What are the coefficients of psi_n for this state?

[SamRoss]

Homework Statement

A particle in the infinite square well has its initial wave function an even mixture of the first two stationary states:
Ψ(x,0) = A[$\psi_1 (x) + \psi_2 (x)$]
If you measured the energy of this particle, what values might you get, and what is the probability of getting each of them?

Relevant Equations

Ψ(x,0) = A[$\psi_1 (x) + \psi_2 (x)$]
$\sum_{n=1}^\infty |c_n|^2=1$

I am working through David Griffiths' "Introduction to Quantum Mechanics". All of the solutions are provided online by Griffiths himself. This is Problem 2.5(e). I understand his solution but I'm confused about one thing. After normalizing Ψ, we find $A=\frac {1}{\sqrt2}$. Griffiths notes that the probability of obtaining each of the energies associated with each $\psi$ is 1/2 which makes sense. What I'm confused about is what the coefficients $c_n$ in front of the $\psi_n$ are (every wave function Ψ is supposed to be a linear combination of the $\psi_n$). If we distribute $A=\frac {1}{\sqrt2}$ in Ψ(x,0) = A[$\psi_1 (x) + \psi_2 (x)$] then it seems like the coefficients $c_n$ are each just $\frac {1}{\sqrt2}$ which would be nice because then $\sum_{n=1}^\infty |c_n|^2$ (all $c_n$ being equal to 0 for n>2) would equal 1 which it should. But aren't the $c_n$ supposed to be different from A? This leads me to look at Ψ(x,0) = A[$\psi_1 (x) + \psi_2 (x)$] again and conclude that each coefficient is simply 1, but this wouldn't work because then $\sum_{n=1}^\infty |c_n|^2$ would be equal to 2. So what are the $c_n$?

 

[vela]

But aren't the $c_n$ supposed to be different from A?

Nope. Your first take was correct.

Note that the equation $\sum_n \lvert c_n \rvert^2 = 1$ is another way of saying that the wave function is normalized.

 

[DrClaude]

The $\psi_n(x)$ form a complete basis for the particle in an infinite square well, such that any wave function can be written as
$$
\Psi(x) = \sum_{n=1}^\infty c_n \psi_n(x)
$$
You have $\Psi(x) = A \left[ \psi_1(x) + \psi_2(x) \right]$, so you get immediately that $c_1 = A$, $c_2 = A$, and all other $c_n = 0$.

 

[SamRoss]

#vela #DrClaude

Thanks for the responses. $c_n = \frac {1} {\sqrt2}$ seems reasonable. I'm still confused about the general method for finding A and $c_n$, though. A few pages back, Griffiths introduced the potential energy function for the infinite square well, V(x) = 0 if x is between 0 and a and $\infty$ otherwise. He then solved the time-independent Schrödinger equation and found that the stationary state $\psi_n = A\sin(\frac {n \pi} {a}x)$ with $A = \sqrt{\frac {2} {a}}$ due to normalization. Later, he says the general solution is $$\Psi(x,t) = \sum_{n=1}^\infty c_n \psi_n (x) e^{\frac {-i E_n t} {\hbar}} = \sum_{n=1}^\infty c_n \sqrt{\frac {2} {a}} \sin(\frac {n \pi} {a}x) e^{\frac {-i E_n t} {\hbar}} $$

So here's where I'm getting tripped up in my mind...
1. The general equation above places $c_n$ next to $\sqrt{\frac {2} {a}}=A$ making it seem like the two are completely separate entities which is why I was confused earlier (and I suppose I still am) when the problem made it seem like we could simply take A as our $c_n$.
2. Something I did not bring up in my earlier post - if Griffiths already found that A is $\sqrt{\frac {2} {a}}$ for a particle in an infinite square well then how is it that A = $\frac {1} {\sqrt{2}}$ for the problem in my original post which was also for an infinite square well? I understand the math behind normalizing $\Psi(x,0) = A[\psi_1(x)+\psi_2(x)]$ in order to obtain A = $\frac {1} {\sqrt{2}}$, I'm just confused regarding the seeming inconsistency in having two different A's.

