What is the meaning of the lower case i in the Schrodinger equation?

[bluestar]

When examining the Schrodinger equation for a particle in a square well there is a lower case “i” that shows up in the equations that never gets defined. I have checked several sources and its usage is somewhat uniform and yet not defined.

Can someone define the “i” ?

One source I am looking at is:
galileo.phys.virginia.edu/classes/252/electron_in_a_box "dot" html
It will be the first equation.
Thanks

 

[DavidWhitbeck]

It's the imaginary number, $$i = \sqrt{-1}$$.

 

[cesiumfrog]

You might alternatively think of it as an operator that advances a wave by one quarter cycle.

Notice how wave-number (multiplied onto the wave-function) is exactly that far out of phase from spatial derivative (operated onto the wave-function)? This (which you should recognise as "the momentum operator") gives an example of why complex numbers are convenient when describing waves.

 

[bluestar]

Thanks for the help

Dang, I forgot about i = sqrt(-1)
I haven’t used that in years. …Thanks,

Very interesting comment CesiumFrog.
I’ll consider that as I proceed with my work.

 

[kahoomann]

You might alternatively think of it as an operator that advances a wave by one quarter cycle.

Notice how wave-number (multiplied onto the wave-function) is exactly that far out of phase from spatial derivative (operated onto the wave-function)? This (which you should recognise as "the momentum operator") gives an example of why complex numbers are convenient when describing waves.

But what does Schrodinger equation mean?
We know what Newton's equation (F=ma) means

 

[jsc314159]

The Schrodinger equation is a form of the wave equation. It can be interpreted like a wave equation. The acceleration at each point on the wave function is due to the Hamiltonian and that the larger the concavity of the wavefunction at point, the larger the acceleration.

Does that make sense or am I full of hot air?

jsc

 

[bluestar]

Possible answer

Great Question Kahoomann

I have been studying the Schrodinger equation and its various spinoffs and here is what I have learned. Remember I am the novice and I am sure the pros may ring-in on the subject.

1st DeBroglie made the connection that atomic and subatomic particles exhibit wave properties.

2nd Schrodinger felt there was a connection between the wave-particle duality and the energy of the particle. Schrodinger derived a formula that described the energy of the particle in terms of a wave equation. The energy of the particle was equal to two parts of the wave equation. The first part results from the kinetic energy of the particle and the second part is the influence of a external field that may have an effect on the particle.

3rd From Schrodinger’s equation several things can be described or predicted in the atomic and subatomic world.

 

[Fredrik]

But what does Schrodinger equation mean?
We know what Newton's equation (F=ma) means

I'll pretend that space is one-dimensional here, just to simplify the notation.

What Newton's second law really says is that position as a function of time satisfies a differential equation that's nice enough to guarantee that there exists a unique solution for each initial condition. This means that we if we know the position and velocity at one time, we can calculate position as a function of time, and that function tells us the position and velocity at all times. So we can think of the pair of numbers (x₀,v₀) in the initial condition x(t₀)=x₀, x'(t₀)=v₀ as representing the "state" of the system, and the differential equation as describing the time evolution of that state.

What quantum mechanics tells us is that a state can't be represented by a pair of numbers (x₀,v₀). Instead it's represented by a function $\psi:\mathbb R^2\rightarrow\mathbb C$. The time evolution of that function is given by the Schrödinger equation.

When you solve it, you find that the solutions are of the form exp(-iEt+ipx). This is an eigenfunction of id/dt with eigenvalue E, and an eigenfunction of -id/dx with eigenvalue p. The identification of this E and p with energy and momentum comes from the fact that if you substitute the energy and momentum in the classical non-relativistic equation E=p²/2m+V(x) with those derivative operators, then the result is the Schrödinger equation.

So the Schrödinger equation can be thought of as the quantum version of E=p²/2m+V(x) because it expresses the relationship between energy and momentum, but it can also be thought of as the quantum version of F=ma, since it tells us the time evolution of the mathematical quantity that represents the state of the system.

 

[Mohandas]

Nice answer Fredrik. Some additional points:

While F=ma determines x(t) for all future time, the Schrödinger equation determines Psi(x,t) for all future time. So while Newtons F=ma returns x(t), the S.E returns a wave function which is spread out in space.

At this point it is hard to see how this can represent a state of a particle, given the fact that particles are not spread out in space, but, rather, they are localized at some point. Enter Statistical Interpretation; all quantum mechanics has to offer (compared to F=ma) is statistical information about the possible whereabouts of a particle at a given time. How do you get that information? You square the absolute value of the wave function and integrate over your desired interval, and what you are left with after that exercise is a probability of finding the particle in that interval at a time t.

 

[Fredrik]

I would say that the particles are spread out in space. The statistical interpretation doesn't contradict that. A particle is spread out until a position measurement "squeezes" its wave function into a shape with a very sharp peak.

This is just something we have to get used to when we learn quantum mechanics. Between measurements, physical systems can be in states that do not correspond to a possible outcome of a measurement. If we e.g. measure the spin component of an electron in a particular direction, the result is always +1/2 or -1/2. But if we assume that the electron is always in one of those states, we can use that assumption to derive Bell inequalities, which have been found not to hold in experiments.

So the assumption that the spin component of an electron always has a definite value is definitely false. There's no reason to think position is any different. We just have to get used to the fact that in quantum mechanics, the answer to the question "where is the particle now?" isn't a number (or 3 numbers), it's a complex-valued function (which also happens to be the answer to many other questions we may want to ask about the particle, like "what's its momentum?").