Let there be a infinite square well where the volume density V(x)V(x) is

V(x)={0,0≤x≤L∞,otherwiseV(x) = \begin{cases} 0, & 0 \le x \le L \\ \infty, & \text{otherwise}\end{cases}

Finding wave function

Inside the well, TISE reduces to a free-particle form

−ℏ22md2ϕdx2=Eϕ  ⟺  d2ϕdx2=−2mEℏ2ϕ(x)-\frac{\hbar^2}{2m}\frac{d^2\phi}{dx^2} = E\phi \iff \frac{d^2\phi}{dx^2} = -\frac{2mE}{\hbar^2}\phi(x)

A general solution to this is (it's a harmonic oscillator)

ϕ(x)=a eikx+b e−ikx=Asin⁡(kx)+Bcos⁡(kx)\phi(x) = a\,e^{ikx} + b\,e^{-ikx} = A\sin(kx) + B\cos(kx)

where

k≜2mE/ℏ2k\triangleq \sqrt{2mE/\hbar^2}

Find BB Let's set the boundary conditions

ϕ(x)=0x≤0,x≥L\phi(x)=0\quad x\leq0,x\geq L

and

ϕ(0)=0⇒B=0\phi(0)=0\Rightarrow B=0

Find knk_n

ϕ(L)=0⇒sin⁡(kL)=0⇒kL=nπn∈Z+\phi(L)=0\Rightarrow \sin(kL)=0\Rightarrow kL=n\pi\quad n\in\mathbb{Z^+}

so

kn=nπLk_n = \frac{n\pi}{L}

Find AA by normalizing

1=An2 L2  ⟹  An=2L1 = A_n^2\,\frac{L}{2} \implies A_n = \sqrt{\frac{2}{L}}

Hence

ϕn(x)=2Lsin⁡ ⁣(nπxL)\boxed{\phi_n(x) = \sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right)}

This wave function means at at nn energies there are n−1n-1 nodes in the sin⁡\sin curve

When the potential is symmetric after recentering box around origin i.e.,

−V(−x)=V(x)-V(-x)=V(x)

The Hamiltonian commutes with parity operator

P^ϕ(x)=ϕ(−x)\hat{P}\phi(x)=\phi(-x)

Then

[H^,P^]=0[\hat{H},\hat{P}]=0

Finding Energy

Recall that we defined kk right above. Rearranging gives

En=ℏ2kn22m=n2π2ℏ22mL2\boxed{ E_n = \frac{\hbar^2 k_n^2}{2m} = \frac{n^2\pi^2\hbar^2}{2mL^2}}

Example

superposition of ground and first excited states

ψ(x,0)=12ϕ1(x)+12ϕ2(x)≡12(∣1⟩+∣2⟩)\psi(x, 0) = \frac{1}{\sqrt{2}}\phi_1(x) + \frac{1}{\sqrt{2}}\phi_2(x) \equiv \frac{1}{\sqrt{2}}\big(\ket{1} + \ket{2}\big)

Substitute

ϕn(x)=2Lsin⁡ ⁣(nπxL)\phi_n(x) = \sqrt{\frac{2}{L}}\sin\!\left(\frac{n\pi x}{L}\right)

to get

ψ(x,0)=1L[sin⁡(k1x)+sin⁡(k2x)]\psi(x, 0) = \frac{1}{\sqrt{L}}\big[\sin(k_1 x) + \sin(k_2 x)\big]

it evolves over time to pick up its own phase

ψ(x,t)=1L[e−iω1tsin⁡(k1x)+e−iω2tsin⁡(k2x)]\psi(x, t) = \frac{1}{\sqrt{L}}\big[e^{-i\omega_1 t}\sin(k_1 x) + e^{-i\omega_2 t}\sin(k_2 x)\big]

where

ωn=En/ℏ\omega_n=E_n/\hbar

Note the probability density is

∣ψ(x,t)∣2=1L[sin⁡2(k1x)+sin⁡2(k2x)+2cos⁡[(ω1−ω2)t]sin⁡(k1x)sin⁡(k2x)]|\psi(x, t)|^2 = \frac{1}{L}\big[\sin^2(k_1 x) + \sin^2(k_2 x) + 2\cos[(\omega_1 - \omega_2)t]\sin(k_1 x)\sin(k_2 x)\big]