Quantum Harmonic Oscillator Coherent States – Complete Notes

Coherent State for the Harmonic Oscillator and its properties

• Discovered by R. J. Glauber in 1963.
• Glauber received the Nobel Prize in 2005 for the relevance of coherent states in quantum optics.

The state describing a laser beam can be briefly characterized as:

1. An indefinite number of photons.
2. A precisely defined phase.

Laser dynamics → coherent state.
Normal light → unpolarized / incoherent.

Uncertainty relation:

\[ \Delta N \, \Delta \Phi \ge \frac{1}{2} \]

Here

• \( \Delta N \) : fluctuation in occupation number
• \( \Delta \Phi \) : fluctuation in phase

For a laser:

• \( \Delta \Phi \) → very small
• \( \Delta N \) → large

For normal light:

• \( \Delta N \) → fixed / small
• \( \Delta \Phi \) → large (not in the same phase)

Definition of Coherent States

A coherent state \( |\alpha\rangle \) (also known as a Glauber state) is defined as an eigenstate of the annihilation operator \( \hat{a} \) with eigenvalue \( \alpha \in \mathbb{C} \).

\[ \hat{a} |\alpha\rangle = \alpha |\alpha\rangle \]

Since \( \hat{a} \) is a non-Hermitian operator, \( \alpha \) must be a complex number, i.e.

\[ \alpha = |\alpha| e^{i\phi} \]

Coherent State for Harmonic Oscillator

We know that for a one-dimensional harmonic oscillator

\[ \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2 \]

The time-independent Schrödinger equation is

\[ \hat{H} |\psi\rangle = E |\psi\rangle \]

Here \( |\psi\rangle \) is an energy eigenstate and \( E \) is the corresponding energy eigenvalue (a real number).

\[ E_n = \hbar\omega \left( n + \frac{1}{2} \right) \]

In the energy basis,

\[ \langle \hat{x} \rangle = 0, \qquad \langle \hat{p} \rangle = 0 \] \[ \langle \hat{x}^2 \rangle = \frac{(2n+1)\hbar}{2m\omega} = \sigma_x^2 \] \[ \langle \hat{p}^2 \rangle = \frac{(2n+1)m\hbar\omega}{2} = \sigma_p^2 \] \[ \sigma_x \sigma_p = (2n+1)\frac{\hbar}{2} \]

For \( n=0 \),

\[ \sigma_x \sigma_p = \frac{\hbar}{2} \]

which is the minimum value and corresponds to the Gaussian wavefunction of the ground state.

The Ladder Operator Method

The ladder operators are defined as

\[ \hat{a} = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} + \frac{i}{m\omega}\hat{p} \right), \qquad \hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}} \left( \hat{x} - \frac{i}{m\omega}\hat{p} \right) \]

and the following are the representations of \( \hat{x} \) and \( \hat{p} \):

\[ \hat{x} = \sqrt{\frac{\hbar}{2m\omega}} \left( \hat{a}^\dagger + \hat{a} \right) \] \[ \hat{p} = i\sqrt{\frac{\hbar m\omega}{2}} \left( \hat{a}^\dagger - \hat{a} \right) \]

Properties of Ladder Operators

The operator \( \hat{a} \) is not Hermitian, since

\[ \hat{a} \ne \hat{a}^\dagger \]

Consider an energy eigenstate \( |n\rangle \). Then

\[ \hat{a}^\dagger |n\rangle = \sqrt{n+1}\,|n+1\rangle \] \[ \hat{a} |n\rangle = \sqrt{n}\,|n-1\rangle \]

The commutation relations are

\[ [\hat{x},\hat{p}] = i\hbar \] \[ [\hat{a},\hat{a}^\dagger] = 1 \]

Number Operator

Let us define the number operator \( \hat{N} \) with the following properties:

\[ \hat{N} = \hat{a}^\dagger \hat{a} \] \[ \hat{N}|n\rangle = n|n\rangle \]

