Quantum Harmonic Oscillator

One Dimensional Case

$m$ is moving in a 1D potential from a force $F = - κ x$ , $V (x) = \frac{1}{2} κ x^{2} = \frac{1}{2} m ω^{2} x^{2}$ , $ω = \sqrt{κ / m}$ .

Then, $\hat{H} Ψ = - \frac{ℏ^{2}}{2 m} Ψ^{″} + \frac{1}{2} m ω^{2} X^{2} Ψ = E Ψ$ . $- \frac{1}{2} α \frac{d^{2}}{d q^{2}} + \frac{1}{2} β q^{2} = E$ .

Dimensionless Formulation

$α = \sqrt{\frac{m ω}{ℏ}}, ξ = α x, λ = \frac{2 E}{ℏ ω}$ . $- \frac{d^{2}}{d ξ^{2}} Ψ + (λ - ξ^{2}) Ψ = 0$ .

For asymptotic behavior, $ξ ≫ λ$ , $(\frac{d}{d ξ^{2}} - ξ^{2}) Ψ = 0$ , $Ψ = ξ^{P} \exp (\pm ξ^{2} / 2)$ . To have our expected behavior, $Ψ = ξ^{P} \exp (- ξ^{2} / 2)$ .

So as an Ansatz, $Ψ (ξ) = \exp (- ξ^{2} / 2) H (ξ)$ . $\frac{d^{2}}{d ξ^{2}} H (ξ) - 2 ξ \frac{d}{d ξ} H + (λ - 1) H = 0$ .

Since the exponential is always even, we can construct even and odd $H (ξ)$ .

Series solution of this differential equation. $H_{e} (ξ) = \sum_{k}^{\infty} c_{k} ξ^{2 k}$ and $H_{o} (ξ) = \sum_{k}^{\infty} d_{k} ξ^{2 k + 1}$ .

Even Solution

For $H_{e}$ , $\sum_{k} c_{k} ((2 k) (2 k - 1) ξ^{2 k - 2} + (λ - 1 - 4 k) ξ^{2 k})$ Shifting by 1, $\sum_{k} (c_{k + 1} 2 (k + 1) (2 k + 1) + c_{k} (λ - 1 - 4 k)) ξ^{2 k} = 0$ , $c_{k + 1} = \frac{4 k + 1 - λ}{2 (k + 1) (2 k + 1)} c_{k}$ .

$\frac{c_{k + 1}}{c_{k}} = \frac{4 k + 1 - λ}{2 (k + 1) (2 k + 1)}$ . For large $k$ , $\sim \frac{1}{k}$ .

Examining, $\exp^{ξ^{2}}$ , $\sum_{n} ξ^{2 n} n!$ , $\frac{c_{n + 1}}{c_{n}} = \frac{n!}{(n + 1)!} \sim \frac{1}{n}$ . Thus, we need to introduce a cutoff. So, we need to make a $N$ such that $4 N + 1 - λ = 0$ . Then $c_{N} \neq 0$ but all future terms are zero.

Then $λ = 4 N + 1$ and $H (ξ) = \sum_{k}^{N} c_{k} ξ^{2 k}$ .

Odd Solution

$λ = 4 N + 3$ .

Combining

$λ = 2 n + 1, n \in N$ . So, $E_{n} = ℏ ω (2 n + 1) / 2$ . $Ψ_{n} (ξ) = N_{n} \exp (- ξ^{2} / 2) H_{n} (ξ)$ .

Hermite Polynomials

$H_{n}^{″} (ξ) - 2 ξ H_{n}^{'} (ξ) + 2 n H_{n} (ξ) = 0$ - derived from the ansatz in the Hamiltonian and removing the terms with the exponential. $H_{n} (ξ) = (- 1)^{n} \exp (ξ^{2}) \frac{d^{n}}{d ξ^{n}} \exp (- ξ^{2})$

$G (ξ, s) = \exp (- s^{2} + 2 s ξ) = \sum_{n} H_{n} (ξ) s^{n} / n!$ gives the power series of the polynomial. So, $H_{n} (ξ) = \frac{\partial^{n}}{\partial s^{n}} G (ξ, s)$ .

