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Finding the Wave Function Eigenfunctions and Energy Eigenvalues in an Infinite Potential Well 📂Quantum Mechanics

Finding the Wave Function Eigenfunctions and Energy Eigenvalues in an Infinite Potential Well

Proposition

When the potential takes the form of an infinite well over the interval [0,a][0, a], the energy (eigenvalue) EnE_{n} and the wave function (eigenstate) ψn\psi_{n} of the wave function are as follows.

En=n2π222ma2ψn(x)=2asin(nπax)n=0,1,2,(0) \begin{align*} E_{n} &=\frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}} \\[1em] \psi_{n}{(x)} &= \textstyle \sqrt{\frac{2}{a}}\sin \left( \frac{n\pi}{a}x \right) \end{align*} \qquad\qquad n = 0, 1, 2, \dots \tag{0}

Explanation

V(x)={,<x<00,0<x<a,a<x< V(x) = \begin{cases} \infty, & -\infty \lt x \lt 0 \\ 0, & 0 \lt x \lt a \\ \infty, & a \lt x \lt \infty \end{cases}

A potential of the form shown above UU is called an infinite potential well. This system models a situation where a particle cannot leave a certain interval, also known as the particle in a box model. It is a very simple model but serves as an important example where it demonstrates results significantly different from those in classical mechanics. In classical mechanics, the position where a particle may be found is uniform within the interval, whereas in quantum mechanics, the probability of finding the particle varies with its position.

The state with the lowest energy level (=n=n) among the wave functions is known as the ground state. Any state that is not the ground state is referred to as an excited state.

ground state: ψ1(x)=2asin(πax)first excited state: ψ2(x)=2asin(2πax)second excited state: ψ3(x)=2asin(3πax) \begin{align*} \text{ground state: } & \textstyle \psi_{1}(x)=\sqrt{\frac{2}{a}}\sin\left( \frac{\pi}{a}x \right) \\[1em] \text{first excited state: } & \textstyle \psi_2(x)=\sqrt{\frac{2}{a}}\sin\left( \frac{2\pi}{a}x \right) \\[1em] \text{second excited state: } & \textstyle \psi_{3}(x)=\sqrt{\frac{2}{a}}\sin\left( \frac{3\pi}{a}x \right) \\ \vdots \end{align*}

The wave function cannot have arbitrary energy, but only energies of the form depicted in equation (0)(0). Energy is proportional to n2n^{2}, and it appears in discrete rather than continuous values. This is called quantization.

Derivation

The time-independent 1-dimensional Schrödinger equation is given as follows.

(22md2dx2+V(x))ψ(x)=Eψ(x) \left(-\frac{\hbar^{2}}{2m}\frac{ d ^{2} }{ d x^{2} }+V(x)\right)\psi(x) = E\psi(x)

V(x)={,<x<00,0<x<a,a<x< V(x) = \begin{cases} \infty, & -\infty \lt x \lt 0 \\ 0, & 0 \lt x \lt a \\ \infty, & a \lt x \lt \infty \end{cases}

E<0E \lt 0

Since the potential is always greater than or equal to 00, a wave function does not exist when the energy is negative.

E>0E \gt 0

[1] For x>a|x| \gt a

In this interval, since 0<E<V=0 \lt E \lt V = \infty, it follows that ψ(x)=0\psi(x) = 0.

[2] For 0<x<a0 \lt x \lt a

In the interval [0,a][0, a], the potential is V=0V = 0, and the Schrödinger equation is given by:

d2ψdx2+2m2Eψ=0 \frac{d^{2} \psi}{dx^{2}} + \frac{2m}{\hbar^{2}}E\psi=0

Since EE is positive, 2m2E\dfrac{2m}{\hbar^{2}}E is also positive. Let us define this as k2k^{2}. Then the Schrödinger equation becomes:

d2ψdx2+k2ψ=0 \frac{d^{2} \psi}{dx^{2}} + k^{2}\psi = 0

The solution to this differential equation is as follows.

ψ(x)=Asinkx+Bcoskx \psi(x) = A\sin kx + B\cos kx

[3] Boundary Conditions

Given that the wave function ψ\psi is continuous, the values of the functions obtained in [1] and [2] must match at the boundary. Thus, when x=0x = 0, the following must hold:

0=ψ(0)=Asin0+Bcos0=B    B=0 0 = \psi (0) = A\sin 0 + B\cos 0 = B \implies B=0

The boundary condition must also hold when x=ax = a, therefore:

0=ψ(a)=Asinka(B=0)(1) 0 = \psi (a) = A\sin ka \qquad (\because B = 0) \tag{1}

The sine function is zero for integer nn, meaning that solutions exist for all integer nn, leading to:

Asinka=0    ka=nπ    k=nπa(2) A \sin ka = 0 \implies ka = n\pi \implies k = \frac{n\pi}{a} \tag{2}

    ψn(x)=Asin(nπax) \implies \psi_{n}(x) = \textstyle A\sin \left( \frac{n\pi}{a}x \right)

The wave function must be square integrable such that 11.

ψnψndx=1=0aA2sin2nπaxdx=A20a12(1cos2nπax)dx=A212[xa2nπsin2nπax]0a=A2a2 \begin{align*} \int\limits_{-\infty}^{\infty} \psi_{n}^{\ast} \psi_{n} dx = 1 &= \int_{0}^{a}\textstyle |A|^{2} \sin^{2} \frac{n\pi}{a}x dx \\ &= |A|^{2} \int_{0}^{a}\textstyle \frac{1}{2}(1-\cos \frac{2n\pi}{a}x)dx \\ &= |A|^{2}\textstyle \frac{1}{2} \left[x-\frac{a}{2n\pi}\sin \frac{2n\pi}{a}x\right]_{0}^a \\ &= |A|^{2}\textstyle \frac{a}{2} \end{align*}

    A2=2a    A=2a(3) \implies |A|^{2} = \frac{2}{a} \implies A = \sqrt{\frac{2}{a}} \tag{3}

In the third equality, the half-angle formula was used. Therefore, the wave function is as follows.

ψn(x)=2asin(nπax) \psi_{n}(x)= \textstyle \sqrt{\frac{2}{a}}\sin \left( \frac{n\pi}{a}x \right)

Also, given that 2m2E=k2\dfrac{2m}{\hbar^{2}}E=k^{2} was defined earlier, the energy corresponding to the wave function ψn\psi_{n} is given by:

En=2k22m=n2π222ma2 E_{n} = \frac{\hbar^{2}k^{2}}{2m} = \frac{n^{2}\pi^{2}\hbar^{2}}{2ma^{2}}