 

[TSny]

When Griffiths uses $A$ in $\Psi(x) = A[\psi_1(x) + \psi_2(x)]$, it is not meant to denote the same constant, $A$, appearing in $\psi_n(x) = A \sin \left ( \frac{n \pi}{a}x \right)$. In each case, $A$ just denotes a normalization constant for that particular state.

 

[vela]

There are two normalizations going on here: normalizing the basis vectors and normalizing the linear combination of basis vectors. The factor of $\sqrt{2/a}$ for the infinite square well is to normalize each of the eigenstates. The $A$ in this problem is to normalize the linear combination of $\psi_1$ and $\psi_2$.

 

[PeroK]

#vela #DrClaude

Thanks for the responses. $c_n = \frac {1} {\sqrt2}$ seems reasonable. I'm still confused about the general method for finding A and $c_n$, though. A few pages back, Griffiths introduced the potential energy function for the infinite square well, V(x) = 0 if x is between 0 and a and $\infty$ otherwise. He then solved the time-independent Schrödinger equation and found that the stationary state $\psi_n = A\sin(\frac {n \pi} {a}x)$ with $A = \sqrt{\frac {2} {a}}$ due to normalization. Later, he says the general solution is $$\Psi(x,t) = \sum_{n=1}^\infty c_n \psi_n (x) e^{\frac {-i E_n t} {\hbar}} = \sum_{n=1}^\infty c_n \sqrt{\frac {2} {a}} \sin(\frac {n \pi} {a}x) e^{\frac {-i E_n t} {\hbar}} $$

So here's where I'm getting tripped up in my mind...
1. The general equation above places $c_n$ next to $\sqrt{\frac {2} {a}}=A$ making it seem like the two are completely separate entities which is why I was confused earlier (and I suppose I still am) when the problem made it seem like we could simply take A as our $c_n$.
2. Something I did not bring up in my earlier post - if Griffiths already found that A is $\sqrt{\frac {2} {a}}$ for a particle in an infinite square well then how is it that A = $\frac {1} {\sqrt{2}}$ for the problem in my original post which was also for an infinite square well? I understand the math behind normalizing $\Psi(x,0) = A[\psi_1(x)+\psi_2(x)]$ in order to obtain A = $\frac {1} {\sqrt{2}}$, I'm just confused regarding the seeming inconsistency in having two different A's.

This is essentially the same situation you have in linear algebra. Let's take 2D vectors. You can make an orthonormal basis from the vectors $(1,1)$ and $(1, -1)$ by normalising them:
$$e_1 = \frac{1}{\sqrt 2}(1, 1), \ e_2 = \frac{1}{\sqrt 2}(1, -1)$$
Now we can express any vector $v$ as a linear combination of $e_1, e_2$. E.g.:
$$v = 3e_1 + 4e_2$$
But, if we want to normalise $v$ we must have:
$$\hat v = \frac 1 5 (3e_1 + 4e_2)$$
In your example, the function $\sin(\frac {n \pi} {a}x)$ is like an un-normalised basis vector like $(1, 1)$. The normalised function $\sqrt{\frac 2 a}\sin(\frac {n \pi} {a}x)$ is a normalised basis vector like $e_1$. And a normalised solution to the wave equation is like the vector $\hat v$.

For example, we could have:
$$\psi = \frac 1 5(3\psi_1 + 4\psi_2) = \frac 1 5 (3\sqrt{\frac 2 a}\sin(\frac {\pi} {a}x) + 4\sqrt{\frac 2 a}\sin(\frac {\pi} {a}x))$$
Which can also be written:
$$\psi = \frac 3 5 \psi_1 + \frac 4 5 \psi_2 = \frac 3 5 \sqrt{\frac 2 a}\sin(\frac {\pi} {a}x) + \frac 4 5 \sqrt{\frac 2 a}\sin(\frac {2\pi} {a}x)$$

 

[SamRoss]

#vela #DrClaude #TSny #PeroK

Okay, I think I got it. There are two A constants - one was for $\Psi$ and one was for $\psi$. Thanks for your help, everybody!