The recurrence relation is then, $2 n H_{n - 1} (ξ) = H_{n}^{'} (ξ)$

Finding normalization: $\int_{R} d x | Ψ_{n} (x) |^{2} = \int d ξ \exp (- ξ^{2}) G (ξ, s) G (ξ, t)$ .

Number Operator

$\hat{a} = \sqrt{\frac{m ω}{2 ℏ}} (X + \frac{i}{m ω} P)$ , annihilation operator
${\hat{a}}^{†} = \sqrt{\frac{m ω}{2 ℏ}} (X - \frac{i}{m ω} P)$ , creation operator
$N = {\hat{a}}^{†} \hat{a}, [\hat{a}, {\hat{a}}^{†}] = I$
$H = ℏ ω (2 N + 1) / 2$
$N | n ⟩ = n | n ⟩ \Rightarrow H | n ⟩ = ℏ ω (2 n + 1) / 2 | n ⟩, E_{n} = ℏ ω (2 n + 1) / 2$ .
$X = \sqrt{\frac{ℏ}{2 m ω}} (a + a^{†})$
$P = i \sqrt{\frac{m ω ℏ}{2}} (a^{†} - a)$

This implies that the eigenstate energies are equally spaced by $ℏ ω$ starting at $ℏ ω / 2$ .

$[N, a] = - a$ $[N, a^{†}] = a^{†}$

Deriving some properties

$a^{†} | n ⟩ = | a^{†} n ⟩ = | ? ⟩$ . $N a^{†} | n ⟩ = (a^{†} N + a^{†}) | n ⟩ = (n + 1) a^{†} | n ⟩$ Thus, $N | a^{†} n ⟩ = (n + 1) | a^{†} n ⟩$ . Thus, $| a^{†} n ⟩ = \tilde{c} | n + 1 ⟩$ .

Similarly, $a | n ⟩ = \tilde{d} | n - 1 ⟩$ .

$⟨ N ⟩ = n = ⟨ n | a^{†} a | n ⟩ = ⟨ a n | a n ⟩ = ⟨ n - 1 | {\tilde{d}}^{*} \tilde{d} | n - 1 ⟩$ Thus, $\tilde{d} = \sqrt{n}$ .

$⟨ n | a a^{†} | n ⟩ = ⟨ n | N + I | n ⟩ = ⟨ a^{†} n | a^{†} n ⟩ = ⟨ n + 1 | | \tilde{c} |^{2} | n + 1 ⟩$ . Thus, $\tilde{c} = \sqrt{n + 1}$ .

General Forms

$a^{2} | n ⟩ = \sqrt{n (n - 1)} | n - 2 ⟩$ . $a^{m} | n ⟩ = \sqrt{\frac{n!}{(n - m)!}} | n - m ⟩$

$a^{† m} | n ⟩ = \sqrt{\frac{(n + m)!}{n!}} | n + m ⟩$ .

$a^{†} | n ⟩ = \sqrt{n + 1} | n ⟩ = \sqrt{(n + 1)!} n!$ .

$| n ⟩ = (a^{†})^{n} / \sqrt{n!} | 0 ⟩$ .

$a | 0 ⟩ = \sqrt{0} | - 1 ⟩ = 0$ .

$⟨ n^{'} | a | n ⟩ = δ_{n}^{n^{'} + 1} \sqrt{n}$

$⟨ n^{'} | a^{†} | n ⟩ = δ_{n}^{n^{'} - 1} n \sqrt{n + 1}$

$⟨ n^{'} | X | n ⟩$ , $n^{'} = n \pm 1$ .

$⟨ n^{'} | P | n ⟩$ , $n^{'} = n \pm 1$ .

Thus, $X$ and $P$ are non-diagonal in this basis. $[H, X] = 2 i ℏ P, [H, P] = - 2 i ℏ X$ (Note check this commutator, this was put on the fly)

Wavefunction Correspondance

$n ⟩ \Leftrightarrow Ψ_{n} (x)$ .

