Wave Function for a Particle in a Symmetric Infinite Square Well

[MatinSAR]

Homework Statement

Consider a particle of mass $m$ confined in a one-dimensional infinite square well potential centered at $x=0$, defined as:$$ V(x) = \begin{cases}
0 & \text{for } -\frac{a}{2} \leq x \leq \frac{a}{2}, \\
\infty & \text{otherwise.}
\end{cases}$$

Relevant Equations

Schrödinger equation.

We want to solve the Schrödinger equation in the region $-a/2 \leq x \leq a/2$ where $V(x) = 0$.

The equation is:$$\frac{d^2\psi(x)}{dx^2} + k^2\psi(x) = 0 \quad \text{where} \quad k^2 = \frac{2mE}{\hbar^2}$$ The general solution is:$$\psi(x) = A\sin(kx) + B\cos(kx)$$Applying Boundary Conditions
We know $\psi(-a/2) = \psi(a/2) = 0$, which gives:$$\begin{cases}
-A\sin(ka/2) + B\cos(ka/2) = 0 \\
A\sin(ka/2) + B\cos(ka/2) = 0
\end{cases} $$
$\textbf{Case 1:}$ If $B = 0$ (sine solutions)
The equations reduce to $A\sin(ka/2) = 0$
Non-trivial solution requires $\sin(ka/2) = 0$
This gives $k = \frac{2n\pi}{a}$for $n=1,2,3,...$
Wave function: $\psi(x) = A\sin\left(\frac{2n\pi x}{a}\right)$

$\textbf{Case 2:}$ If $\cos(ka/2) = 0$ (cosine solutions)
Then $ka/2 = \frac{n\pi}{2}$ for odd $n=1,3,5,...$
Wave function has both sine and cosine functions ...

Where I'm Stuck:
My book gives different solutions than what I derived here.
For Case 2, It is has only cosine function.

Would appreciate any insight into this discrepancy.

 

[vela]

Look at what you wrote for the first line of case 1. What would the analogous line be for case 2?

 

[MatinSAR]

Look at what you wrote for the first line of case 1. What would the analogous line be for case 2?

I see.
I just added both sides of the equations derived from boundary conditions to get B=0. If I subtract the two sides, the result should be A=0. Just one question: Why do the coefficients become zero? Why don't the sine and cosine terms vanish instead? Because then we have no wave?

 

[kuruman]

Perhaps it is simpler to consider that the probability density $\psi^*(x)\psi(x)$ must also vanish at the boundaries. In its general form, $$\psi^*(x)\psi(x) = A^2\sin^2(kx)+B^2\cos^2(kx)+2AB\sin(kx)\cos(kx).$$If $\sin^2(ka)=0$, implying that $\cos^2(ka)=1$, it follows that $~\psi^*(a)\psi(a) = B^2$ which vanishes only if $B=0$. Similarly, if the cosine term vanishes at the boundary, it follows that $A$ must be set equal to zero.

This says that, with the symmetric potential about 0, there can be even eigenstates (cosines) or odd eigenstates (sines) but not both. In summary, $$
\psi(x) =
\begin{cases}
\cos\left[(2n+1)\dfrac{\pi x}{2a}\right] &\text{(even)} \\
\\
\sin\left[(2n)\dfrac{\pi x}{2a}\right] &\text{(odd)}
\end{cases}$$ where $n=0,1,2,\dots~$ Note that even/odd values of $n$ correspond to even/odd eigenstates ($0$ is even).

Of course the trivial case $\psi(x)=0$ is also a solution of the Schrodinger equation but that means that there is no particle which makes it much less interesting.

 

[MatinSAR]

there can be even eigenstates (cosines) or odd eigenstates (sines) but not bot

, $$
\psi(x) =
\begin{cases}
\cos\left[(2n+1)\dfrac{\pi x}{2a}\right] &\text{(even)} \\
\\
\sin\left[(2n)\dfrac{\pi x}{2a}\right] &\text{(odd)}
\end{cases}$$

Thank you for your response, @kuruman.

I’ve noticed a concern: In every book I’ve encountered, eigenfunctions are typically discussed as having a definite parity (either odd or even). However, the approach seems to shift when addressing components rather than eigenstates, as seen in your explanation above. Could you clarify why this distinction is made?
I mean why components have known parity instead of eigenstates?

 

[kuruman]

Thank you for your response, @kuruman.

I’ve noticed a concern: In every book I’ve encountered, eigenfunctions are typically discussed as having a definite parity (either odd or even). However, the approach seems to shift when addressing components rather than eigenstates, as seen in your explanation above. Could you clarify why this distinction is made?
I mean why components have known parity instead of eigenstates?

I am not sure what you mean by "components". The general solution of the Schrodinger equation for a particle in a box is $$\psi(x) = C_1e^{ikx} + C_2e^{-ikx}$$ where $C_1$ and $C_2$ are generally complex. Having written this, one applies the boundary conditions and finds specific values for the constant coefficients.

I am not sure that I have answered your question because I don't really understand it.

 

[MatinSAR]

I am not sure what you mean by "components". The general solution of the Schrodinger equation for a particle in a box is $$\psi(x) = C_1e^{ikx} + C_2e^{-ikx}$$ where $C_1$ and $C_2$ are generally complex. Having written this, one applies the boundary conditions and finds specific values for the constant coefficients.

I am not sure that I have answered your question because I don't really understand it.

Apologies for the confusion earlier. By components, I was referring to $\psi (x)$ in your first post. There exists a general ket that contains all the system's information, and to extract position-related details, we expand this ket in the position basis: $$\psi (r)=⟨𝑟∣𝜓⟩$$, which represents the components.

Known parity is a property of eigenvectors($∣-r⟩=\pm ∣r⟩$), yet it is commonly applied to components by yourself and others in similar discussions.

I'm coming to the conclusion that Prof. Zettili could have written his otherwise good book much better if he hadn't skimmed over some things so quickly. After rereading it, I see that he applied this to components too... Completely confusing.

 

[vela]

I just added both sides of the equations derived from boundary conditions to get B=0. If I subtract the two sides, the result should be A=0. Just one question: Why do the coefficients become zero? Why don't the sine and cosine terms vanish instead? Because then we have no wave?

Zetteli likely established earlier that if the potential is even, then the solutions will be even or odd. That allows you to conclude either $A=0$ or $B=0$.

You can also look at it this way which doesn't rely on parity. You have a system of equations
$$
\begin{bmatrix}
-\sin \tfrac {ka}2 & \cos \tfrac{ka}2 \\
\sin \tfrac {ka}2 & \cos \tfrac{ka}2
\end{bmatrix} \begin{bmatrix} A \\ B \end{bmatrix} = 0.$$ In general, the only solution is the trivial solution $A=B=0$ unless the determinant of the matrix $-\sin ka$ vanishes. This gives you $ka=n\pi$. When $n$ is odd, the second column vanishes, and you get $A=0$. When $n$ is even, the first column vanishes, and you get $B=0$.