Time-Independent Schrödinger Equation

This is aka. Schrödinger equation for wave functions.

Depends on Schrödinger equation, Hamiltonian II.

Let's derive the Schrödinger equation for wave functions.

Recall Schrödinger equation where

iℏddt∣ψ(t)⟩=H^∣ψ(t)⟩i\hbar \frac{d}{dt}\ket{\psi(t)} = \hat{H}\ket{\psi(t)}

Formal solution via the time evolution operator is

∣ψ(t)⟩=U^(t)∣ψ(0)⟩,ψ(x,t)=⟨x∣ψ(t)⟩\ket{\psi(t)} = \hat{U}(t)\ket{\psi(0)}, \qquad \psi(x,t) = \braket{x | \psi(t)}

The question is: how do we find U^(t)\hat{U}(t) and ψ(x,t)\psi(x,t) given H^\hat{H}

Method 1

This is already proved in uniform dynamics

U^(t)=e−iH^t/ℏ\hat{U}(t)=e^{-i\hat{H}t/\hbar}

Let's try a few Hamiltonians H^\hat{H} out This is a translation operator by x0x_0

H^=1tp^x0\hat{H}=\frac{1}{t}\hat{p}x_0

This works cleanly

U^(t)=e−ip^x0/ℏ\hat{U}(t)=e^{-i\hat{p}x_0/\hbar}

However what if the Hamiltonian is the full kinetic energy + potential energy?

H^=p^22m+V(x^)\hat{H}=\frac{\hat{p}^2}{2m}+V(\hat{x})

This doesn't work cleanly. Via BCH we get

U^(t)=e−i[p^22m+V(x^)]t/ℏ≈exp⁡[−iH^tℏ+it24mℏ[p^,V′(x^)] + ⋯ ]\hat{U}(t)=e^{-i\left[\frac{\hat{p}^2}{2m} + V(\hat{x})\right]t/\hbar} \approx \exp\left[-\frac{i\hat{H}t}{\hbar} + \frac{it^2}{4m\hbar}[\hat{p},V'(\hat{x})] ~+ ~\cdots\right]

This is a numerical approximation, not an analytical solution.

Method 2

This is the reverse of Schrödinger equation, Hamiltonian I. There, we postulated that particle energies are quantified and derived the Schrödinger equation from that. Here, we postulate the Schrödinger equation and derive that particle energies by applying it.

Given

First we must diagonalize H^\hat{H}. We should find the eigenstates of H^\hat{H} Note ϕn(x)\phi_n(x) is shortened to ϕn\phi_n but basically ϕ(x)\phi(x) is an eigenfunction representing definite energy EE -- it is a wave function.

H^∣ϕn⟩=En∣ϕn⟩\hat{H}\ket{\phi_n} = E_n\ket{\phi_n}

So

⇒U^(t)∣ϕn⟩=e−iH^t/ℏ∣ϕn⟩\Rightarrow\quad\hat{U}(t)\ket{\phi_n} = e^{-i\hat{H}t/\hbar}\ket{\phi_n} =e−iEnt/ℏ∣ϕn⟩ = e^{-iE_n t/\hbar}\ket{\phi_n}

If we decompose this as we know any state ∣ψ(0)⟩\ket{\psi(0)} is a superposition. Note that cnc_n are probability amplitudes.

∣ψ(0)⟩=∑ncn(0)∣ϕn⟩\ket{\psi(0)}=\sum_n c_n(0)\ket{\phi_n}

By linearity, U^(t)\hat{U}(t) acts for all terms in ∣ψ(t)⟩=U^(t)∣ψ(0)⟩\ket{\psi(t)}=\hat{U}(t)\ket{\psi(0)}

∣ψ(0)⟩=∑ncn(0)∣ϕn⟩\ket{\psi(0)}=\sum_nc_n(0) \ket{\phi_n} ∣ψ(t)⟩=∑ncn(t) e−iEnt/ℏ∣ϕn⟩\boxed{\ket{\psi(t)} = \sum_n c_n(t)\,e^{-iE_n t/\hbar}\ket{\phi_n}}

where cn(t)c_n(t) is how much of ∣ϕn⟩\ket{\phi_n} position is in the state ∣ψ(t)⟩\ket{\psi(t)} at time tt

and

TISE

To get TISE from method 2 we start from

H^∣ϕn⟩=En∣ϕn⟩\hat{H}|\phi_n\rangle = E_n|\phi_n\rangle

Project onto position basis ∣x⟩\ket{x}

⟨x∣H^∣ϕn⟩=En⟨x∣ϕn⟩\langle x|\hat{H}|\phi_n\rangle = E_n\langle x|\phi_n\rangle

Note we just said above that

H^=p^22m+V(x^),⟨x∣p^∣ϕ⟩=−iℏdϕdx\hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x}), \qquad \langle x|\hat{p}|\phi\rangle = -i\hbar\frac{d\phi}{dx}

so we arrive at the Time-Independent Schrödinger Equation (TISE)

[−ℏ22md2dx2+V(x)]ϕ(x)=E ϕ(x)\boxed{\left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\right]\phi(x) = E\,\phi(x)}

where

(U^(t)ϕ)(x)=e−iEt/ℏϕ(x)(\hat{U}(t)\phi)(x)=e^{-iEt/\hbar}\phi(x)   ⟹  ⟨x∣U^(t)∣ϕn⟩=e−iEnt/ℏ⟨x∣ϕn⟩\implies\langle x|\hat{U}(t)|\phi_n\rangle = e^{-iE_n t/\hbar}\langle x|\phi_n\rangle   ⟹  U^(t)∣ϕn⟩=e−iEnt/ℏ∣ϕn⟩\implies\hat{U}(t)|\phi_n\rangle = e^{-iE_n t/\hbar}|\phi_n\rangle

Note that ϕ(x)\phi(x) are stationary states. so

  ⟹  U^(t)=∑ne−iEnt/ℏ ∣ϕn⟩⟨ϕn∣\implies\boxed{\hat{U}(t) = \sum_n e^{-iE_n t/\hbar}\,|\phi_n\rangle\langle\phi_n|}

This is sort of similar to separation of variables for PDEs.

Energy

if we solve this out then

d2ϕ(x)dx2=−k2ϕ(x)\frac{d^2\phi(x)}{dx^2}=-k^2\phi(x)

where

k2≜2mEℏ2k^2 \triangleq \frac{2mE}{\hbar^2}