What do imaginary numbers practically represent in the Schrödinger equation?

Question

I know that the $i$ that appears in the Schrödinger equation, which is the imaginary unit, is used to solve problems that arise with roots of negative numbers. But what is the meaning of that negative root in that equation? Does it tell anything about the wave-function itself or is it something added just to make the LHS equal?

Answer 1

The Schrödinger equation tells you how a state evolves with time. Suppose the system is at $t=0$ in a normalised eigenstate $\psi(0)$ of the Hamiltonian with eigenvalue $E$. Because the Hamiltonian is an hermitian operator we know that $E$ is real. To obtain the time evolution you must solve the equation $$ E\,\psi = i \partial_t \psi, $$ which gives $\psi(t)=e^{-iEt}\psi(0)$. As you can see, $|\psi(t)|=|\psi(0)|=1$, satisfying the rule that the sum of all the probabilities must be equal to one. This is called unitarity.

If the imaginary unit $i$ wasn't there, the solutions would be $\psi(t)=e^{Et}\psi(0)$, and we wouldn't have constant $|\psi(t)|$ breaking this basic principle.

Answer 2

Complex numbers have an amplitude and phase. The i is one way of describing the phase. Wave-functions just map a coordinate to a complex number, from which you can get the amplitude and phase of a wave. Phase is what allows interference in QM. In a classical description of the two slit interferometer you'd have real amplitudes going through each, no phase or interference.

Answer 3

The classical plane wave solution is written as: $\psi (x,t)=\phi(x) e^{-i\omega t}$. When the time derivative is applied, it yields a factor $-i\omega$.

Then the operator $i\hbar \frac{\partial}{\partial t}$ applied to a plane wave solution is like multiplying it by $i\hbar(-i\omega)=\hbar \omega=E$.

I mean, the origin of the '$i$' is the '$e^{-i\omega t}$' factor in the plane wave.

Answer 4

What $i$ does here is simplify a system of two real differential equations into a more compact form.

The complex numbers should be thought of more as a two dimensional vector space over the real numbers rather than as some sort of mystical imaginary thing.

Complex numbers are especially good at simplifying two dimensional equations when there is some sort of symmetry involved, which we have here. Most equations in physics involve some sort of symmetry, and they have their own examples of special notation (such as dot products, cross products, and spherical coordinates) that allow several equations to be folded into 1, so imaginary numbers aren’t particularly special here.

Finally, the imaginary factor corresponds to a 90 degree rotation of some sort. In a completely non rigorous sense, this shows that the ‘movement’ of the field is perpendicular to its position/strength. In classical physics, things that move perpendicularly to their position stay bounded forever, which is what many solutions to Schrodinger’s equation do. This is non-rigorous because the Hamiltonian can alter the phase on the right side.

Answer 5

You're looking for a physical reason to use imaginary numbers? Such a reason may not exist. However, we use mathematics to model the universe all the time (e.g. we use irrational numbers like $\pi$ which also may not "exist") and here using $\sqrt{-1}$ is likely no different. It is a number that can be used within a theoretical framework that is tested with experiment. If the predictions are consistent with experiment, we label the theory's predictions correct.

With all that said, I normally think of using $i$ as a convenient (maybe necessary?) way of dealing with interference of the eigenfunctions of Schrodinger's equation during propagation, so that $|\Psi|^2$ is correct.

Answer 6

The Schrödinger equation describes the behavior of an entity called the wave-function. The wave-function is not real; it does not represent any physical phenomenon. Instead, it represents a mathematical simplification which transforms a heinous differential equation into a merely wretched one that describes far more than the probability distribution we get from $\left<\psi | \psi \right>$.

In some sense, the Schrödinger equation and the wave-function it describes are the result of a square root, but the equation squared is of far less value because it does not allow us to use operators. For example, to find an expectation value of a particular wave-function, we typically find

$$\left< \psi |\mathcal O |\psi \right>$$

where $\psi$ represents the wave-function and $\mathcal O$ represents some operator. For example, if we want to know the expectation value for the momentum of $\left| \psi \right>$, we would choose an appropriate basis for $\left| \psi \right>$ and then find $\left<\psi | p | \psi \right>$.

In differential equations, imaginary coefficients on first order linear equations change exponential behavior into sinusoidal behavior. In the simplest second order linear differential equation, $f(x) + a f''(x) = 0$, complex $a$ changes $f$ from a sinusoid to some product of sinusoidal and exponential functions.

Answer 7

Its well known QM behaves strangely. This arises from the super-position of the wave functions. The amplitude of the wave-function is complex. This involves the imaginary $i$ which is the square root of -1. The square of the amplitude is probability.

Another way of putting this is that the weirdness of QM comes from taking the square root of probability.