Going from 1+1=2 to Quantum Mechanics Chaimongkol, 2026 QHO Coherent States
Let all the variables in Quantum Harmonic Oscillator
From definition, define z z z eigenvalue of annihilation operator a ^ \hat{a} a ^
a ^ ∣ z ⟩ = z ∣ z ⟩ , z ∈ C \hat{a}\ket{z} = z\ket{z}, \quad z \in \mathbb{C} a ^ ∣ z ⟩ = z ∣ z ⟩ , z ∈ C Note that a ^ † \hat{a}^\dagger a ^ † Hilbert Space is infinite dimension.
Expanding in the numbers basis where c n c_n c n Probability amplitude
∣ z ⟩ = ∑ n = 0 ∞ c n ∣ n ⟩ \ket{z} = \sum_{n=0}^{\infty} c_n \ket{n} ∣ z ⟩ = n = 0 ∑ ∞ c n ∣ n ⟩ Apply a ^ \hat{a} a ^ QHO Quantum Dynamics we proved a ^ ∣ n ⟩ = n ∣ n − 1 ⟩ \hat{a}\ket{n}=\sqrt{n}\ket{n-1} a ^ ∣ n ⟩ = n ∣ n − 1 ⟩
a ^ ∣ z ⟩ = ∑ n c n n ∣ n − 1 ⟩ = z ∣ z ⟩ \hat{a}\ket{z} = \sum_n c_n \sqrt{n}\ket{n-1} = z\ket{z} a ^ ∣ z ⟩ = n ∑ c n n ∣ n − 1 ⟩ = z ∣ z ⟩ Let m = n − 1 m=n-1 m = n − 1
c m + 1 m + 1 = z c m ⟹ c n n = z c n − 1 c_{m+1}\sqrt{m+1} = z\, c_m \Longrightarrow c_n\sqrt{n} = z\, c_{n-1} c m + 1 m + 1 = z c m ⟹ c n n = z c n − 1 So
c n = z n c n − 1 = z n z n − 1 c n − 2 = ⋯ = z n n ! c 0 c_n = \frac{z}{\sqrt{n}}c_{n-1} = \frac{z}{\sqrt{n}}\frac{z}{\sqrt{n-1}}c_{n-2} = \cdots = \frac{z^n}{\sqrt{n!}}c_0 c n = n z c n − 1 = n z n − 1 z c n − 2 = ⋯ = n ! z n c 0 ⟹ ∣ z ⟩ = c 0 ∑ n = 0 ∞ z n n ! ∣ n ⟩ \implies\boxed{\ket{z} = c_0 \sum_{n=0}^{\infty} \frac{z^n}{\sqrt{n!}}\ket{n}} ⟹ ∣ z ⟩ = c 0 n = 0 ∑ ∞ n ! z n ∣ n ⟩ Normalize to get c 0 c_0 c 0
⟨ z ∣ z ⟩ = ∣ c 0 ∣ 2 ∑ n ∣ z ∣ 2 n n ! = ∣ c 0 ∣ 2 e ∣ z ∣ 2 = ! 1 \braket{z|z} = |c_0|^2 \sum_n \frac{|z|^{2n}}{n!} = |c_0|^2 e^{|z|^2} \stackrel{!}{=} 1 ⟨ z ∣ z ⟩ = ∣ c 0 ∣ 2 n ∑ n ! ∣ z ∣ 2 n = ∣ c 0 ∣ 2 e ∣ z ∣ 2 = ! 1 ⟹ c 0 = e − ∣ z ∣ 2 / 2 \Longrightarrow \boxed{c_0 = e^{-|z|^2/2}} ⟹ c 0 = e − ∣ z ∣ 2 /2 Coherent State Observable
The Observable for a coherent state becomes
Where N ^ \hat{N} N ^ Number Operator
⟨ z ∣ N ^ ∣ z ⟩ = ⟨ z ∣ a ^ † a ^ ∣ z ⟩ = ∣ z ∣ 2 \langle z|\hat{N}|z\rangle = \langle z|\hat{a}^\dagger \hat{a}|z\rangle = |z|^2 ⟨ z ∣ N ^ ∣ z ⟩ = ⟨ z ∣ a ^ † a ^ ∣ z ⟩ = ∣ z ∣ 2 P ( n ) = ∣ ⟨ n ∣ z ⟩ ∣ 2 = ∣ c n ∣ 2 = e − ∣ z ∣ 2 ∣ z ∣ 2 n n ! P(n) = |\braket{n|z}|^2 = |c_n|^2 = e^{-|z|^2}\frac{|z|^{2n}}{n!} P ( n ) = ∣ ⟨ n ∣ z ⟩ ∣ 2 = ∣ c n ∣ 2 = e − ∣ z ∣ 2 n ! ∣ z ∣ 2 n this is a Poisson Distribution with mean E [ n ] = ∣ z ∣ 2 \mathbb{E}[n]=|z|^2 E [ n ] = ∣ z ∣ 2