The commutation relations with ladder operators are

\[ [\hat{N},\hat{a}^\dagger] = \hat{a}^\dagger \] \[ [\hat{N},\hat{a}] = -\hat{a} \]

The Hamiltonian can be written in terms of the number operator as

\[ \hat{H} = \hbar\omega \left(\hat{N} + \frac{1}{2}\right) \]

Hence the eigenstates of \( \hat{N} \) are also eigenstates of the Hamiltonian.

\[ \hat{H}|n\rangle = \hbar\omega\left(\hat{N}+\frac{1}{2}\right)|n\rangle = \hbar\omega\left(n+\frac{1}{2}\right)|n\rangle \]

Thus the energy eigenvalues are

\[ E_n = \hbar\omega \left(n+\frac{1}{2}\right) \]

Action of the Number Operator

And,

\[ \hat{N}(\hat{a}^\dagger |n\rangle) = (n+1)\hat{a}^\dagger |n\rangle \] \[ \hat{N}(\hat{a}|n\rangle) = (n-1)\hat{a}|n\rangle \]

Note: The smallest eigenvalue of \( \hat{N} \) is \(0\), therefore

\[ \hat{a}|0\rangle = 0 \]

Hence,

\[ n = \langle n|\hat{N}|n\rangle = \langle n|\hat{a}^\dagger \hat{a}|n\rangle = (\hat{a}|n\rangle)^\dagger (\hat{a}|n\rangle) \ge 0 \]

Arbitrary Eigenstate

Since

\[ \hat{a}^\dagger |n\rangle = \sqrt{n+1}\,|n+1\rangle \]

Now,

\[ |n\rangle = \frac{\hat{a}^\dagger}{\sqrt{n}}|n-1\rangle \] \[ = \frac{(\hat{a}^\dagger)^2}{\sqrt{n(n-1)}}|n-2\rangle \]

or

\[ |n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle \]

Solving the Eigenvalue Equation

We have the eigenvalue equation for the coherent state \( |\alpha\rangle \):

\[ \hat{a}|\alpha\rangle = \alpha|\alpha\rangle \]

Let us assume the following expansion:

\[ |\alpha\rangle = \sum_{n=0}^{\infty} C_n |n\rangle \]

Determination of the Coefficients

Starting from

\[ \hat{a}|\alpha\rangle = \hat{a}\sum_{n=0}^{\infty} C_n |n\rangle = \alpha |\alpha\rangle \]

we get

\[ \sum_{n=1}^{\infty} C_n \sqrt{n}\,|n-1\rangle = \alpha \sum_{n=0}^{\infty} C_n |n\rangle \]

Now change the index by setting \( n-1 = m \). Then

\[ \sum_{m=0}^{\infty} C_{m+1}\sqrt{m+1}\,|m\rangle = \alpha \sum_{m=0}^{\infty} C_m |m\rangle \]

Comparing the coefficients of \( |m\rangle \), we obtain

\[ C_{m+1}\sqrt{m+1} = \alpha C_m \]

Therefore,

\[ C_{m+1} = \frac{\alpha}{\sqrt{m+1}}\,C_m \]

Iteratively,

\[ C_1 = \alpha C_0 \] \[ C_2 = \frac{\alpha}{\sqrt{2}} C_1 = \frac{\alpha^2}{\sqrt{2}} C_0 \] \[ C_3 = \frac{\alpha^3}{\sqrt{3!}} C_0 \]

Hence, in general,

\[ C_n = \frac{\alpha^n}{\sqrt{n!}}\,C_0 \]

Therefore,

\[ |\alpha\rangle = \sum_{n=0}^{\infty} C_n |n\rangle = \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} C_0 |n\rangle \]

Normalization

We have obtained

\[ |\alpha\rangle = \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} C_0 |n\rangle \]

Now compute the normalization:

\[ \langle \alpha | \alpha \rangle = \sum_{n=0}^{\infty} \frac{|\alpha|^{2n}}{n!} |C_0|^2 \langle n|n\rangle \]

Since \( \langle n|n\rangle = 1 \),

\[ 1 = |C_0|^2 \sum_{n=0}^{\infty} \frac{|\alpha|^{2n}}{n!} \]

Using the exponential series

\[ \sum_{n=0}^{\infty} \frac{|\alpha|^{2n}}{n!} = e^{|\alpha|^2} \]

we obtain

\[ 1 = |C_0|^2 e^{|\alpha|^2} \] \[ C_0 = e^{-|\alpha|^2/2} \]

Normalized Coherent State

Hence the normalized coherent state is

\[ |\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n\rangle \]

Thus a coherent state is a superposition of number states (Fock states).

