QM with one continuous degree of freedom

Consider the Hilbert space $𝐿 2 (ℝ)$ with vectors represented in the position basis $Ψ (𝑥, 𝑡) = ⟨ 𝑥 | 𝜓 (𝑡) ⟩$ . The momentum operator is given by

ˆ 𝑝 = - 𝑖 ℏ 𝜕 𝜕 𝑥

and thus the Hamiltonian operator by

ˆ 𝐻 (𝑡) = - ℏ 2 𝑚 𝜕 2 𝜕 𝑥 2 + 𝑉 (𝑥, 𝑡)

and the Schrödinger equation becomes

𝑖 ℏ 𝜕 𝜕 𝑡 Ψ (𝑥, 𝑡) = - ℏ 2 𝑚 𝜕 2 𝜕 𝑥 2 + 𝑉 (𝑥, 𝑡)

Time independent Schrödinger equation

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

Ψ 𝑛 (𝑥, 𝑡) = 𝜓 𝑛 (𝑥) 𝑒 - 𝑖 𝐸 𝑛 𝑡 / ℏ 𝐸 𝑛 𝜓 = - ℏ 2 𝑚 𝜕 2 𝜕 𝑥 2 𝜓 + 𝑉 (𝑥) 𝜓

and thus general solutions are given by¹

Ψ (𝐫, 𝑡) = \sum 𝑐 𝑛 𝜓 𝑛 (𝑥) 𝑒 - 𝑖 𝐸 𝑛 𝑡 / ℏ

Properties of solutions

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

𝐸 𝜓 (𝑥) = ˆ 𝐻 𝜓 (𝑥) = - ℏ 2 𝑚 𝜕 2 𝜕 𝑥 2 𝜓 (𝑥) + 𝑉 (𝑥) 𝜓 (𝑥)

it follows immediately that

𝐸 𝜓 (- 𝑥) = - ℏ 2 𝑚 𝜕 2 𝜕 𝑥 2 𝜓 (- 𝑥) + 𝑉 (- 𝑥) 𝜓 (- 𝑥) = - ℏ 2 𝑚 𝜕 2 𝜕 𝑥 2 𝜓 (- 𝑥) + 𝑉 (𝑥) 𝜓 (- 𝑥) = ˆ 𝐻 𝜓 (- 𝑥)

Let $𝜓 \pm (𝑥) = 𝜓 (𝑥) \pm 𝜓 (- 𝑥)$ , so

𝜓 \pm (- 𝑥) = 𝜓 (- 𝑥) \pm 𝜓 (𝑥) = \pm (𝜓 (𝑥) \pm 𝜓 (- 𝑥)) = \pm 𝜓 \pm (𝑥)

hence $𝜓 +$ is even and $𝜓 -$ is odd. Then $ˆ 𝐻 𝜓 \pm = 𝐸 𝜓 \pm$ by linearity, and $𝜓 = 12 𝜓 + + 12 𝜓 -$ , so any solution is a linear combination of even and odd eigenstates. Hence $𝜓 \pm$ may be chosen as eigenstates. Note in cases of energy degeneracy there is always a choice.

General properties

The canonical commutation relations is

[ˆ 𝑥, ˆ 𝑝] = 𝑖 ℏ

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

Particular potentials

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

QM with one continuous degree of freedom

Time independent Schrödinger equation

Properties of solutions

General properties

Particular potentials

Footnotes