any number $c$ such that $|c|^2=1$) without changing the physical properties of the wavefunction. Multiplication by a number of unit modulus is called changing the phase of the wavefunction. The only time the phase of a wavefunction is relevant is when you are comparing the phases of two wavefunctions. In this case, a relative phase difference can cause interference. You can still multiply by a number of unit modulus, but you have to be consistent and multiply all wavefunctions by the same factor so that the relative phases remain the same.

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  • $\begingroup$ Thank you. This is exactly what I was looking for. Given what I already knew I could have deduced that this is possible as e^ipi/2 is i and I knew I could do that. but the underlying principle is that it doesn't change the physics. $\endgroup$ – Sean O'Gary Aug 4 at 1:50

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