Checking the Eigenvalue Equation

Now that we have derived the coherent state \( |\alpha\rangle \), let us verify that it satisfies the eigenvalue equation.

\[ |\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n\rangle \]

Let us check that the state satisfies the eigenvalue equation.

\[ \hat{a}|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} \hat{a}|n\rangle \]

Using \( \hat{a}|n\rangle = \sqrt{n}|n-1\rangle \), we obtain

\[ \hat{a}|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{n=1}^{\infty} \frac{\alpha^n \sqrt{n}}{\sqrt{n!}} |n-1\rangle \]

Changing the index \( n-1=m \),

\[ \hat{a}|\alpha\rangle = e^{-|\alpha|^2/2} \sum_{m=0}^{\infty} \frac{\alpha^{m+1}}{\sqrt{m!}} |m\rangle \]

Factoring out \( \alpha \),

\[ \hat{a}|\alpha\rangle = \alpha e^{-|\alpha|^2/2} \sum_{m=0}^{\infty} \frac{\alpha^m}{\sqrt{m!}} |m\rangle \] \[ \hat{a}|\alpha\rangle = \alpha |\alpha\rangle \]

Probability Distribution

\[ \langle \hat{N} \rangle = \langle \alpha | \hat{a}^\dagger \hat{a} |\alpha\rangle = (\hat{a}|\alpha\rangle)^\dagger(\hat{a}|\alpha\rangle) \] \[ \bar{n} = (\alpha^* \alpha) \langle \alpha | \alpha \rangle \] \[ \bar{n} = |\alpha|^2 \]

Thus the photon number is proportional to the square of the amplitude.

Classical Electromagnetic Wave

Average energy density

\[ u = \frac{\epsilon_0}{2} |E_0|^2 \]

Total energy

\[ U = \frac{\epsilon_0 V}{2} |E_0|^2 \]

Quantum Mechanical Analog

\[ \bar{n} = |\alpha|^2 = \frac{U}{\hbar\omega} = \frac{\epsilon_0 V}{2\hbar\omega} |E_0|^2 \]

Probability Distribution

The probability distribution is

\[ P(n) = |\langle n|\alpha\rangle|^2 \]

Using the coherent state expansion,

\[ P(n) = \left| \left\langle n \left| e^{-|\alpha|^2/2} \sum_{m=0}^{\infty} \frac{\alpha^m}{\sqrt{m!}} |m\rangle \right.\right\rangle \right|^2 \] \[ = \left| e^{-|\alpha|^2/2} \frac{\alpha^m}{\sqrt{m!}} \delta_{mn} \right|^2 \] \[ = \left| e^{-|\alpha|^2/2} \frac{\alpha^n}{\sqrt{n!}} \right|^2 \] \[ = \frac{e^{-|\alpha|^2}|\alpha|^{2n}}{n!} \]

Since \( \bar{n} = |\alpha|^2 \), this becomes

\[ P(n) = \frac{e^{-\bar{n}}\bar{n}^{\,n}}{n!} \]

Thus the photon number distribution for a coherent state is a Poisson distribution:

\[ P(X=n)=\frac{\mu^n e^{-\mu}}{n!} \]

with mean \( \mu = \bar{n} \).

Hence, the probability of finding the Fock state \( |n\rangle \) in a coherent state is

\[ P_n(\alpha) = |\langle n|\alpha\rangle|^2 = \frac{e^{-|\alpha|^2}|\alpha|^{2n}}{n!} \]