Module 3: Ladder Operators — The Algebraic Shortcut
The Most Powerful Tool in Quantum Mechanics
← Module 2 | ↑ Course Index | Module 4 →
1 Motivation: There Must Be a Better Way
In Module 2, we solved the QHO by substituting a power series into a differential equation and requiring it to terminate. It worked, but it was:
- Tedious (lots of algebra)
- Not generalizable (a new calculation for each system)
- Not transparent (it’s hard to see why the energy is quantized)
In 1940, Paul Dirac found an algebraic method that derives the entire spectrum of the QHO without solving any differential equation. This method:
- Reveals the deep structure of the problem
- Generalizes to quantum field theory, quantum optics, and quantum computing
- Is the language in which 95% of modern quantum physics is written
Let’s learn it.
Textbook reference: Shankar, Section 7.4; Griffiths Section 2.3.1; Cohen-Tannoudji Vol. 1, Chapter V.
2 Defining the Ladder Operators
The Hamiltonian is:
\[ \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2 \]
Notice that classically, \(p^2/2m + m\omega^2 x^2/2\) looks like the sum of two squares. We can try to “factor” it. Define:
\[ \boxed{\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right)} \]
\[ \boxed{\hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)} \]
These are called the annihilation operator (\(\hat{a}\)) and creation operator (\(\hat{a}^\dagger\)). Note: \(\hat{a}^\dagger\) is the Hermitian conjugate (adjoint) of \(\hat{a}\).
As we’ll prove shortly, \(\hat{a}^\dagger\) raises the quantum number \(n\) by 1 (creates a quantum of energy), and \(\hat{a}\) lowers \(n\) by 1 (annihilates a quantum). In quantum optics, this corresponds to creating and destroying photons.
2.1 Expressing \(\hat{x}\) and \(\hat{p}\) in Terms of \(\hat{a}\) and \(\hat{a}^\dagger\)
Solving the two equations above for \(\hat{x}\) and \(\hat{p}\):
\[ \boxed{\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}\left(\hat{a} + \hat{a}^\dagger\right)} \]
\[ \boxed{\hat{p} = i\sqrt{\frac{m\omega\hbar}{2}}\left(\hat{a}^\dagger - \hat{a}\right)} \]
Using the oscillator length \(x_0 = \sqrt{\hbar/m\omega}\):
\[ \hat{x} = \frac{x_0}{\sqrt{2}}\left(\hat{a} + \hat{a}^\dagger\right), \qquad \hat{p} = \frac{i\hbar}{\sqrt{2}\,x_0}\left(\hat{a}^\dagger - \hat{a}\right) \]
3 Key Commutation Relations
The fundamental commutator \([\hat{x}, \hat{p}] = i\hbar\) implies:
Computing \([\hat{a}, \hat{a}^\dagger]\):
\[ [\hat{a}, \hat{a}^\dagger] = \frac{m\omega}{2\hbar}\left[\hat{x} + \frac{i\hat{p}}{m\omega},\, \hat{x} - \frac{i\hat{p}}{m\omega}\right] \]
\[ = \frac{m\omega}{2\hbar}\left([\hat{x}, -\frac{i\hat{p}}{m\omega}] + [\frac{i\hat{p}}{m\omega}, \hat{x}]\right) \]
\[ = \frac{m\omega}{2\hbar}\left(-\frac{i}{m\omega}[\hat{x}, \hat{p}] + \frac{i}{m\omega}[\hat{p}, \hat{x}]\right) \]
\[ = \frac{m\omega}{2\hbar}\left(-\frac{i}{m\omega}(i\hbar) + \frac{i}{m\omega}(-i\hbar)\right) = \frac{m\omega}{2\hbar} \cdot \frac{2\hbar}{m\omega} = 1 \]
\[ \boxed{[\hat{a},\, \hat{a}^\dagger] = 1} \]
This single commutation relation encodes all the physics of the harmonic oscillator.