$0 = ⟨ x^{'} | a | 0 ⟩ = ⟨ x^{'} \sqrt{\frac{m ω}{2 ℏ}} (X + i P / m ω) | 0 ⟩ = ⟨ x^{'} ⟨ x^{'} | 0 ⟩ + \frac{i}{m ω} (- i ℏ) \frac{d}{d x^{'}} ⟨ x^{'} | 0 ⟩$ . Thus, $x^{'} Ψ_{0} (x^{'}) + \frac{ℏ}{m ω} Ψ_{0}^{'} (x^{'})$ . This gives, $Ψ_{0} (x^{'}) = \sqrt[4]{\frac{m ω}{π ℏ}} e^{- \frac{{(\frac{x^{'}}{\sqrt{\frac{ℏ}{m ω}}})}^{2}}{2}}$ .

Uncertainty Relations

$Δ X^{2} = ⟨ X^{2} ⟩ - ⟨ X ⟩^{2} = ⟨ X^{2} ⟩, Δ p^{2} = ⟨ P^{2} ⟩$ .

$⟨ X^{2} ⟩ = \frac{ℏ}{2 m ω} ⟨ n | a^{2} + a a^{†} + a^{†} a + a^{† 2} | n ⟩ = \frac{ℏ}{2 m ω} ⟨ 2 N + 1 ⟩ = \frac{ℏ}{2 m ω} (2 N + 1)$

$(Δ X)^{2} (Δ P)^{2} \geq \frac{ℏ}{4}$ .

What is Oscillating

Time Evolution of QM HO States

Heisenberg Picture

$\frac{d A_{H}}{d t} = \frac{i}{ℏ} [H, A_{H}]$ .

For the QMHO, $P_{H}, X_{H}$ .

$\frac{d P_{H}}{d t} = \frac{i}{ℏ} [\frac{P_{H}^{2}}{2 m}, + \frac{m ω^{2}}{2} X^{2}, P_{H}] = - m ω^{2} X_{H}$

Similarly for $X$ : $\frac{d X_{H}}{d t} = \frac{P_{H}}{m}$ .

$\frac{d^{2} X_{H}}{d t^{2}} = \frac{1}{m} \frac{d P_{H}}{d t} = - ω^{2} X_{H}$ . Thus, $X_{H} (t) = \frac{P_{H} (0)}{m ω} \sin (ω t) + X_{H} (0) \cos (ω t)$ . $P_{H} (t) = - m ω X_{H} (0) \sin (ω t) + P_{H} (0) \cos (ω t)$ .

$X_{H} (t) = U^{†} (t, t_{0}) X_{H} (0) U (t, t_{0}) = \exp (i H t / ℏ) X_{H} (0) \exp (- i H t / ℏ)$

Recall: $e^{B} A e^{- B} = A + [B, A] + \frac{1}{2!} [B, [B, A]] + \dots$ .

Then, $X_H(t) = X_H(0) + [iHt/ℏ,X_H(0)] + ⋯ = $ same thing from a series expansion of a sine and cosine.

Expectation Value

$⟨ n | X (t) | n ⟩ = ⟨ n | \frac{P_{H} (0)}{m ω} \sin ω t + X_{H} (0) \cos ω t | n ⟩ = \cos ω t ⟨ n | X_{H} (0) | n ⟩ + \frac{1}{m ω} ⟨ n | P_{H} (0) | n ⟩ = 0$ .

$⟨ α | X (t) | α ⟩ = \cos ω t \sum_{n m} ⟨ n | X_{H} (0) | m ⟩ c_{n}^{*} c_{m} + \frac{1}{m ω} \sin ω t \sum_{n m} ⟨ n | P_{H} (0) | m ⟩ c_{n}^{*} c_{m}$ .

$Δ X (0) Δ X (t) \geq \frac{1}{2} | ⟨ [X (0), X (t)] ⟩ | = \frac{1}{2} | ⟨ \frac{1}{m ω} \sin ω t \cdot i ℏ ⟩ | = \frac{ℏ}{2 m ω} | \sin ω t |$ .

Nobel Prize in 2005

Glauber, Hall, Hansch. $a | α ⟩ = α | α ⟩$ , eigenstate of annihilation operator, $| c_{n} |^{2} = \frac{{\overset{―}{n}}^{n}}{n!} \exp (- \overset{―}{n})$ . So if your coefficients are poisson-distributed then you have an eigenstate of annihilation operator and you have a coherent system. Sakurai 2.21.