Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 4 / b

Table of contents
- Solution b

Solution

The density operator is diagonal in the oscillator number basis, so

⟨ a^{2} ⟩ = ⟨ a^{† 2} ⟩ = 0

. With

\overset{n}{ˉ} = ⟨ a^{†} a ⟩

and

a a^{†} = a^{†} a + 1

,

⟨ x^{2} ⟩ = \frac{ℏ}{2 mω} (2 \overset{n}{ˉ} + 1), ⟨ p^{2} ⟩ = \frac{m ℏ ω}{2} (2 \overset{n}{ˉ} + 1) .

Hence

⟨ \frac{p ^{2}}{2 m} ⟩ = ⟨ \frac{m ω ^{2} x ^{2}}{2} ⟩ = \frac{ℏ ω}{4} (2 \overset{n}{ˉ} + 1) .

Equality holds even at zero temperature, when each contribution is

ℏ ω /4

. Classical equipartition alone would miss this zero-point energy.

 Ancestors (14)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version