Note this is the energy expectation
⟨ E ⟩ = ⟨ H ^ ⟩ = ℏ ω ( ⟨ N ^ ⟩ + 1 2 ) = ℏ ω ( ∣ z ∣ 2 + 1 2 ) \langle E \rangle = \langle \hat{H} \rangle = \hbar\omega\!\left(\langle \hat{N}\rangle + \tfrac{1}{2}\right) = \hbar\omega\!\left(|z|^2 + \tfrac{1}{2}\right) ⟨ E ⟩ = ⟨ H ^ ⟩ = ℏ ω ( ⟨ N ^ ⟩ + 2 1 ) = ℏ ω ( ∣ z ∣ 2 + 2 1 ) Coherent State Wavefunction
The Wave function for a coherent state becomes
a ^ ∣ z ⟩ = z ∣ z ⟩ \hat{a}\ket{z}=z\ket{z} a ^ ∣ z ⟩ = z ∣ z ⟩ Let variables from Quantum Harmonic Oscillator
⟹ 1 2 ( x α + α ∂ ∂ x ) ϕ z ( x ) = z ϕ z ( x ) \implies\frac{1}{\sqrt{2}}\left(\frac{x}{\alpha} + \alpha\frac{\partial}{\partial x}\right)\phi_z(x) = z\,\phi_z(x) ⟹ 2 1 ( α x + α ∂ x ∂ ) ϕ z ( x ) = z ϕ z ( x ) rearranging gives
∂ ϕ z ∂ x = ( 2 z α − x α 2 ) ϕ z \frac{\partial \phi_z}{\partial x} = \left(\frac{\sqrt{2}\,z}{\alpha} - \frac{x}{\alpha^2}\right)\phi_z ∂ x ∂ ϕ z = ( α 2 z − α 2 x ) ϕ z ln ϕ z = 2 z α x − x 2 2 α 2 + const \ln\phi_z = \frac{\sqrt{2}\,z}{\alpha}x - \frac{x^2}{2\alpha^2} + \text{const} ln ϕ z = α 2 z x − 2 α 2 x 2 + const ϕ z ( x ) = N z e − ( x − b ) 2 / 2 α 2 e i k x \phi_z(x) = N_z\, e^{-(x-b)^2/2\alpha^2}\, e^{ikx} ϕ z ( x ) = N z e − ( x − b ) 2 /2 α 2 e ik x b = 2 α Re [ z ] , k = 2 α Im [ z ] b = \sqrt{2}\,\alpha\,\text{Re}[z], \qquad k = \frac{\sqrt{2}}{\alpha}\text{Im}[z] b = 2 α Re [ z ] , k = α 2 Im [ z ] Note Expected Value of an Observable are
⟨ z ∣ x ^ ∣ z ⟩ = b , ⟨ z ∣ p ^ ∣ z ⟩ = ℏ k \braket{z|\hat{x}|z} = b, \qquad \braket{z|\hat{p}|z} = \hbar k ⟨ z ∣ x ^ ∣ z ⟩ = b , ⟨ z ∣ p ^ ∣ z ⟩ = ℏ k Note Gaussian property
Δ x = α 2 , Δ p = ℏ α 2 ⟹ Δ x Δ p = ℏ 2 \Delta x = \frac{\alpha}{\sqrt{2}}, \qquad \Delta p = \frac{\hbar}{\alpha\sqrt{2}} \Longrightarrow \Delta x\,\Delta p = \frac{\hbar}{2} Δ x = 2 α , Δ p = α 2 ℏ ⟹ Δ x Δ p = 2 ℏ Note
z = 1 2 ( ⟨ x ^ ⟩ α + i α ℏ ⟨ p ^ ⟩ ) \boxed{z = \frac{1}{\sqrt{2}}\left(\frac{\braket{\hat{x}}}{\alpha} + i\frac{\alpha}{\hbar}\braket{\hat{p}}\right)} z = 2 1 ( α ⟨ x ^ ⟩ + i ℏ α ⟨ p ^ ⟩ ) This can be proven but I don't think the professor proved it.
QHO Observables