Quantum Harmonic Oscillator
The Classic Harmonic Oscillator applies to quantum physics.
H ^ = p ^ 2 2 m + 1 2 m ω 2 x ^ 2 , [ x ^ , p ^ ] = i ℏ \hat{H} = \frac{\hat{p}^2}{2m} + \tfrac{1}{2}m\omega^2\hat{x}^2, \qquad [\hat{x}, \hat{p}] = i\hbar H ^ = 2 m p ^ 2 + 2 1 m ω 2 x ^ 2 , [ x ^ , p ^ ] = i ℏ
= 1 2 m ω 2 ( x ^ 2 + p ^ 2 m 2 ω 2 ) \boxed{= \tfrac{1}{2}m\omega^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2}\right)} = 2 1 m ω 2 ( x ^ 2 + m 2 ω 2 p ^ 2 )
This looks like a sum of two squares.
Note that
a 2 + b 2 = ( a − i b ) ( a + i b ) a^2 + b^2 = (a - ib)(a + ib) a 2 + b 2 = ( a − ib ) ( a + ib )
applies to operators
( x ^ − i p ^ m ω ) ( x ^ + i p ^ m ω ) = x ^ 2 + p ^ 2 m 2 ω 2 + i m ω ( i ℏ ) \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) ( x ^ − mω i p ^ ) ( x ^ + mω i p ^ ) = x ^ 2 + m 2 ω 2 p ^ 2 + mω i ( i ℏ )
where [ x ^ , p ^ ] = i ℏ [\hat{x},\hat{p}]=i\hbar [ x ^ , p ^ ] = i ℏ
Let
A ^ = x ^ − i p ^ m ω \hat{A}^ =\hat{x}-\frac{i\hat{p}}{m\omega} A ^ = x ^ − mω i p ^
A ^ † = x ^ + i p ^ m ω \hat{A}^\dagger =\hat{x}+\frac{i\hat{p}}{m\omega} A ^ † = x ^ + mω i p ^
so
A ^ † A ^ = x ^ 2 + p ^ 2 m 2 ω 2 − ℏ m ω \hat{A}^\dagger\hat{A} = \hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} - \frac{\hbar}{m\omega} A ^ † A ^ = x ^ 2 + m 2 ω 2 p ^ 2 − mω ℏ
We substitute this back into H ^ \hat{H} H ^
H ^ = 1 2 m ω 2 ( A ^ † A ^ + ℏ m ω I ) \hat{H} = \tfrac{1}{2}m\omega^2\left(\hat{A}^\dagger\hat{A} + \frac{\hbar}{m\omega}\mathbb{I}\right) H ^ = 2 1 m ω 2 ( A ^ † A ^ + mω ℏ I )
Therefore even if A ^ = 0 \hat{A}=0 A ^ = 0 , then the lower bound for hamiltonian is
H ^ = E ≥ 1 2 ℏ ω \hat{H}=E\geq \frac{1}{2}\hbar \omega H ^ = E ≥ 2 1 ℏ ω
Let
α ≜ ℏ / m ω \boxed{\alpha\triangleq \sqrt{\hbar/m\omega}} α ≜ ℏ/ mω
so
A ^ = x ^ + i α 2 ℏ p ^ \hat{A} = \hat{x} + i\,\frac{\alpha^2}{\hbar}\,\hat{p} A ^ = x ^ + i ℏ α 2 p ^
= α ( x ^ α + i α ℏ p ^ ) = \alpha\left(\frac{\hat{x}}{\alpha} + i\,\frac{\alpha}{\hbar}\,\hat{p}\right) = α ( α x ^ + i ℏ α p ^ )
Let lowering/annihilation operator a ^ \hat{a} a ^ be
a ^ ≜ 1 a 2 A ^ = 1 2 ( x ^ α + i α ℏ p ^ ) \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)} a ^ ≜ a 2 1 A ^ = 2 1 ( α x ^ + i ℏ α p ^ )
and raising/creation operator be a ^ † \hat{a}^\dagger a ^ †
