技术问询：为何定态波函数是哈密顿算符的本征函数？

Great question—this is a core, foundational concept in quantum mechanics, so let's unpack it step by step starting from first principles to make it concrete.

First, let's clarify what a stationary state actually means: it's a quantum state where all measurable properties (like the probability of finding the particle in a specific region, or the average value of energy) don't change over time. No time dependence in observable quantities—that's the defining feature of a stationary state.

Now, let's tie this to the time-dependent Schrödinger equation, which governs how quantum states evolve over time:

iℏ ∂ψ(x,t)/∂t = Ĥψ(x,t)

Here, ℏ is the reduced Planck constant, ψ(x,t) is the full time-dependent wave function, and Ĥ is the Hamiltonian operator (it encodes the total energy of the system: kinetic energy + potential energy).

To find stationary states, we assume the wave function can be separated into a spatial component and a time component: ψ(x,t) = ψ(x)φ(t). Plugging this into the Schrödinger equation gives us:

iℏ ψ(x) dφ(t)/dt = φ(t) Ĥψ(x)

If we divide both sides by ψ(x)φ(t), we get a key split:

iℏ/φ(t) * dφ(t)/dt = (Ĥψ(x))/ψ(x)

The left side only depends on time, and the right side only depends on space. The only way these two can be equal for all values of time and space is if both sides equal the same constant. Let's call this constant E—we'll soon see this is the total energy of the state.

Solving the time-dependent part #

The time equation simplifies to:

dφ(t)/dt = -iE/ℏ φ(t)

The solution is φ(t) = e^(-iEt/ℏ). A critical detail here: the modulus squared of this time-dependent term is |e^(-iEt/ℏ)|² = 1 (since the exponential of an imaginary number has a magnitude of 1, it doesn't affect the size of the wave function over time).

Solving the spatial part #

The spatial equation becomes:

Ĥψ(x) = Eψ(x)

This is exactly the definition of an eigenfunction equation! Here, ψ(x) is an eigenfunction of the Hamiltonian operator Ĥ, and E is the corresponding eigenvalue (which we now confirm is the fixed energy of the state).

Why this makes it a stationary state #

Now let's verify that this state meets the stationary state criteria:

• Probability density: |ψ(x,t)|² = |ψ(x)|² * |φ(t)|² = |ψ(x)|² * 1. This doesn't depend on time at all—so the probability of finding the particle in any region stays constant over time.
• Observable expectations: For any observable operator Â, the expectation value ⟨Â⟩ = ∫ψ*(x,t)Âψ(x,t) dx = ∫ψ*(x)Âψ(x) dx * |φ(t)|². Again, since |φ(t)|² = 1, the average value of the observable doesn't change with time.

A quick note on superpositions #

You might ask: what if we take a superposition of two Hamiltonian eigenfunctions with different energies, like ψ(x,t) = aψ₁(x)e^(-iE₁t/ℏ) + bψ₂(x)e^(-iE₂t/ℏ)? The probability density here would include an interference term: 2Re(a*bψ₁*ψ₂e^(i(E₁-E₂)t/ℏ)) that oscillates with time. This means the state isn't stationary—only single eigenstates (or degenerate eigenstates with the same energy) produce the time-independent behavior that defines a stationary state.

To wrap it up: stationary states require no time dependence in observables, which forces the wave function to separate into spatial and time components that satisfy the Hamiltonian's eigenfunction equation. It's a direct, logical consequence of the Schrödinger equation and the core definition of what makes a quantum state "stationary."

内容的提问来源于stack exchange，提问作者user178466