1. If we solve the time independent Schrödinger equation for a particle in a box of length $L$, we get:

    $$\psi_n\left(x\right)=A\sin\left(\frac{\pi n}{L}x\right)$$ I then see that we normalize $A$ such that: $$A=\sqrt{\frac{2}{L}}$$ so that the sum of the probabilities is $1$. My question is, why is the amplitude the same for each eigenvector? Why can't it be dependent upon $n$?

  2. The general solution should be (please correct me if I'm mistaken):

    $$\psi\left(x,t\right)=\sum_{n=0}^{\infty}C_{n}\sqrt{\frac{2}{L}}\sin\left(\frac{\pi n}{L}x\right)e^{-i\frac{E_{n}}{\hbar}t}$$

    So if I have the following initial condition: $$\psi\left(x,0\right)=P_0\sin\left(\frac{3\pi x}{L}\right)\cos\left(\frac{\pi x}{L}\right)$$

    I get:


    But the sum of probabilities doesn't come out right: $$\int_{0}^{L}|\psi\left(x\right)|^{2}dx=2L\neq1$$

TL:DR; I think I have a misunderstanding concerning the normalization when working with continuous functions, I will be happy to hear what I'm doing wrong


2 Answers 2


The normalisation constant $A$ could depend on $n$ (and indeed in most problems it does, for example, the Harmonic Oscillator and the Hydrogen atom, to name two famous ones). It just turns out that it doesn't in this particular case.

More importantly, the normalisation constant is in general different for different wavefunctions, since it is chosen in order for the total integral of the mod-square of the wavefunction to be 1. (In other words, the particle exits somewhere in space, so that the probability of finding it somewhere $\int_{-\infty}^\infty \text{d}x \,\, |\psi|^2 = 1$.) Thus, this constant depends heavily on the state that you choose: you need to calculate it whenever you're starting off with a new state.

However, a nice thing about the way Quantum Mechanics works is that once you normalise a wavefunction, if it's allowed to evolve in time, it will remain normalised. This is related to something called unitary time evolution.

The problem with the example you've stated is that you have started off with a state which is not normalised, which I assume is because you seem to think that $P_0 = \sqrt{2/L}$. This is not true! So the first thing to do when you start solving a problem like this is choose the constant $P_0$ such that $\psi(x,0)$ is normalised. In this case, you should be able to see that if:

$$\psi(x,0) = P_0 \sin{\left(\frac{3\pi x}{L}\right)}\cos{\left(\frac{\pi x}{L}\right)},$$ then $$\int_{-\infty}^\infty |\psi(x,0)|^2 = 1 \quad \implies \quad |P_0|^2 = \frac{4}{L}. $$

One choice (the simplest) could be that $P_0 = \sqrt{\dfrac{4}{L}}$.

Then, you could do what you had earlier done and decompose the state in terms of the two stationary states (you've got the second $\sin$ term of this expansion wrong in your question, but I assume it's a typo or something):

$$\psi (x,0) = \frac{1}{\sqrt{L}}\left(\sin{\left(\frac{2\pi x}{L}\right)} + \sin{\left(\frac{4\pi x}{L}\right)}\right)$$

From this you can easily find the time-evolved state, and you are guaranteed to have it remain normalised.


Just because a specific wavefunction is normalised doesn't mean that any linear combination will also be normalised. You found the factor by which your composite function is off to be a proper wavefunction, so just divide by $\sqrt{2L}$ and it'd be normalised.

That's why you'll often find a factor of $\frac{1}{\sqrt{2}}$ for a combination of 2 wavefunctions. If you have two orthonormal wavefunctions $\psi(x)$ and $\phi(x)$ and you add them to form a new wavefunction $f(x) = C(\psi + \phi)$ you need \begin{align*} 1 &= \int |C(\psi + \phi)|^2\mathrm{d}x = |C|^2 \int \psi^* \psi + \psi^*\phi + \psi\phi^* + \phi^* \phi \ \mathrm{d}x = |C^2| \int |\psi|^2 + |\phi|^2 \mathrm{d}x = 2 |C|^2 \end{align*} so $C = \frac{1}{\sqrt{2}}$ for the wavefunction to be normalised. (Assuming the cross terms vanish, because the wavefunctions are choosen to be orthogonal)

The general solution you wrote down is indeed a general solution for the given PDE, that doesn't mean that it's a valid wavefunction yet. You have to normalise it just like you had to normalise the $\psi_n$ individually. As for why those normalization don't depend on $n$ just do the integral, you'll find that the $n$ will cancel out.


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