4 The Hamiltonian in Terms of \(\hat{a}\) and \(\hat{a}^\dagger\)
Computing \(\hat{a}^\dagger\hat{a}\):
\[ \hat{a}^\dagger\hat{a} = \frac{m\omega}{2\hbar}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right) \]
\[ = \frac{m\omega}{2\hbar}\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} + \frac{i}{m\omega}[\hat{x}, \hat{p}]\right) \]
\[ = \frac{m\omega}{2\hbar}\left(\hat{x}^2 + \frac{\hat{p}^2}{m^2\omega^2} - \frac{\hbar}{m\omega}\right) \]
\[ = \frac{1}{\hbar\omega}\left(\frac{\hat{p}^2}{2m} + \frac{m\omega^2\hat{x}^2}{2}\right) - \frac{1}{2} = \frac{\hat{H}}{\hbar\omega} - \frac{1}{2} \]
Solving for \(\hat{H}\):
\[ \boxed{\hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right) = \hbar\omega\left(\hat{N} + \frac{1}{2}\right)} \]
where we defined the number operator:
\[ \hat{N} = \hat{a}^\dagger\hat{a} \]
The number operator is Hermitian (\(\hat{N}^\dagger = \hat{N}\)) and its eigenvalues give the quantum number \(n\).
5 Deriving the Spectrum Algebraically
5.1 Step 1: The Number Operator Has Non-Negative Eigenvalues
Let \(|\nu\rangle\) be an eigenstate of \(\hat{N}\) with eigenvalue \(\nu\): \(\hat{N}|\nu\rangle = \nu|\nu\rangle\).
Then: \[ \nu = \langle\nu|\hat{N}|\nu\rangle = \langle\nu|\hat{a}^\dagger\hat{a}|\nu\rangle = \|\hat{a}|\nu\rangle\|^2 \geq 0 \]
So all eigenvalues of \(\hat{N}\) are non-negative: \(\nu \geq 0\).
5.2 Step 2: \(\hat{a}\) Lowers the Eigenvalue by 1
Show \([\hat{N}, \hat{a}] = -\hat{a}\):
\[ [\hat{N}, \hat{a}] = [\hat{a}^\dagger\hat{a},\, \hat{a}] = \hat{a}^\dagger[\hat{a}, \hat{a}] + [\hat{a}^\dagger, \hat{a}]\hat{a} = 0 + (-1)\hat{a} = -\hat{a} \]
Similarly, \([\hat{N}, \hat{a}^\dagger] = +\hat{a}^\dagger\).
Now, apply \(\hat{N}\) to \(\hat{a}|\nu\rangle\):
\[ \hat{N}\left(\hat{a}|\nu\rangle\right) = (\hat{a}\hat{N} - \hat{a})|\nu\rangle = \hat{a}(\nu|\nu\rangle) - \hat{a}|\nu\rangle = (\nu - 1)\left(\hat{a}|\nu\rangle\right) \]
So \(\hat{a}|\nu\rangle\) is an eigenstate of \(\hat{N}\) with eigenvalue \(\nu - 1\):
\[ \hat{a}|\nu\rangle \propto |\nu - 1\rangle \]
\(\hat{a}\) lowers the quantum number by 1. It is the annihilation operator.
5.3 Step 3: \(\hat{a}^\dagger\) Raises the Eigenvalue by 1
Similarly: \(\hat{a}^\dagger|\nu\rangle \propto |\nu + 1\rangle\).
\(\hat{a}^\dagger\) raises the quantum number by 1. It is the creation operator.
5.4 Step 4: The Spectrum Must Be \(n = 0, 1, 2, \ldots\)
Starting from any eigenstate \(|\nu\rangle\), applying \(\hat{a}\) repeatedly gives:
\[ |\nu\rangle \to |\nu - 1\rangle \to |\nu - 2\rangle \to \cdots \]
But all eigenvalues must be \(\geq 0\). So there must be a lowest state \(|0\rangle\) where the ladder terminates:
\[ \hat{a}|0\rangle = 0 \quad \text{(the ground state is annihilated by } \hat{a}\text{)} \]
From this, \(\hat{N}|0\rangle = \hat{a}^\dagger\hat{a}|0\rangle = 0\), so \(\nu_0 = 0\).
