Transition Amplitudes

A transition amplitude is a complex number that answers the question:

if the system starts in one quantum alternative, what is the amplitude for finding another alternative after time evolution?

For normalized states $\lvert a\rangle$ and $\lvert b\rangle$, the basic object is the matrix element

$$\mathcal A_{b\leftarrow a}(t_b,t_a) = \langle b\rvert U(t_b,t_a)\lvert a\rangle.$$

Here $U(t_b,t_a)$ is the time-evolution operator. The amplitude is not itself a probability. It is the quantity whose phase and magnitude combine with other amplitudes before the Born rule is applied.

What Is a Transition Amplitude?

Suppose a closed system is prepared at time $t_a$ in the state $\lvert a\rangle$. The state at the later time $t_b$ is

$$\lvert\psi(t_b)\rangle = U(t_b,t_a)\lvert a\rangle.$$

The amplitude for a later measurement to find the normalized state $\lvert b\rangle$ is the projection of this evolved state onto $\lvert b\rangle$:

$$\langle b\rvert\psi(t_b)\rangle = \langle b\rvert U(t_b,t_a)\lvert a\rangle.$$

The notation $b\leftarrow a$ is only a reminder of the experimental story: start with $a$, evolve, test for $b$. The operator acts on the ket to its right:

$$\lvert a\rangle \xrightarrow{\;U(t_b,t_a)\;} U(t_b,t_a)\lvert a\rangle \xrightarrow{\;\langle b\rvert\;} \mathcal A_{b\leftarrow a}.$$

If $t_b=t_a$, then $U=I$ and the transition amplitude reduces to the ordinary overlap

$$\mathcal A_{b\leftarrow a}(t_a,t_a) = \langle b\vert a\rangle.$$

Thus ordinary probability amplitudes are equal-time transition amplitudes.

From Amplitudes to Probabilities

For a normalized initial state $\lvert a\rangle$ and a final nondegenerate outcome represented by $\lvert b\rangle$, the transition probability is

$$P(b,t_b\mid a,t_a) = \left| \langle b\rvert U(t_b,t_a)\lvert a\rangle \right|^2.$$

This is the Born rule applied after time evolution. The amplitude contains phase information; the probability does not.

For a final subspace represented by a projector $P_B$, the probability is

$$P(B,t_b\mid a,t_a) = \left\| P_BU(t_b,t_a)\lvert a\rangle \right\|^2.$$

Equivalently,

$$P(B,t_b\mid a,t_a) = \langle a\rvert U^\dagger(t_b,t_a)P_BU(t_b,t_a) \lvert a\rangle.$$

For an initial density operator $\rho_a$, the same statement is

$$P(B,t_b) = \operatorname{Tr}\!\left[ P_BU(t_b,t_a)\rho_a U^\dagger(t_b,t_a) \right].$$

The single matrix element $\langle b\rvert U\lvert a\rangle$ is the cleanest formula, but projectors and density operators are the safer language for degeneracy, mixed preparation, and coarse-grained outcomes.

Basis Dependence

A transition amplitude depends on the initial and final kets used to define it. A change of basis changes the collection of matrix elements, although physical probabilities for the same projectors do not change.

Even a phase convention changes an amplitude. If

$$\lvert a'\rangle=e^{i\phi_a}\lvert a\rangle, \qquad \lvert b'\rangle=e^{i\phi_b}\lvert b\rangle,$$

then

$$\langle b'\rvert U\lvert a'\rangle = e^{i(\phi_a-\phi_b)} \langle b\rvert U\lvert a\rangle.$$

The probability is invariant:

$$\left| \langle b'\rvert U\lvert a'\rangle \right|^2 = \left| \langle b\rvert U\lvert a\rangle \right|^2.$$

This is why amplitudes are not directly observed in isolation. Relative phases are physical when amplitudes are combined, but an arbitrary phase convention for a single basis ket is not.

Position-Space Transition Amplitudes

The coordinate-space propagator kernel is a transition amplitude between position eigenkets:

$$K(x_b,t_b;x_a,t_a) = \langle x_b\rvert U(t_b,t_a)\lvert x_a\rangle.$$

This is the canonical home of the notation $K$. The detailed kernel properties are developed in Propagator Kernel.

Because $\lvert x\rangle$ is delta-normalized rather than square-normalized, $K$ is not a probability amplitude for a normalizable initial state by itself. It is an integral kernel. Given an initial wavefunction $\psi(x_a,t_a)$,

$$\psi(x_b,t_b) = \int dx_a\, K(x_b,t_b;x_a,t_a)\psi(x_a,t_a).$$

Only after this integral has produced a wavefunction should one form the final position probability density:

$$p(x_b,t_b) = \lvert\psi(x_b,t_b)\rvert^2.$$

For configuration spaces other than the line, the measure must match the Hilbert-space inner product. On a circle, sphere, half-line, box, or many-particle configuration space, the symbol $dx_a$ is replaced by the appropriate measure and boundary conditions.

