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I am relatively new to quantum mechanics and I've stumbled upon an issue. Assume that we have an infinite square well that follows this figure:enter image description here

Now I am trying to solve for the wave function, using the time independent Schrodinger's Equation.

$$\frac{-h^2}{2m} \frac{d^2}{dx^2} \Psi(x) = E \Psi(x)$$

The solution to the ODE is:

$$\Psi(x) = Ae^{ikt} + Be^{-ikt} $$ where $k = \frac{\sqrt{2mE}}{\hbar}$

Next, I solve for the boundary conditions, first by solving $\Psi(0)=0$ and I arrive at $A+B=0$, which implies that $B = -A$ and that the wave function is $$\Psi(x) = Ae^{ikx} - Ae^{-ikx}$$

Now my attempt to simplify the new wave function led me to $$\Psi(x) = A2i\sin(kx)$$ However, when I view the infinite square well problem from other sources, I always see that the simplification leads to: $$\Psi(x) = A\sin(kx)$$

I am wondering what properties are there that enables us to neglect the imaginary constant $2i$. As I am very new to the subject I might be missing some seemingly trivial concepts.

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    $\begingroup$ They have simply redefined their constant $A$. Since $A2i$ is also a constant we can rename it $\tilde A = A2i$ $\endgroup$ – Turbotanten Apr 15 '20 at 10:58
  • $\begingroup$ Related: physics.stackexchange.com/q/77894/2451 and links therein. $\endgroup$ – Qmechanic Apr 15 '20 at 14:13
  • $\begingroup$ What happened to the negative sign on the second exponent? $\endgroup$ – R.W. Bird Apr 15 '20 at 14:38
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Two ways to see this, one is that the $2i$ can simply be absorbed in the constant $A$, but I presume that the fact that the constant is complex is bothering you.

The only thing we care about the wavefunction is its square modulus, which represent a probability density. If you have a wave function $\psi(x)$ and you multiply it by a complex constant of modulus $1$, $|\psi(x)|^2$ doesn't change, hence the wave function makes equivalent physical predictions. So multiplying a wave function by $i$ doesn't give a physically different result, and you can just drop the phase.

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  • $\begingroup$ I see, but assuming if we keep the constant then $|\Psi(x)|^2 = |2i\sin(kx)|^2 = 4|\sin(kx)|^2 $. If we were to remove the constant then the 4 would also be removed. $\endgroup$ – Pun3rs Apr 15 '20 at 11:04
  • $\begingroup$ the 2 gets absorbed in the $A$ constant, remember that the wave function must also be normalized! $\endgroup$ – user2723984 Apr 15 '20 at 11:06
  • $\begingroup$ Yep, disregard that reply my bad :/, if we were to absorb the constant into our normalisation constant then we will get the traditional $\sqrt{\frac{2}{L}}$ but if we do otherwise, then the we would have a different normalisation constant and its all the same since the norm squared is what matters. Is this correct? $\endgroup$ – Pun3rs Apr 15 '20 at 11:57
  • $\begingroup$ Not precisely, if you kept the 2 you'd get $1/L$, and the final result is always the same because the factor 2 is already there. You could keep the whole $2i$ and get $-i/L$ $\endgroup$ – user2723984 Apr 15 '20 at 12:06
  • $\begingroup$ Yep, it would not precisely be $\frac{2}{L}$ but all of them works at the end. I see, thank you. $\endgroup$ – Pun3rs Apr 15 '20 at 12:41
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The time-independent Schrodinger equation for the infinite square well is

(1)$$\frac{-\hbar}{2m} \frac{d^2}{dx^2} \Psi(x) = E \Psi(x)$$

We can substitute in your solution:

(2)$$\Psi(x) = A2i\sin(kx)$$

(3)$$\frac{-\hbar}{2m} \frac{d^2}{dx^2} A2i\sin(kx) = E A2i\sin(kx)$$ Since $A2i$ is a constant the differentiation doesn't affect it, so we can move it to the left

(4)$$-A2i\frac{\hbar}{2m} \frac{d^2}{dx^2} \sin(kx) = E A2i\sin(kx)$$

If we now differentiate twice we get

(5)$$A2ik^2\frac{\hbar}{2m}\sin(kx)=EA2i\sin(kx)$$

Now we can divide both sides by $$

(6)$$k^2\frac{\hbar}{2m}=E$$ We can see that the $2Ai$, which are constants, do not affect the solution of the differential equation because they cancel out on both sides. Therefore $A$ can be any number, including complex, because it can be removed on both sides, in the differential equation. If now we say $\bar{A}=2iA$ we can substitute it into equation 2 and we get

(5)$$\Psi(x) = \bar{A}\sin(kx)$$

Now if we solve the Schrodinger equation in this form we can see that it becomes

(6)$$k^2\frac{\hbar}{2m}=E$$

which is equivalent to what we got before. This means we can absorb the constants into one constant $\bar{A}$ and not affect the solution of the Schrodinger equation. Therefore this means that $\Psi(x) = \bar{A}\sin(kx)$ is also a valid solution to the Schrodinger equation.

Then we see that for the equation to have a probability of one, over all of space, the equation becomes:

$$\Psi(x) = \sqrt{\frac{2}{L}}\sin(kx)$$

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  • $\begingroup$ Thank you, I now understand that we can combine the constants into a new variable. $\endgroup$ – Pun3rs Apr 15 '20 at 12:42

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