Quantum Harmonic Oscillator

The Classic Harmonic Oscillator applies to quantum physics.

H^=p^22m+12mω2x^2,[x^,p^]=i\hat{H} = \frac{\hat{p}^2}{2m} + \tfrac{1}{2}m\omega^2\hat{x}^2, \qquad [\hat{x}, \hat{p}] = i\hbar =12mω2(x^2+p^2m2ω2)\boxed{= \tfrac{1}{2}m\omega^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2}\right)}

This looks like a sum of two squares. Note that

a2+b2=(aib)(a+ib)a^2 + b^2 = (a - ib)(a + ib)

applies to operators

(x^ip^mω)(x^+ip^mω)=x^2+p^2m2ω2+imω(i)\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right) = \hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} + \frac{i}{m\omega}(i\hbar)

where [x^,p^]=i[\hat{x},\hat{p}]=i\hbar Let

A^=x^ip^mω\hat{A}^ =\hat{x}-\frac{i\hat{p}}{m\omega} A^=x^+ip^mω\hat{A}^\dagger =\hat{x}+\frac{i\hat{p}}{m\omega}

so

A^A^=x^2+p^2m2ω2mω\hat{A}^\dagger\hat{A} = \hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} - \frac{\hbar}{m\omega}

We substitute this back into H^\hat{H}

H^=12mω2(A^A^+mωI)\hat{H} = \tfrac{1}{2}m\omega^2\left(\hat{A}^\dagger\hat{A} + \frac{\hbar}{m\omega}\mathbb{I}\right)

Therefore even if A^=0\hat{A}=0, then the lower bound for hamiltonian is

H^=E12ω\hat{H}=E\geq \frac{1}{2}\hbar \omega

Let

α/mω\boxed{\alpha\triangleq \sqrt{\hbar/m\omega}}

so

A^=x^+iα2p^\hat{A} = \hat{x} + i\,\frac{\alpha^2}{\hbar}\,\hat{p} =α(x^α+iαp^)= \alpha\left(\frac{\hat{x}}{\alpha} + i\,\frac{\alpha}{\hbar}\,\hat{p}\right)

Let lowering/annihilation operator a^\hat{a} be

a^1a2A^=12(x^α+iαp^)\boxed{\hat{a} \triangleq \frac{1}{a\sqrt{2}}\hat{A} =\frac{1}{\sqrt{2}}\left(\frac{\hat{x}}{\alpha} + i\,\frac{\alpha}{\hbar}\,\hat{p}\right)}

and raising/creation operator be a^\hat{a}^\dagger

The 2\sqrt{2} in the denominator keeps [a^,a^]=1[\hat{a},\hat{a}^\dagger]=1 which is clean

[a^,a^]=12[x^α+iαp^,x^αiαp^][\hat{a}, \hat{a}^\dagger] = \tfrac{1}{2}\left[\frac{\hat{x}}{\alpha} + i\frac{\alpha}{\hbar}\hat{p}, \frac{\hat{x}}{\alpha} - i\frac{\alpha}{\hbar}\hat{p}\right] =i2[x^,p^]+i2[p^,x^]=1= -\frac{i}{2\hbar}[\hat{x}, \hat{p}] + \frac{i}{2\hbar}[\hat{p}, \hat{x}] = 1 [a^,a^]=1\boxed{[\hat{a}, \hat{a}^\dagger] = 1}

Thus

a^=12(x^α+iαp^),a^=12(x^αiαp^)\hat{a} = \frac{1}{\sqrt{2}}\left(\frac{\hat{x}}{\alpha} + i\frac{\alpha}{\hbar}\hat{p}\right), \qquad \hat{a}^\dagger = \frac{1}{\sqrt{2}}\left(\frac{\hat{x}}{\alpha} - i\frac{\alpha}{\hbar}\hat{p}\right)

Adding them causes the p^\hat{p} term to cancel

x^=α2(a^+a^)\boxed{\hat{x} = \frac{\alpha}{\sqrt{2}}(\hat{a} + \hat{a}^\dagger)}

Subtracting them causes x^\hat{x} term to cancel

p^=iα2(a^a^)\boxed{\hat{p} = \frac{\hbar}{i\alpha\sqrt{2}}(\hat{a} - \hat{a}^\dagger)}

where

H^=12mω2(x^2+p^2m2ω2)\hat{H}= \tfrac{1}{2}m\omega^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2}\right) =12mω2(2mωa^a^+mω)=\tfrac{1}{2}m\omega^2\left(\frac{2\hbar}{m\omega}\hat{a}^\dagger\hat{a} + \frac{\hbar}{m\omega}\right) =ωa^a^+12ω= \hbar\omega\,\hat{a}^\dagger\hat{a} + \tfrac{1}{2}\hbar\omega =ω(N^+12)= \hbar\omega\left(\hat{N} + \tfrac{1}{2}\right)

where number operator N^\hat{N}

N^=a^a^\boxed{\hat{N}=\hat{a}^\dagger\hat{a}}

Note from Ladder of States that

NE=Eω12N_E=\frac{E}{\hbar \omega}-\frac{1}{2}

NEN_E must be a non-negative integer -- the chain reaches 0\ket{0} after NEN_E steps.

If NEN_E is not a int then subtracting will never get to zero and overshoot to a negative decimal number. EE=0.3<0\braket{E_-|E_-}=-0.3<0 is not possible because norm squared are always 0\geq 0