Energy-Basis Transition Amplitudes

For a time-independent Hamiltonian with energy eigenstates

$$H\lvert n\rangle=E_n\lvert n\rangle,$$

the evolution operator acts diagonally:

$$U(t_b,t_a)\lvert n\rangle = e^{-iE_nT/\hbar}\lvert n\rangle, \qquad T=t_b-t_a.$$

Therefore

$$\langle m\rvert U(t_b,t_a)\lvert n\rangle = \delta_{mn}e^{-iE_nT/\hbar}.$$

In a stationary Hamiltonian, energy eigenstates do not transition into different energy eigenstates. They acquire phases.

More general amplitudes are obtained by inserting an energy resolution of identity. For arbitrary normalized states,

$$\langle b\rvert U(T)\lvert a\rangle = \sum_n \langle b\vert n\rangle e^{-iE_nT/\hbar} \langle n\vert a\rangle,$$

with integrals added for continuous spectra. The same idea gives the spectral form of a propagator kernel:

$$K(x_b,T;x_a,0) = \sum_n \psi_n(x_b)\psi_n^*(x_a) e^{-iE_nT/\hbar}.$$

Nontrivial transitions between energy eigenstates require something else: a time-dependent Hamiltonian, an interaction picture split, a perturbation, or a comparison between different initial and final Hamiltonians.

Composition of Amplitudes

Time-evolution operators compose:

$$U(t_b,t_a) = U(t_b,t_c)U(t_c,t_a).$$

Insert a complete orthonormal basis $\{\lvert c\rangle\}$ at the intermediate time:

$$I=\sum_c\lvert c\rangle\langle c\rvert.$$

Then

$$\begin{aligned} \langle b\rvert U(t_b,t_a)\lvert a\rangle &= \sum_c \langle b\rvert U(t_b,t_c)\lvert c\rangle \langle c\rvert U(t_c,t_a)\lvert a\rangle. \end{aligned}$$

The rule is sum amplitudes over unobserved alternatives. If the intermediate alternative is actually measured and the result is recorded, then probabilities are combined according to the measurement protocol instead. This distinction is the source of much quantum interference.

For continuous intermediate labels, the sum becomes an integral. In position space,

$$K(x_b,t_b;x_a,t_a) = \int dx_c\, K(x_b,t_b;x_c,t_c) K(x_c,t_c;x_a,t_a).$$

This is the kernel composition law.

Path-Integral Preview

The path integral starts from the composition rule for transition amplitudes. Split the time interval into many short steps and insert many position resolutions of identity:

$$I=\int dx_j\,\lvert x_j\rangle\langle x_j\rvert.$$

One obtains an integral over intermediate positions,

$$\int dx_{N-1}\cdots dx_1 \prod_{j=0}^{N-1} K(x_{j+1},t_{j+1};x_j,t_j).$$

For ordinary nonrelativistic systems, taking a carefully regulated continuum limit leads formally to

$$K(x_b,t_b;x_a,t_a) = \int_{x(t_a)=x_a}^{x(t_b)=x_b} \mathcal D x\, e^{iS[x]/\hbar}.$$

This expression should be read as an amplitude construction, not a probability distribution over paths. The path-integral chapter explains the approximation, limiting procedure, and measure issues in From Propagators to Path Integrals.

Example: Spin Precession Amplitudes

Let

$$H=\frac{\hbar\omega}{2}\sigma_z, \qquad U(t)=e^{-i\omega t\sigma_z/2}.$$

Prepare the spin in $\lvert +x\rangle$ and later measure in the $x$ basis. Since

$$\lvert +x\rangle = \frac{1}{\sqrt2} \left( \lvert +z\rangle+\lvert -z\rangle \right),$$

one finds

$$\langle +x\rvert U(t)\lvert +x\rangle = \cos\frac{\omega t}{2},$$

and

$$\langle -x\rvert U(t)\lvert +x\rangle = -i\sin\frac{\omega t}{2}.$$

Therefore

$$P(+x,t\mid +x,0) = \cos^2\frac{\omega t}{2}, \qquad P(-x,t\mid +x,0) = \sin^2\frac{\omega t}{2}.$$

The relative phase between the $z$-basis energy amplitudes becomes an observable oscillation in the $x$-basis probabilities.

