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.

  • 4
    $\begingroup$ They have simply redefined their constant $A$. Since $A2i$ is also a constant we can rename it $\tilde A = A2i$ $\endgroup$ Apr 15, 2020 at 10:58
  • $\begingroup$ Related: physics.stackexchange.com/q/77894/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 15, 2020 at 14:13
  • $\begingroup$ What happened to the negative sign on the second exponent? $\endgroup$
    – R.W. Bird
    Apr 15, 2020 at 14:38

2 Answers 2


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.

  • $\begingroup$ 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, 2020 at 11:04
  • $\begingroup$ the 2 gets absorbed in the $A$ constant, remember that the wave function must also be normalized! $\endgroup$ Apr 15, 2020 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, 2020 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 $-i/L$ $\endgroup$ Apr 15, 2020 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, 2020 at 12:41

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


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


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:

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

  • $\begingroup$ Thank you, I now understand that we can combine the constants into a new variable. $\endgroup$
    – Pun3rs
    Apr 15, 2020 at 12:42

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