Applying \(\hat{a}\) to \(|\nu\rangle\) must land on \(|0\rangle\) after \(\nu\) steps, meaning \(\nu\) must be a non-negative integer: \(\nu = n \in \{0, 1, 2, \ldots\}\).
The energy eigenvalues are therefore:
\[ \boxed{E_n = \hbar\omega\left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \ldots} \]
We derived this with zero differential equations! Just algebra.
6 Matrix Elements and Normalization
6.1 Normalization of Ladder Operations
We need to find the proportionality constants:
Computing \(\|\hat{a}|n\rangle\|^2\):
\[ \|\hat{a}|n\rangle\|^2 = \langle n|\hat{a}^\dagger\hat{a}|n\rangle = \langle n|\hat{N}|n\rangle = n \]
So \(\hat{a}|n\rangle = \sqrt{n}|n-1\rangle\).
Computing \(\|\hat{a}^\dagger|n\rangle\|^2\):
\[ \|\hat{a}^\dagger|n\rangle\|^2 = \langle n|\hat{a}\hat{a}^\dagger|n\rangle = \langle n|(\hat{N} + 1)|n\rangle = n + 1 \]
So \(\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle\).
\[ \boxed{\hat{a}|n\rangle = \sqrt{n}\,|n-1\rangle, \qquad \hat{a}^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle} \]
6.2 Building All States from the Ground State
Applying \(\hat{a}^\dagger\) repeatedly to \(|0\rangle\):
\[ |n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}\,|0\rangle \]
The state \(|n\rangle\) is a Fock state — a state with exactly \(n\) quanta of energy. In quantum optics, \(|n\rangle\) is a state with exactly \(n\) photons. The operation \(\hat{a}^\dagger\) “creates” one photon, and \(\hat{a}\) “destroys” one.
6.3 Matrix Elements of \(\hat{x}\) and \(\hat{p}\)
Using \(\hat{x} = \frac{x_0}{\sqrt{2}}(\hat{a} + \hat{a}^\dagger)\):
\[ \langle m|\hat{x}|n\rangle = \frac{x_0}{\sqrt{2}}\left(\sqrt{n}\,\delta_{m,n-1} + \sqrt{n+1}\,\delta_{m,n+1}\right) \]
So \(\hat{x}\) only connects adjacent levels. Similarly for \(\hat{p}\).
Using \(\hat{p} = \frac{i\hbar}{\sqrt{2}\,x_0}(\hat{a}^\dagger - \hat{a})\):
\[ \langle m|\hat{p}|n\rangle = \frac{i\hbar}{\sqrt{2}\,x_0}\left(\sqrt{n+1}\,\delta_{m,n+1} - \sqrt{n}\,\delta_{m,n-1}\right) \]
Important: \(\langle n|\hat{x}|n\rangle = 0\) and \(\langle n|\hat{p}|n\rangle = 0\) for all \(n\). (No expectation value of position or momentum in energy eigenstates — they oscillate symmetrically.)