Common Mistakes

• Treating a transition amplitude as a probability rather than taking an absolute square after amplitudes have been combined.
• Summing probabilities over unobserved intermediate alternatives instead of summing amplitudes.
• Forgetting that amplitudes depend on basis and phase conventions, while probabilities for fixed projectors do not.
• Treating $K(x_b,t_b;x_a,t_a)$ as a probability density for a particle with exact initial position.
• Ignoring degeneracy and coarse graining; projectors are safer than individual basis kets when outcomes are not one-dimensional.
• Assuming a time-independent Hamiltonian causes transitions between its own energy eigenstates. It only gives phase evolution in that basis.

Cross-Links

• Probability Amplitudes
• Born Rule
• Transition Probabilities
• Time-Evolution Operator
• Unitarity and Conservation of Probability
• Change of Basis
• Coordinate Representation
• Composition Law
• Propagator Kernel
• Propagator Table
• From Propagators to Path Integrals

References

• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, 1965.
• L. S. Schulman, Techniques and Applications of Path Integration, Wiley, 1981.

Exercises

1. Show that rephasing the initial and final kets changes a transition amplitude by a phase but leaves the transition probability unchanged.

Solution

Let

$$\lvert a'\rangle=e^{i\phi_a}\lvert a\rangle, \qquad \lvert b'\rangle=e^{i\phi_b}\lvert b\rangle.$$

Then

$$\langle b'\rvert U\lvert a'\rangle = e^{-i\phi_b}e^{i\phi_a} \langle b\rvert U\lvert a\rangle.$$

Taking the absolute square removes the phase:

$$\left| \langle b'\rvert U\lvert a'\rangle \right|^2 = \left| \langle b\rvert U\lvert a\rangle \right|^2.$$

2. For a time-independent Hamiltonian, show that energy eigenstates only acquire phases.

Solution

If

$$H\lvert n\rangle=E_n\lvert n\rangle,$$

then

$$U(T)\lvert n\rangle = e^{-iHT/\hbar}\lvert n\rangle = e^{-iE_nT/\hbar}\lvert n\rangle.$$

Therefore

$$\langle m\rvert U(T)\lvert n\rangle = e^{-iE_nT/\hbar}\langle m\vert n\rangle = \delta_{mn}e^{-iE_nT/\hbar}.$$

3. Derive the composition law for transition amplitudes using an intermediate orthonormal basis.

Solution

Start from

$$\langle b\rvert U(t_b,t_a)\lvert a\rangle = \langle b\rvert U(t_b,t_c)U(t_c,t_a)\lvert a\rangle.$$

Insert

$$I=\sum_c\lvert c\rangle\langle c\rvert$$

between the two time-evolution operators:

$$\langle b\rvert U(t_b,t_a)\lvert a\rangle = \sum_c \langle b\rvert U(t_b,t_c)\lvert c\rangle \langle c\rvert U(t_c,t_a)\lvert a\rangle.$$

The intermediate alternatives are summed as amplitudes because no intermediate outcome has been recorded.

4. In the spin-precession example, verify the two $x$-basis amplitudes.

Solution

Use

$$U(t)\lvert +z\rangle = e^{-i\omega t/2}\lvert +z\rangle, \qquad U(t)\lvert -z\rangle = e^{i\omega t/2}\lvert -z\rangle.$$

With

$$\lvert +x\rangle = \frac{1}{\sqrt2}(\lvert +z\rangle+\lvert -z\rangle), \qquad \lvert -x\rangle = \frac{1}{\sqrt2}(\lvert +z\rangle-\lvert -z\rangle),$$

one obtains

$$\langle +x\rvert U(t)\lvert +x\rangle = \frac12 \left( e^{-i\omega t/2}+e^{i\omega t/2} \right) = \cos\frac{\omega t}{2},$$

and

$$\langle -x\rvert U(t)\lvert +x\rangle = \frac12 \left( e^{-i\omega t/2}-e^{i\omega t/2} \right) = -i\sin\frac{\omega t}{2}.$$

5. Explain why $K(x_b,t_b;x_a,t_a)$ should not be interpreted as a probability density.

Solution

The kernel is a matrix element between delta-normalized position kets:

$$K(x_b,t_b;x_a,t_a) = \langle x_b\rvert U(t_b,t_a)\lvert x_a\rangle.$$

The state $\lvert x_a\rangle$ is not a normalizable physical state, and $K$ is an integral kernel. For a normalizable initial wavefunction,

$$\psi(x_b,t_b) = \int dx_a\, K(x_b,t_b;x_a,t_a)\psi(x_a,t_a).$$

The probability density is then $\lvert\psi(x_b,t_b)\rvert^2$, not $\lvert K(x_b,t_b;x_a,t_a)\rvert^2$ by itself.