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Solving the Schrodinger equation with the time-independent Schrödinger equation  Updated 2024-12-15  +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":
$ψ_{t} (E, t) = e^{- i E t / ℏ}$
(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:
$\sum_{i = 0}^{\infty} e^{- i E 0 / ℏ} E_{i} ψ_{i} (x) = \sum_{i = 0}^{\infty} E_{i} ψ_{i} (x) = i n i t i a l c o n d i t i o n$
(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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Uncertainty principle  Updated 2024-12-15  +Created 1970-01-01

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The wave equation contains the entire state of a particle.

From mathematical formulation of quantum mechanics remember that the wave equation is a vector in Hilbert space.

And a single vector can be represented in many different ways in different basis, and two of those ways happen to be the position and the momentum representations.

More importantly, position and momentum are first and foremost operators associated with observables: the position operator and the momentum operator. And both of their eigenvalue sets form a basis of the Hilbert space according to the spectral theorem.

When you represent a wave equation as a function, you have to say what the variable of the function means. And depending on weather you say "it means position" or "it means momentum", the position and momentum operators will be written differently.

This is well shown at: Video "Visualization of Quantum Physics (Quantum Mechanics) by udiprod (2017)".

Furthermore, the position and momentum representations are equivalent: one is the Fourier transform of the other: position and momentum space. Remember that notably we can always take the Fourier transform of a function in

L^{2}

due to Carleson's theorem.

Then the uncertainty principle follows immediately from a general property of the Fourier transform: en.wikipedia.org/w/index.php?title=Fourier_transform&oldid=961707157#Uncertainty_principle

In precise terms, the uncertainty principle talks about the standard deviation of two measures.

We can visualize the uncertainty principle more intuitively by thinking of a wave function that is a real flat top bump function with a flat top in 1D. We can then change the width of the support, but when we do that, the top goes higher to keep probability equal to 1. The momentum is 0 everywhere, except in the edges of the support. Then:

to localize the wave in space at position 0 to reduce the space uncertainty, we have to reduce the support. However, doing so makes the momentum variation on the edges more and more important, as the slope will go up and down faster (higher top, and less x space for descent), leading to a larger variance (note that average momentum is still 0, due to to symmetry of the bump function)
to localize the momentum as much as possible at 0, we can make the support wider and wider. This makes the bumps at the edges smaller and smaller. However, this also obviously delocalises the wave function more and more, increasing the variance of x

Bibliography:

www.youtube.com/watch?v=bIIjIZBKgtI&list=PL54DF0652B30D99A4&index=59 "K2. Heisenberg Uncertainty Relation" by doctorphys (2011)
physics.stackexchange.com/questions/132111/uncertainty-principle-intuition Uncertainty Principle Intuition on Physics Stack Exchange

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