6.4 Matrix Elements of \(\hat{x}^2\) and \(\hat{p}^2\)
\[ \hat{x}^2 = \frac{x_0^2}{2}(\hat{a} + \hat{a}^\dagger)^2 = \frac{x_0^2}{2}\left(\hat{a}^2 + \hat{a}^\dagger{}^2 + 2\hat{N} + 1\right) \]
Taking the diagonal (\(m = n\)) matrix element:
\[ \langle n|\hat{x}^2|n\rangle = \frac{x_0^2}{2}(2n + 1) = \left(n + \frac{1}{2}\right)x_0^2 = \frac{\hbar}{m\omega}\left(n + \frac{1}{2}\right) \]
Similarly:
\[ \langle n|\hat{p}^2|n\rangle = m\hbar\omega\left(n + \frac{1}{2}\right) \]
7 The Uncertainty Principle in the QHO
Using \(\langle\hat{x}\rangle = 0\) and \(\langle\hat{x}^2\rangle = \left(n + \frac{1}{2}\right)x_0^2\):
\[ (\Delta x)_n = \sqrt{\langle\hat{x}^2\rangle - \langle\hat{x}\rangle^2} = x_0\sqrt{n + \frac{1}{2}} \]
\[ (\Delta p)_n = \frac{\hbar}{x_0}\sqrt{n + \frac{1}{2}} \]
Therefore:
\[ \Delta x\, \Delta p = \hbar\left(n + \frac{1}{2}\right) \geq \frac{\hbar}{2} \]
The ground state \(n = 0\) gives:
\[ (\Delta x)_0\, (\Delta p)_0 = \frac{\hbar}{2} \]
The ground state of the QHO saturates the Heisenberg uncertainty principle!
\[(\Delta x)_0\,(\Delta p)_0 = \frac{\hbar}{2}\]
This is the minimum uncertainty product. The ground state is a minimum uncertainty state.
8 Commutators Involving \(\hat{H}\)
Using \(\hat{H} = \hbar\omega(\hat{N} + 1/2)\) and \([\hat{N}, \hat{a}] = -\hat{a}\):
\[ [\hat{H},\, \hat{a}] = -\hbar\omega\,\hat{a} \]
\[ [\hat{H},\, \hat{a}^\dagger] = +\hbar\omega\,\hat{a}^\dagger \]
\[ [\hat{H},\, \hat{x}] = \frac{i\hbar}{m}\hat{p} \qquad [\hat{H},\, \hat{p}] = -im\omega^2\hbar\,\hat{x} \]
These commutators are the quantum analog of Hamilton’s equations of motion. They will be crucial for time evolution.
9 Summary: The Ladder Operator Toolbox
| Object | Definition / Formula |
|---|---|
| Annihilation operator | \(\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right)\) |
| Creation operator | \(\hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)\) |
| Number operator | \(\hat{N} = \hat{a}^\dagger\hat{a}\) |
| Hamiltonian | \(\hat{H} = \hbar\omega(\hat{N} + \frac{1}{2})\) |
| Fundamental commutator | \([\hat{a}, \hat{a}^\dagger] = 1\) |
| Action on Fock states | \(\hat{a}|n\rangle = \sqrt{n}|n-1\rangle\), \(\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle\) |
| \(\hat{x}\) in terms of \(\hat{a}\) | \(\hat{x} = \frac{x_0}{\sqrt{2}}(\hat{a} + \hat{a}^\dagger)\) |
| \(\hat{p}\) in terms of \(\hat{a}\) | \(\hat{p} = \frac{i\hbar}{\sqrt{2}x_0}(\hat{a}^\dagger - \hat{a})\) |
| Build \(|n\rangle\) from \(|0\rangle\) | \(|n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle\) |
Before moving on, make sure you can:
- Define \(\hat{a}\) and \(\hat{a}^\dagger\) and invert to get \(\hat{x}\) and \(\hat{p}\).
- Prove \([\hat{a}, \hat{a}^\dagger] = 1\) from \([\hat{x}, \hat{p}] = i\hbar\).
- Express \(\hat{H}\) in terms of \(\hat{N}\).
- Prove that \(\hat{a}|n\rangle \propto |n-1\rangle\) using commutators alone.
- Write down \(\hat{a}|n\rangle = \sqrt{n}|n-1\rangle\) and \(\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle\).
- Compute \(\langle n|\hat{x}^2|n\rangle\) and use it to find \(\Delta x\) for eigenstate \(|n\rangle\).
Bonus challenge: Compute \(\langle n|\hat{x}^4|n\rangle\) using the ladder operators. This appears in perturbation theory.