Example of Finding Hamiltonian

Let there be a particle in space where energy is kinetic energy plus potential energy.

In classical physics, we know that

KE=12mv2=p22mKE = \frac{1}{2}mv^2 = \frac{p^2}{2m}

where KEKE is kinetic energy, mm is mass, vv is velocity and p=mvp=mv is momentum. We know that potential energy V(x)V(x) is a function of position xx.

Hence in classical physics, energy is

E=KE+V(x)=p22m+V(x)E=KE+V(x)=\frac{p^2}{2m}+V(x)

A wavelength λ\lambda travelling in the direction xx can be written as eikxe^{ikx} where wave number kk is defined as

k≜2πλk\triangleq \frac{2\pi}{\lambda}

de Broglie in 1924 hypothesized that every particle has wave-like behavior with wavelength λ\lambda

λ=hp\lambda = \frac{h}{p} ⇒p=h2π/k=ℏk\Rightarrow p=\frac{h}{2\pi/k}=\hbar k

But we want this as an operator where the plane wave is an eigenstate of p^\hat{p} with eigenvalue ℏk\hbar k (its momentum).

p^eikx=ℏkeikx\hat{p}e^{ikx}=\hbar ke^{ikx}

Now differentiate eikxe^{ikx} w.r.t xx

⇒∂∂xeikx=ikeikx\Rightarrow\frac{\partial}{\partial x}e^{ikx}=ike^{ikx}

Multiply by ℏ/i\hbar/i

⇒ℏi∂∂xeikx=ℏkeikx\Rightarrow\frac{\hbar}{i}\frac{\partial}{\partial x}e^{ikx}=\hbar ke^{ikx}

Comparing like terms we find

p^=ℏi∂∂x\hat{p}=\frac{\hbar}{i}\frac{\partial}{\partial x}

Anyways, after doing all of that, we’re going to guess the Hamiltonian. Guess that the Hamiltonian is H(t)=EH(t)=E

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

Expand the state into basis via Spectral Decomposition

∣ψ(t)⟩=∑k⟨xk∣ψ(t)⟩∣xk⟩=∫dx⟨x∣ψ(t)⟩∣x⟩\left\lvert \psi(t) \right\rangle=\sum_k\left\langle x_k|\psi(t) \right\rangle\left\lvert x_k \right\rangle=\int dx \left\langle x|\psi(t) \right\rangle\left\lvert x \right\rangle

The intergral is the continuum limit of the discretized version on the left. Let’s take the Schrödinger equation and apply it to the state.

⟨x∣iℏddt∣ψ(t)⟩=⟨x∣H(t)∣ψ(t)⟩\left\langle x \right\rvert i\hbar \frac{d}{dt}\left\lvert \psi(t) \right\rangle=\left\langle x \right\rvert H(t)\left\lvert \psi(t) \right\rangle

LHS and RHS are both manipulated to get

iℏ∂∂t⟨x∣ψ(t)⟩=H(t)⟨x∣ψ(t)⟩i\hbar \frac{\partial}{\partial t}\left\langle x|\psi(t) \right\rangle =H(t)\left\langle x|\psi(t) \right\rangle ⇒iℏ∂∂t⟨x∣ψ(t)⟩=⟨x∣p^22m∣ψ(t)⟩+V(x)⟨x∣ψ(t)⟩\Rightarrow i\hbar \frac{\partial}{\partial t}\left\langle x|\psi(t) \right\rangle =\left\langle x \right\rvert\frac{\hat{p}^2}{2m}\left\lvert \psi(t) \right\rangle+V(x)\left\langle x|\psi(t) \right\rangle ⇒iℏ∂∂t⟨x∣ψ(t)⟩=−ℏ22m∂2∂x2⟨x∣ψ(t)⟩+V(x)⟨x∣ψ(t)⟩\Rightarrow i\hbar \frac{\partial}{\partial t}\left\langle x|\psi(t) \right\rangle =-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left\langle x|\psi(t) \right\rangle+V(x)\left\langle x|\psi(t) \right\rangle ⇒iℏ∂∂t⟨x∣ψ(t)⟩=(−ℏ22m∂2∂x2+V(x))⟨x∣ψ(t)⟩\Rightarrow i\hbar \frac{\partial}{\partial t}\left\langle x|\psi(t) \right\rangle =\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)\right)\left\langle x|\psi(t) \right\rangle

If we solve this for specific systems and check against measurements, it will confirm whether or not our guess for H(t)H(t) was right