The 2 \sqrt{2} 2 in the denominator keeps [ a ^ , a ^ † ] = 1 [\hat{a},\hat{a}^\dagger]=1 [ a ^ , a ^ † ] = 1
which is clean
[ a ^ , a ^ † ] = 1 2 [ x ^ α + i α ℏ p ^ , x ^ α − i α ℏ p ^ ] [\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] [ a ^ , a ^ † ] = 2 1 [ α x ^ + i ℏ α p ^ , α x ^ − i ℏ α p ^ ]
= − i 2 ℏ [ x ^ , p ^ ] + i 2 ℏ [ p ^ , x ^ ] = 1 = -\frac{i}{2\hbar}[\hat{x}, \hat{p}] + \frac{i}{2\hbar}[\hat{p}, \hat{x}] = 1 = − 2ℏ i [ x ^ , p ^ ] + 2ℏ i [ p ^ , x ^ ] = 1
[ a ^ , a ^ † ] = 1 \boxed{[\hat{a}, \hat{a}^\dagger] = 1} [ a ^ , a ^ † ] = 1
Thus
a ^ = 1 2 ( x ^ α + i α ℏ p ^ ) , a ^ † = 1 2 ( x ^ α − i α ℏ p ^ ) \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) a ^ = 2 1 ( α x ^ + i ℏ α p ^ ) , a ^ † = 2 1 ( α x ^ − i ℏ α p ^ )
Adding them causes the p ^ \hat{p} p ^ term to cancel
x ^ = α 2 ( a ^ + a ^ † ) \boxed{\hat{x} = \frac{\alpha}{\sqrt{2}}(\hat{a} + \hat{a}^\dagger)} x ^ = 2 α ( a ^ + a ^ † )
Subtracting them causes x ^ \hat{x} x ^ term to cancel
p ^ = ℏ i α 2 ( a ^ − a ^ † ) \boxed{\hat{p} = \frac{\hbar}{i\alpha\sqrt{2}}(\hat{a} - \hat{a}^\dagger)} p ^ = i α 2 ℏ ( a ^ − a ^ † )
where
H ^ = 1 2 m ω 2 ( x ^ 2 + p ^ 2 m 2 ω 2 ) \hat{H}= \tfrac{1}{2}m\omega^2\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2}\right) H ^ = 2 1 m ω 2 ( x ^ 2 + m 2 ω 2 p ^ 2 )
= 1 2 m ω 2 ( 2 ℏ m ω a ^ † a ^ + ℏ m ω ) =\tfrac{1}{2}m\omega^2\left(\frac{2\hbar}{m\omega}\hat{a}^\dagger\hat{a} + \frac{\hbar}{m\omega}\right) = 2 1 m ω 2 ( mω 2ℏ a ^ † a ^ + mω ℏ )
= ℏ ω a ^ † a ^ + 1 2 ℏ ω = \hbar\omega\,\hat{a}^\dagger\hat{a} + \tfrac{1}{2}\hbar\omega = ℏ ω a ^ † a ^ + 2 1 ℏ ω
= ℏ ω ( N ^ + 1 2 ) = \hbar\omega\left(\hat{N} + \tfrac{1}{2}\right) = ℏ ω ( N ^ + 2 1 )
where number operator N ^ \hat{N} N ^
N ^ = a ^ † a ^ \boxed{\hat{N}=\hat{a}^\dagger\hat{a}} N ^ = a ^ † a ^
Note from Ladder of States that
N E = E ℏ ω − 1 2 N_E=\frac{E}{\hbar \omega}-\frac{1}{2} N E = ℏ ω E − 2 1
N E N_E N E must be a non-negative integer -- the chain reaches ∣ 0 ⟩ \ket{0} ∣ 0 ⟩ after N E N_E N E steps.
If N E N_E N E is not a int then subtracting will never get to zero and overshoot to a negative decimal number.
⟨ E − ∣ E − ⟩ = − 0.3 < 0 \braket{E_-|E_-}=-0.3<0 ⟨ E − ∣ E − ⟩ = − 0.3 < 0
is not possible because norm squared are always ≥ 0 \geq 0 ≥ 0
Classic Harmonic Oscillator