# A Simple Explanation for the Schrödinger Equation and Model of Atom? [closed]

I tried reading the Wikipedia article to no avail - I simply cannot understand the Schrödinger Equation (what does each of the variables mean, especially the wave function), and the Schrödinger Model of the Atom. Could someone explain in simple words how this entire thing works and its consequences/conclusions in Physics? • this is a very general question, it is not clear for me where is the starting point for answering... perhaps it is better to start from a standard textbook of quantum mechanics, e.g. Introduction to Quantum Mechanics by David J. Griffiths. And ask question from book rather than top down from wikipedia... Sep 5 '13 at 15:46
• Agree with @user26143. Are you ok with wavefunctions? What about the Born rule? If you have no problem with these then WetSavannaAnimal aka Rod Vance's answer is sufficient. If you don't understand (or accept) these things then you need to go further back: quantum mechanics is the (in some broad sense) the only possible extension of ordinary probability theory that allows "probabilities" (now called amplitudes) to be negative. Sep 6 '13 at 3:44

Try this explanation on for size: it's how I like to think about the Schrödinger equation and pretty near to how Richard Feynman introduces it in his discussion of the Hamiltonian in the "Feynman Lectures on Physics" in chapter 8 "The Hamiltonian Matrix" of the third volume. This would be a good reference for you to read if your overwhelmed by the Wikipedia page.

## Background

Suppose we accept that a system's "state" is encoded as a vector in some Hilbert space (i.e. essentially a vector space one wherein inner products and norms are defined): let's for example consider a quantum harmonic oscillator, so we shall encode the state as a discrete sequence of complex numbers $\Psi = \{\psi_0, \psi_1, \cdots\}$, such that $\sum_j |\psi_j|^2 = 1$. $\psi_0$ is the probability amplitude that the system will be detected quantum ground state, i.e. as close as one can get to "unenergised" without violating the Heisenberg inequality, $\psi_1$ the probability amplitude that the oscillator is in a one photon state, i.e. its energy is $\hbar \omega$, $\psi_2$ the amplitude that it is two photon state, and in general $\psi_N$ the attitude that is in an $N$-photon state; or, if you like, the amplitude that it has had $N$-photons added to its ground state from somewhere outside the oscillator system. More generally, the $\psi_j$ are the probability amplitudes that the system will be detected as being in the $j^{th}$ basis state: one of the basis vectors for the state Hilbert space and they don't have to be the equispaced states of the harmonic oscillator - it could be another system altogether. Obviously, the system must always be in some state, so the relationship $\sum_j |\psi_j|^2 = 1$ always holds.

## The Basic Ideas

The Schrödinger equation is very general: it simply says that a quantum system's makeup and working is in some sense "constant" when the system is sundered from the rest of the World. This vague statement makes more sense in symbols: the mathematical description has to be invariant with respect to time shifts: if I begin with a quantum state at 12 o'clock and evolve it until 1 o'clock, then my state evolution is going to be the same as if I began with the same state at 4 o'clock and waited until five. Now, we assume linearity, so that our state vector (now written as a column vector) is going to evolve following some matrix equation: $\psi(t) = U(t) \psi(0)$, where state transition matrix $U(t)$ must:

1. Fulfil $U(t+s) = U(t) U(s) = U(s) U(t)$ for any time intervals $t$ and $s$. This is simply our discussion about time shift invariance above. Straight away we know $U(t) = \exp(A t)$, for some constant matrix $A$ as the exponential is the only continuous function with this time shift invariance property;
2. It must be unitary: this means it must conserve norms, so that $\sum_j |\psi_j|^2 = 1$ holds at all times: this simply says that the system has to be in some state, owing to the probability interpretation of the squared magnitudes. What justification do we have for this? Well, if the state wandered outside our Hilbert state space, then that Hilbert space would not be a good description of all the system's possible states. So we would simply add basis vectors to our Hilbert space and expand it until it was a good description of all the system's possible states.

So the most general state evolution possible is $\psi(t) = \exp(-\hbar^{-1} i\, \hat{H}\, t)\,\psi(0)$, where $\hat{H}$ is a constant, Hermitian matrix (this is equivalent to the unitaryhood statement). This in turn is equivalent to:

$$i\,\hbar\,\mathrm{d}_t \psi = \hat{H}\,\psi$$

which is the Schrödinger equation. Hopefully the Schrödinger's equation's essential nature should now be clear:

The Schrödinger equation for a quantum system asserts (i) the system's time shift invariance and (ii) that the system must always be in some state in the state Hilbert space when that system is sundered from the rest of the World

For the sake of this argument, simply think of $i$ and $\hbar$ as constants I have arbitrarily pulled out of the right hand side. They make the observables - the operators that define measurement outcomes given a system state $\psi$ - easier to interpret. We pull the constant $i$ out so that our unitaryhood condition is that our $\hat{H}$ matrix is Hermitian rather than skew Hermitian (i.e. its eigenvalues, and thus possible measurement outcomes are real rather than imaginary) and the $\hbar$ has two functions:

1. $\hat{H}$ is a basic time-constant observable; conserved quantities - i.e. those that do not vary with time - are those whose observables commute with $\hat{H}$. You can derive this statement with a bit more work from the Schrödinger equation (i.e. by transforming to the Heisenberg Picture). So you can postulate that $\hat{H}$ is the energy observable. Therefore we need a unit-balancing scaling constant to make the exponent dimensionless in in $\psi(t) = \exp(-\hbar^{-1} i\, \hat{H}\, t)\,\psi(0)$;
2. It gives the expression of the Heisenberg uncertainty principle in its neatest form. But that is another story aside from the Schrödinger equation: see my answer here.

One often chooses to transform the state space coordinates and relax the time shift invariance condition. In this case we get the time varying Schrödinger equation as I describe here.

A last thing that might seem mysterious to you is that the Wiki page deals with continuous wavefunctions instead of discrete state vectors. This is simply a change of co-ordinates: if you like, think of discrete Fourier components representing an equivalent continuous function as an example. The above arguments about the Schrödinger equation work equally well in principle whether $\psi$ may be a discrete column vector $\{\psi_j(t)\}_{j=0}^\infty$ or a continuous function $\psi(\mathbf{r},\,t)$ of some vector of variables $\mathbf{r}$, for example position. Subject to appropriate conditions, continuous functions can also be thought of as living in a countably infinite dimensional Hilbert space. It simply depends on the most convenient description for the problem at hand.

• Do you mean phonon, not photon, for your example of the quantum harmonic oscillator? Sep 6 '13 at 19:30
• @user50229 It could be either. Sorry for the confusion. I'm from a quantum optics background and one thinks of the EM field as a collection of quantum harmonic oscillators, one for each mode in space. I shall try to think about a better word and change Sep 6 '13 at 21:56