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A state of a particle bounded by infinite potential walls at x=0 and x=L is described by a wave function $\psi = a\phi_1 + b\phi_2 $ where $\phi_i$ are the stationary states.

So let's say we want to normalize this wave function. As I understand it the procedure is as follows:

The probability of the particle being at any point from 0 to L is 1. So I need to integrate the wave functions squared over that interval. By the superposition principle it is OK to just add them. On top of that, any $\psi$ can also be expressed as $\psi \psi^*$

$\psi = a\phi_1 + b\phi_2 $

$\psi = (a\phi_1\phi_1^* + b\phi_2\phi_2^*)$

We want to integrate $\vert\psi\vert ^2$

$(a\phi_1\phi_1^*)^2 + 2ab\phi_2\phi_2^*\phi_1\phi_1^*+(b\phi_2\phi_2^*)^2 = (a^2 + b^2)$

Since the phi functions are eigenvalues, the ones on the diagonal of the matrix are the only ones not zero, which is why the cross terms in the middle disappear (they are zero) and the end terms $(\phi_i\phi_i^*)$ are equal to 1. So we get

$\psi = \int_0^L\vert\psi(x)\vert ^2dx = \int_0^L\vert(a^2 + b^2)\vert^2dx=1 $

and therefore

$\int_0^L\vert(a^2 + b^2)\vert^2dx=1 $ and

$(a^2+b^2)^2x\vert^L_0 = 1 \rightarrow (a^2+b^2)^2L=1\rightarrow L=1/(a^2+b^2)^2$

The conceptual question I had was that if we have the probability squared here, is it that or the square root of that probability that is your normalization constant? Further, would it also be permissible to treat each of the wavefunctions as $A\sin\frac{n\pi x}{L}$ where $A_1=a$ and $A_2=b$, and try the integration that way? Given that the wave functions are supposedly different that seemed like it would be wrong, but we also know they are stationary states so they go to zero at either end of the potential well and are sinusoidal, correct?

I know that this area doesn't always cotton to HW type questions. But this is the kind of thing that I think could help a lot of people get their heads around this concept, because I can't be the only one who is a bit lost on how to actually use these techniques.

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  • $\begingroup$ I think what you were trying for was $|\Psi|^2=\Psi\Psi^*$. $\Psi \ne \Psi\Psi^*$. $\endgroup$ – Ari Ben Canaan Apr 17 '14 at 5:28
  • $\begingroup$ yes you are right.. $\endgroup$ – Jesse Apr 17 '14 at 10:30
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Honestly, the argument you're making here is a mess - the question is based on bad premises. So let me show you how to do it properly, and hopefully that will resolve your confusion.

You're right that in order for a wavefunction to be normalized, it must satisfy

$$\int_\text{all space} P(x)\mathrm{d}x = \int_\text{all space} \psi^*(x)\psi(x)\mathrm{d}x = 1\tag{1}$$

But this statement:

On top of that, any $\psi$ can also be expressed as $\psi\psi^∗$

is not correct. Given a function $\psi(x)$, you can write $\psi(x)\psi^*(x)$, but that's a different function.

Anyway, given that your wavefunction can be written

$$\psi(x) = a\phi_1(x) + b\phi_2(x)$$

then you just plug that into the normalization condition (1) and get

$$\int_0^L \bigl(a^* \phi_1^*(x) + b^* \phi_2^*(x)\bigr)\bigl(a \phi_1(x) + b \phi_2(x)\bigr)\mathrm{d}x = 1$$

which expands to

$$\begin{multline} a^*a \int_0^L \phi_1^*(x)\phi_1(x)\mathrm{d}x + a^*b \int_0^L \phi_1^*(x)\phi_2(x)\mathrm{d}x \\ + b^*a \int_0^L \phi_2^*(x)\phi_1(x)\mathrm{d}x + b^*b \int_0^L \phi_2^*(x)\phi_2(x)\mathrm{d}x = 1 \end{multline}\tag{2}$$

Now you can use the identity

$$\int_0^L\phi_1^*(x)\phi_2(x)\mathrm{d}x = \int_0^L\phi_2^*(x)\phi_1(x)\mathrm{d}x = 0$$

which follows from the fact that $\phi_1$ and $\phi_2$ are orthogonal functions (it's not enough that they are eigenfunctions of an operator, they have to be orthogonal), and the identity

$$\int_0^L\phi_1^*(x)\phi_1(x)\mathrm{d}x = \int_0^L\phi_2^*(x)\phi_2(x)\mathrm{d}x = 1$$

which simply reflects the fact that $\phi_1$ and $\phi_2$ are normalized. (Check for yourself that this is the same as the normalization condition, equation (1).) With these two identities, equation (2) reduces to

$$\lvert a\rvert^2 + \lvert b\rvert^2 = 1$$

The conceptual question I had was that if we have the probability squared here, is it that or the square root of that probability that is your normalization constant?

That all depends, how do you define your normalization constant? It depends on what you're normalizing and how exactly you express it as a function. However you do it, the end requirement for normalization is just that $\lvert a\rvert^2 + \lvert b\rvert^2 = 1$.

As far as using the specific sinusoidal form for the $\phi_i$, you can do that in this case, because you're given enough information to figure out that the eigenfunctions are in fact sinusoidal. But you don't really need to know that they are sinusoidal for the preceding argument to work; all you need to know is that the $\phi_i$s are orthonormal.

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  • $\begingroup$ Thanks! to be absolutely clear, the reason (2) reduces to $a^2 + b^2 = 1$ is because $\phi_1(x)\phi_1^*(x) = 1$ and $a^*a = a^2$ and the same applies to b. And if I wanted to normalize the $\psi$ I started with I should have $\frac{1}{(a^2+B^2)}(a\phi_1 + b\phi_2)$ right? $\endgroup$ – Jesse Apr 17 '14 at 3:38
  • $\begingroup$ (1) It's not true that $\phi_1(x)\phi_1^*(x) = 1$. The product $\phi_1(x)\phi_1^*(x)$ is a function, which will in general vary with $x$. Only when you integrate it do you get 1. (2) I had made a mistake by omitting the absolute value signs; $a^*a = \lvert a^2\rvert$, not $a^2$, but accounting for that, yes that is the reason you get $\lvert a\rvert^2 + \lvert b\rvert^2 = 1$. (3) Exercise for the reader :-) Try the procedure I showed you on that function and see if it's normalized. $\endgroup$ – David Z Apr 17 '14 at 4:39

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