Unitarity (physics)

In quantum mechanics, every state is described as a vector in Hilbert space. When a measurement is performed, it is convenient to describe this space using a vector basis in which every basis vector has a defined result of the measurement - e.g., a vector basis of defined momentum in case momentum is measured. The measurement operator is diagonal in this basis.[3]

The probability to get a particular measured result depends on the probability amplitude, that is given by the inner product of the physical state with the basis vectors that diagonalize the measurement operator. For a physical state that is measured after it has evolved in time, the probability amplitude can be described either by the inner product of the physical state ${\displaystyle |\psi \rangle }$ after time evolution with the relevant basis vectors ${\displaystyle \{|\phi _{i}\rangle \}}$, or equivalently by the inner product of the physical state with the basis vectors that are evolved backwards in time. Using the time evolution operator ${\displaystyle e^{-i{\hat {H}}t/\hbar }}$, we have:[4]

${\displaystyle \left\langle \phi _{i}\left|e^{-i{\hat {H}}t/\hbar }\psi \right.\right\rangle =\left\langle \left.e^{-i{\hat {H}}(-t)/\hbar }\phi _{i}\right|\psi \right\rangle }$

But by definition of Hermitian conjugation, this is also:

${\displaystyle \left\langle \phi _{i}\left|e^{-i{\hat {H}}t/\hbar }\psi \right.\right\rangle =\left\langle \left.\phi _{i}\left(e^{-i{\hat {H}}t/\hbar }\right)^{\dagger }\right|\psi \right\rangle =\left\langle \left.\phi _{i}e^{-i{\hat {H}}^{\dagger }(-t)/\hbar }\right|\psi \right\rangle }$

Since these equalities are true for every two vectors, we get

${\displaystyle {\hat {H}}^{\dagger }={\hat {H}}}$

This means that the Hamiltonian is Hermitian and the time evolution operator ${\displaystyle e^{-i{\hat {H}}t/\hbar }}$ is unitary.

Since by Born rule the norm determines the probability to get a particular result in a measurement, unitarity together with Born rule guarantees the sum of probabilities is always one. Furthermore, unitarity together with Born rule implies that the measurement operators in Heisenberg picture indeed describe how the measurement results are expected to evolve in time. This point is further stressed by a hypothetical counter-example: Consider a case of non-unitarity, when one gets a different probability by measuring some operator ${\displaystyle {\cal {{O}(t_{1})}}}$ (in the Heisenberg picture) at time t₁, compared to taking the same measurement, taking into account the time evolution, at time t₂, so that at this time ${\displaystyle {\cal {{O}(t_{2})}}}$ is measured. Through multiple such measurements, one may then construct an experiment where the probability of one result R₁ would be arbitrarily close to 100% if taken at time t₁, but the probability of a different result R₂ would be arbitrarily close to 100% if taken at time t₂. This leads to inconsistency, at least in some interpretations of quantum mechanics.

For example, say Alice and Bob are performing measurements on the same system at different times. Alice is measuring at time t₁ and Bob at time t₂. according to the many-worlds interpretation, Bob will almost surely find himself in a world where the result was R₂. But then, when Bob meets Alice, Alice must have also measured R₂. Thus Alice would tell Bob she had measured a very unrealistic result, with probability arbitrarily close to 0%. Thus in such a scenario, physicists report they have had very unrealistic results, and the notion of probability breaks down.

That the time evolution operator is unitary, is equivalent to the Hamiltonian being Hermitian. Equivalently, this means that the possible measured energies, which are the eigenvalues of the Hamiltonian, are always real numbers.

Unitarity of the S-matrix implies, among other things, the optical theorem. This can be seen as follows:[5]

The S-matrix can be written as:

${\displaystyle S=1+iT}$

where ${\displaystyle T}$ is the part of the S-matrix that is due to interactions; e.g. ${\displaystyle T=0}$ just implies the S-matrix is 1, no interaction occur and all states remain unchanged.

Unitarity of the S-matrix:

${\displaystyle S^{\dagger }S=1}$

is then equivalent to:

${\displaystyle -i\left(T-T^{\dagger }\right)=T^{\dagger }T}$

The left-hand side is twice the imaginary part of the S-matrix. In order to see what the right-hand side is, let us look at any specific element of this matrix, e.g. between some initial state ${\displaystyle I\rangle }$ and final state ${\displaystyle \langle F}$, each of which may include many particles. The matrix element is then:

${\displaystyle \left\langle F\left|T^{\dagger }T\right|I\right\rangle =\sum _{i}\left\langle F\left|T^{\dagger }A_{i}\right|\right\rangle \left\langle A_{i}|T|I\right\rangle }$

where {A_i} is the set of possible on-shell states - i.e. momentum states of particles (or bound complex of particles) at infinity.

Thus, twice the imaginary part of the S-matrix, is equal to a sum representing products of contributions from all the scatterings of the initial state of the S-matrix to another physical state at infinity, with the scatterings of the latter to the final state of the S-matrix. Since the imaginary part of the S-matrix can be calculated by virtual particles appearing in intermediate states of the Feynman diagrams, it follows that these virtual particles must only consist of real particles that may also appear as final states. The mathematical machinery which is used to ensure this includes gauge symmetry and sometimes also Faddeev–Popov ghosts.

According to the optical theorem, the probability amplitude M for any scattering process must obey

${\displaystyle |M|^{2}=2\operatorname {Im} (M)}$

Similar unitarity bounds imply that the amplitudes and cross section cannot increase too much with energy or they must decrease as quickly as a certain formula dictates.