Ciro Santilli @cirosantilli 37

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Schrödinger picture  Updated 2025-05-07  +Created 1970-01-01

To better understand the discussion below, the best thing to do is to read it in parallel with the simplest possible example: Schrödinger picture example: quantum harmonic oscillator.

The state of a quantum system is a unit vector in a Hilbert space.

"Making a measurement" for an observable means applying a self-adjoint operator to the state, and after a measurement is done:

the state collapses to an eigenvector of the self adjoint operator
the result of the measurement is the eigenvalue of the self adjoint operator
the probability of a given result happening when the spectrum is discrete is proportional to the modulus of the projection on that eigenvector.
For continuous spectra such as that of the position operator in most systems, e.g. Schrödinger equation for a free one dimensional particle, the projection on each individual eigenvalue is zero, i.e. the probability of one absolutely exact position is zero. To get a non-zero result, measurement has to be done on a continuous range of eigenvectors (e.g. for position: "is the particle present between x=0 and x=1?"), and you have to integrate the probability over the projection on a continuous range of eigenvalues.
In such continuous cases, the probability collapses to an uniform distribution on the range after measurement.
The continuous position operator case is well illustrated at: Video "Visualization of Quantum Physics (Quantum Mechanics) by udiprod (2017)"

Those last two rules are also known as the Born rule.

Self adjoint operators are chosen because they have the following key properties:

their eigenvalues form an orthonormal basis
they are diagonalizable

Perhaps the easiest case to understand this for is that of spin, which has only a finite number of eigenvalues. Although it is a shame that fully understanding that requires a relativistic quantum theory such as the Dirac equation.

The next steps are to look at simple 1D bound states such as particle in a box and quantum harmonic oscillator.

This naturally generalizes to Schrödinger equation solution for the hydrogen atom.

The solution to the Schrödinger equation for a free one dimensional particle is a bit harder since the possible energies do not make up a countable set.

This formulation was apparently called more precisely Dirac-von Neumann axioms, but it because so dominant we just call it "the" formulation.

Quantum Field Theory lecture notes by David Tong (2007) mentions that:

if you were to write the wavefunction in quantum field theory, it would be a functional, that is a function of every possible configuration of the field $ϕ$ .

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Solving the Schrodinger equation with the time-independent Schrödinger equation  Updated 2025-05-07  +Created 1970-01-01

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Before reading any further, you must understand heat equation solution with Fourier series, which uses separation of variables.

Once that example is clear, we see that the exact same separation of variables can be done to the Schrödinger equation. If we name the constant of the separation of variables

E

for energy, we get:

a time-only part that does not depend on space and does not depend on the Hamiltonian at all. The solution for this part is therefore always the same exponentials for any problem, and this part is therefore "boring":
$ψ_{t} (E, t) = e^{- i Et /ℏ}$
(1)
a space-only part that does not depend on time, bud does depend on the Hamiltonian:
$\hat{H} [ψ_{x} (E, x)] = E ψ_{x} x$
(2)
Since this is the only non-trivial part, unlike the time part which is trivial, this spacial part is just called "the time-independent Schrodinger equation".
Note that the $ψ$ here is not the same as the $ψ$ in the time-dependent Schrodinger equation of course, as that psi is the result of the multiplication of the time and space parts. This is a bit of imprecise terminology, but hey, physics.

Because the time part of the equation is always the same and always trivial to solve, all we have to do to actually solve the Schrodinger equation is to solve the time independent one, and then we can construct the full solution trivially.

Once we've solved the time-independent part for each possible

E

, we can construct a solution exactly as we did in heat equation solution with Fourier series: we make a weighted sum over all possible

E

to match the initial condition, which is analogous to the Fourier series in the case of the heat equation to reach a final full solution:

if there are only discretely many possible values of $E$ , each possible energy $E_{i}$ . we proceed
$\sum_{i = 0}^{\infty} = ψ_{t} (E_{i}, t) ψ_{x} (E_{i}, x) = e^{- i E_{i} t /ℏ} ψ_{x} (E_{i}, x)$
(3)
Equation 3.
Solution of the Schrodinger equation in terms of the time-independent and time dependent parts
.
and this is a solution by selecting $E_{i}$ such that at time $t = 0$ we match the initial condition:
$\sum_{i = 0}^{\infty} e^{- i E 0/ℏ} E_{i} ψ_{i} (x) = \sum_{i = 0}^{\infty} E_{i} ψ_{i} (x) = ini t ia l co n d i t i o n$
(4)
A finite spectrum shows up in many incredibly important cases:
if there are infinitely many values of E, we do something analogous but with an integral instead of a sum. This is called the continuous spectrum. One notable

The fact that this approximation of the initial condition is always possible from is mathematically proven by some version of the spectral theorem based on the fact that The Schrodinger equation Hamiltonian has to be Hermitian and therefore behaves nicely.

It is interesting to note that solving the time-independent Schrodinger equation can also be seen exactly as an eigenvalue equation where:

the Hamiltonian is a linear operator
the value of the energy E is an eigenvalue

The only difference from usual matrix eigenvectors is that we are now dealing with an infinite dimensional vector space.

Furthermore:

we immediately see from the equation that the time-independent solutions are states of deterministic energy because the energy is an eigenvalue of the Hamiltonian operator
by looking at Equation 3. "Solution of the Schrodinger equation in terms of the time-independent and time dependent parts", it is obvious that if we take an energy measurement, the probability of each result never changes with time, because it is only multiplied by a constant

Bibliography:

quantummechanics.ucsd.edu/ph130a/130_notes/node124.html from quantum physics by Jim Branson (2003)

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