Vacuum State

The vacuum state is the normalized no-particle vector in Fock space. It is usually written

$$\lvert0\rangle,$$

or, when one wants to avoid confusion with the zero vector, $\lvert\Omega\rangle$.

The vacuum is a state, not the absence of a state. It has norm one:

$$\langle0\vert0\rangle=1.$$

It belongs to the $N=0$ sector of Fock space and serves as the reference vector from which creation operators build occupation-number states.

Vacuum in Fock Space

For a one-particle Hilbert space $\mathcal h$, bosonic and fermionic Fock spaces both contain a zero-particle sector:

$$\operatorname{Sym}^0\mathcal h \cong \mathbb C, \qquad \wedge^0\mathcal h \cong \mathbb C.$$

The vacuum spans this one-dimensional sector. Schematically,

$$\mathcal F(\mathcal h) = \mathbb C\lvert0\rangle \oplus \mathcal h \oplus \mathcal H_{2} \oplus \cdots,$$

where $\mathcal H_2$ is the symmetric or antisymmetric two-particle sector depending on whether the particles are bosons or fermions.

The vacuum has no occupied modes:

$$\lvert0\rangle = \lvert0_1,0_2,0_3,\ldots\rangle.$$

For every mode number operator $N_i$,

$$N_i\lvert0\rangle=0,$$

and therefore

$$N_{\mathrm{tot}}\lvert0\rangle=0.$$

Not the Zero Vector

The zero vector is written $0$ and has norm zero:

$$\lVert 0\rVert=0.$$

It does not represent a physical state. The vacuum vector $\lvert0\rangle$ has norm one and does represent a physical state: the state with no particles or excitations in the chosen Fock-space description.

The distinction is visible through creation operators. For a bosonic creation operator $a_i^\dagger$,

$$a_i^\dagger\lvert0\rangle = \lvert1_i\rangle,$$

while

$$a_i^\dagger 0=0.$$

Acting on the vacuum creates a one-particle state. Acting on the zero vector gives the zero vector again.

Annihilation Operators

An annihilation operator removes one quantum from a mode. Since the vacuum has no particles to remove,

$$a_i\lvert0\rangle=0$$

for bosonic modes, and

$$c_i\lvert0\rangle=0$$

for fermionic modes.

This equation does not mean the vacuum is the zero vector. It means that trying to annihilate a particle from an empty mode gives the zero vector as the result.

The vacuum is also the starting point for normalized number states. For bosons,

$$\lvert n_i\rangle_B = \frac{(a_i^\dagger)^n}{\sqrt{n!}} \lvert0\rangle,$$

and for fermions in a fixed ordered mode basis,

$$\lvert n_1,\ldots,n_M\rangle_F = (c_1^\dagger)^{n_1} \cdots (c_M^\dagger)^{n_M} \lvert0\rangle,$$

with $n_i\in\{0,1\}$.

Vacuum and Mode Basis

Occupation numbers depend on a chosen mode basis. The vacuum is special because it is empty in every mode basis obtained by a unitary change of one-particle modes.

If

$$b_\alpha^\dagger = \sum_i U_{i\alpha}a_i^\dagger$$

defines a new orthonormal bosonic mode basis, then the corresponding annihilation operators satisfy

$$b_\alpha = \sum_i U_{i\alpha}^*a_i.$$

Since every $a_i$ annihilates the vacuum,

$$b_\alpha\lvert0\rangle=0.$$

Thus the no-particle state remains the no-particle state under ordinary one-particle basis changes. What changes is how non-vacuum states are decomposed into mode occupations.

Vacuum Versus Ground State

The vacuum is the no-particle state. The ground state is the lowest-energy state of a Hamiltonian. They are not the same concept.

For a simple number-conserving nonrelativistic Hamiltonian with positive one-particle energies,

$$H = \sum_i \epsilon_i\,a_i^\dagger a_i, \qquad \epsilon_i>0,$$

the vacuum is an energy eigenstate:

$$H\lvert0\rangle=0.$$

In that model it is also the ground state.

But this is not automatic. A Hamiltonian may include a constant vacuum energy,

$$H = E_{\mathrm{vac}}I + \sum_i \epsilon_i\,a_i^\dagger a_i,$$

so that

$$H\lvert0\rangle = E_{\mathrm{vac}}\lvert0\rangle.$$

The vacuum is still the no-particle state, even if its assigned energy is not zero.

