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.

Helpful background. Bosonic Fock Space and Fermionic Fock Space define the zero-particle sector.

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

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

A normalized vector annihilated by lowering operators

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.

Creation operators build normalized number states from this reference vector.

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 empty under a unitary rotation of one-particle modes. A transformation mixing creation with annihilation is different. For example,

$$b=\cosh r\,a+\sinh r\,a^\dagger, \qquad [b,b^\dagger]=1$$

has $b\lvert0_a\rangle=\sinh r\lvert1_a\rangle\ne0$ for $r\ne0$. Its normalized $b$-vacuum is a squeezed state with $\langle a^\dagger a\rangle=\sinh^2r$. This finite-mode example illustrates why a quasiparticle vacuum can contain original particles; it is not merely a relabeling of the old empty state.

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.

Adding a constant changes the vacuum energy without changing whether it minimizes the Hamiltonian:

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

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.

The empty state need not even be an eigenstate of a number-changing Hamiltonian. A drive $H_{\rm drive}=\eta a^\dagger+\eta^*a$ takes it to $\eta\lvert1\rangle$ on application. Here $\eta$ has units of energy.

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. It also has fluctuations: with $X=(a+a^\dagger)/\sqrt2$, $\langle X\rangle=0$ but $\langle X^2\rangle=1/2$. The word “vacuum” here means no quanta of that oscillator mode, not zero energy or vanishing correlations.

Exercises

1. 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.

2. 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.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• 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.
• M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison-Wesley, 1995.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.