Schrödinger Picture

The Schrödinger picture is the formulation in which quantum states carry the main time dependence:

$$\lvert\psi_S(t)\rangle =U(t,t_0)\lvert\psi_S(t_0)\rangle.$$

Observables are usually fixed unless they have explicit time dependence.

For a compact Core Formalism comparison with the other pictures, see Pictures of Motion Overview.

State evolution in a fixed representation

The state satisfies

$$i\hbar\frac{d}{dt}\lvert\psi_S(t)\rangle =H_S(t)\lvert\psi_S(t)\rangle.$$

An observable $A_S$ has no picture-induced time dependence. If it changes with time, that dependence is explicit in the observable itself, not caused by moving to another picture.

Expectation Values

The expectation value is

$$\langle A\rangle_t = \langle\psi_S(t)\rvert A_S\lvert\psi_S(t)\rangle.$$

The Heisenberg picture gives the same number by moving the time dependence into $A_H(t)$.

Coordinate Representation

In position representation,

$$\psi(\mathbf r,t) =\langle \mathbf r\lvert\psi_S(t)\rangle.$$

For a spinless nonrelativistic particle with a scalar potential and no vector potential,

$$i\hbar\frac{\partial\psi}{\partial t} = \left[ -\frac{\hbar^2}{2m}\nabla^2 +V(\mathbf r,t) \right]\psi.$$

This is why the Schrödinger picture is often the first language of wave mechanics.

When It Is Most Useful

Use the Schrödinger picture when:

• the state vector or wavefunction itself is the object of interest;
• the Hamiltonian is diagonalized and states evolve by phases;
• numerical propagation of states is the main task;
• boundary conditions are easiest to impose on wavefunctions;
• the problem is a canonical wave-mechanics system.

Example: Energy-Basis Evolution

For a time-independent Hamiltonian,

$$\lvert\psi_S(0)\rangle =\sum_n c_n\lvert E_n\rangle$$

evolves as

$$\lvert\psi_S(t)\rangle = \sum_n c_n e^{-iE_nt/\hbar}\lvert E_n\rangle.$$

The basis vectors are fixed. The coefficients acquire time-dependent phases. In a two-level example with $H=\hbar\omega\sigma_z/2$ and initial state $\lvert+x\rangle$, this gives

$$\lvert\psi_S(t)\rangle=\frac{1}{\sqrt2} \left(e^{-i\omega t/2}\lvert+z\rangle+e^{i\omega t/2}\lvert-z\rangle\right).$$

The fixed projector $\Pi_{+x}=\lvert+x\rangle\langle+x\rvert$ then gives $P(+x;t)=\cos^2(\omega t/2)$. A relative phase has changed a measurable probability.

Common Mistakes

• Thinking the Schrödinger picture is more physically real than the Heisenberg picture.
• Treating stationary-state phases as unimportant in superpositions.
• Forgetting explicit time dependence in an observable.
• Confusing a wavefunction representation with the whole state.
• Assuming every problem is easiest in the Schrödinger picture.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.

Exercises

For the two-level state above, calculate $\langle\sigma_x\rangle$ and the probability of $+z$. Explain why one changes while the other does not.

Solution

The off-diagonal entries of $\sigma_x$ connect the two energy amplitudes, giving $\langle\sigma_x\rangle=\cos\omega t$. The $+z$ probability is the squared magnitude of its coefficient, $1/2$. Energy-basis populations stay fixed while interference in another basis changes.