Quantum mechanics MOC

QM with one continuous degree of freedom

Consider the Hilbert space 𝐿2(ℝ) with vectors represented in the position basis Ψ(𝑥,𝑡) =⟨𝑥|𝜓(𝑡)⟩. The momentum operator is given by

ˆ𝑝=−𝑖ℏ𝜕𝜕𝑥

and thus the Hamiltonian operator by

ˆ𝐻(𝑡)=−ℏ2𝑚𝜕2𝜕𝑥2+𝑉(𝑥,𝑡)

and the Schrödinger equation becomes

𝑖ℏ𝜕𝜕𝑡Ψ(𝑥,𝑡)=−ℏ2𝑚𝜕2𝜕𝑥2+𝑉(𝑥,𝑡)

Time independent Schrödinger equation

If 𝑉 is time-independent the stationary states are given by the time-independent Schrödinger equation

Ψ𝑛(𝑥,𝑡)=𝜓𝑛(𝑥)𝑒−𝑖𝐸𝑛𝑡/ℏ𝐸𝑛𝜓=−ℏ2𝑚𝜕2𝜕𝑥2𝜓+𝑉(𝑥)𝜓

and thus general solutions are given by¹

Ψ(𝐫,𝑡)=∑𝑐𝑛𝜓𝑛(𝑥)𝑒−𝑖𝐸𝑛𝑡/ℏ

Properties of solutions

1. If 𝑉(𝑥) is an even function then 𝜓(𝑥) is either odd or even

Proof of 1

Let 𝑉(𝑥) be even, i.e. 𝑉(𝑥) =𝑉( −𝑥) Let 𝜓 be a solution to the TISE so that

𝐸𝜓(𝑥)=ˆ𝐻𝜓(𝑥)=−ℏ2𝑚𝜕2𝜕𝑥2𝜓(𝑥)+𝑉(𝑥)𝜓(𝑥)

it follows immediately that

𝐸𝜓(−𝑥)=−ℏ2𝑚𝜕2𝜕𝑥2𝜓(−𝑥)+𝑉(−𝑥)𝜓(−𝑥)=−ℏ2𝑚𝜕2𝜕𝑥2𝜓(−𝑥)+𝑉(𝑥)𝜓(−𝑥)=ˆ𝐻𝜓(−𝑥)

Let 𝜓±(𝑥) =𝜓(𝑥) ±𝜓( −𝑥), so

𝜓±(−𝑥)=𝜓(−𝑥)±𝜓(𝑥)=±(𝜓(𝑥)±𝜓(−𝑥))=±𝜓±(𝑥)

hence 𝜓+ is even and 𝜓− is odd. Then ˆ𝐻𝜓± =𝐸𝜓± by linearity, and 𝜓 =12𝜓+ +12𝜓−, so any solution is a linear combination of even and odd eigenstates. Hence 𝜓± may be chosen as eigenstates. Note in cases of energy degeneracy there is always a choice.

General properties

1. The canonical commutation relations is

[ˆ𝑥,ˆ𝑝]=𝑖ℏ

2. The energy of a normalizable solution must exceed the the infimum of the potential.

Particular potentials

• QM of a particle in a 1D infinite square well
• QM of a particle in a 1D harmonic oscillator
• QM of a free particle in 1D
• QM of a particle in a 1D Dirac delta potential
• QM of particle in a 1D finite square well

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develop | en | SemBr

Footnotes

1. 2018. Introduction to Quantum Mechanics, §2.1, p. 26 ↩