In many-body physics, the ground state at fixed density is usually not the empty vacuum. A filled Fermi sea, a Bose condensate, or an interacting many-particle ground state contains particles relative to the original particle operators. Sometimes one redefines quasiparticle operators so that the many-body ground state is a quasiparticle vacuum. That is a new vacuum relative to new excitations, not the original no-particle state.

Harmonic-Oscillator Connection

For one harmonic oscillator,

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right).$$

The oscillator ground state satisfies

$$a\lvert0\rangle=0,$$

and

$$H\lvert0\rangle = \frac12\hbar\omega \lvert0\rangle.$$

Thus the oscillator vacuum has nonzero zero-point energy. In quantum optics and field-mode language, this state is often called the vacuum of that mode. The word “vacuum” here means no quanta of that oscillator mode, not zero energy.

Vacuum in QFT Preview

In relativistic quantum field theory, the vacuum is far richer than the empty state of a finite nonrelativistic Fock space. It is usually defined as the lowest-energy state, often with symmetry requirements such as translation or Poincare invariance. It can have nonzero correlation functions and zero-point contributions.

The particle interpretation of a QFT state can also depend on the field modes used, especially in curved spacetime or for accelerated observers. Those issues belong to QFT and mathematical-physics treatments. The important lesson for this volume is more modest:

• in nonrelativistic Fock space, the vacuum is the normalized $N=0$ state;
• creation operators build particles or excitations from it;
• annihilation operators kill it;
• its energy depends on the Hamiltonian convention;
• a many-body ground state need not be the no-particle vacuum.

Common Mistakes

• Confusing the vacuum vector $\lvert0\rangle$ with the zero vector $0$.
• Assuming the vacuum must have zero energy.
• Assuming the vacuum is always the physical ground state of the problem.
• Forgetting that a filled Fermi sea can be a ground state with many particles.
• Calling a coherent state or condensate a vacuum of the original particle operators.
• Forgetting that a quasiparticle vacuum is defined relative to new creation and annihilation operators.
• Importing QFT vacuum intuitions into elementary nonrelativistic Fock space without stating the context.

Cross-Links

• Occupation-Number Basis
• Bosonic Fock Space
• Fermionic Fock Space
• Number States
• Mode Occupations
• Particle-Number Superselection Preview
• Fock Space Examples
• Creation and Annihilation Operators
• Number Operators
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Entanglement in QFT Preview
• Ladder-Operator Solution
• Harmonic Oscillator to Fields

References

• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.

Exercises

1. Vacuum versus zero vector. Explain why $\lvert0\rangle$ and $0$ cannot mean the same thing.

Solution

The vacuum vector is normalized:

$$\langle0\vert0\rangle=1.$$

The zero vector has norm zero and does not represent a physical state. Also, a creation operator can create a one-particle state from the vacuum,

$$a^\dagger\lvert0\rangle=\lvert1\rangle,$$

whereas

$$a^\dagger0=0.$$

2. Annihilation of the vacuum. If $a\lvert0\rangle=0$, why does that not imply $\lvert0\rangle=0$?

Solution

The equation says that the result of applying $a$ to the vacuum is the zero vector. Linear operators can map nonzero vectors to zero. Here the physical meaning is that there is no particle in the mode for the annihilation operator to remove.

3. Two-mode vacuum. For two bosonic modes $a$ and $b$, compute $N_a\lvert0_a,0_b\rangle$ and $N_b\lvert0_a,0_b\rangle$.

Solution

The vacuum has zero occupation in both modes. Therefore

$$N_a\lvert0_a,0_b\rangle=0, \qquad N_b\lvert0_a,0_b\rangle=0.$$

4. Oscillator zero-point energy. For

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right),$$

find the vacuum energy.

Solution

Since $a^\dagger a\lvert0\rangle=0$,

$$H\lvert0\rangle = \frac12\hbar\omega\lvert0\rangle.$$

The vacuum energy is $\hbar\omega/2$.

5. Ground state versus vacuum. Why is a filled Fermi sea not the vacuum of the original electron operators?

Solution

The original electron vacuum has no electrons. A filled Fermi sea contains many occupied electron modes. It may be the ground state at fixed density or chemical potential, but it is not the no-electron state. One can define quasiparticle operators for which the filled sea behaves as a quasiparticle vacuum, but that is a different vacuum relative to different excitations.