Why does the mathematical constant $e$ enter into quantum mechanics so much?

Question

In A. Zee's book Quantum Field Theory in a Nutshell, he mentions on pages 11-12 the following formula which he assumes reader had encountered before:

\begin{equation} \langle q | p \rangle ~=~ \frac{e^{iqp/\hbar}}{\sqrt{2\pi\hbar}}. \end{equation}

I keep seeing $e$ come up in quantum mechanics. For instance, in that same chapter he writes $e^{-iH\delta t}$.

Why is $e$ useful in quantum mechanics? What does it signify in this context relative to the inner product of the bra and ket vector on the left?

Answer 1

It's not $e$ per se, but rather the exponential function

$$ \exp(x) \equiv \mathrm{e}^{x} = \sum_{n \in \mathbb{N}} \frac{x^n}{n!}$$

and the exponential functions appears naturally in various contexts, the most relevant in QM being the Fourier transform and the map between self-adjoint and Hermitian operators, or, in another guise, as the natural map from a Lie algebra to its Lie group.

It also appears almost anywhere where differential equations occur, since the solutions to wave equations, among others, can be expressed by it (this is also related to it being the kernel of the Fourier transform).

Answer 2

Why is e useful in quantum mechanics?

Essentially, it is because $e^{ax}$ is an eigenfunction for the $\frac{d}{dx}$ operator:

$$\frac{d}{dx}e^{ax}= ae^{ax}$$

So, as an almost trivial example, it follows from Schrodinger's equation

$$\hat H |\psi\rangle = i\hbar \frac{\partial}{\partial t}|\psi\rangle$$

that an energy eigenstate, with energy eigenvalue $E_n$, can be written as

$$|\psi_n(t)\rangle = |\psi_n(0)\rangle e^{-i\frac{E_n}{\hbar}t}$$

which can be directly checked:

$$\hat H |\psi_n(t)\rangle=i\hbar \frac{\partial}{\partial t}|\psi_n(t)\rangle = i\hbar |\psi_n(0)\rangle \frac{\partial}{\partial t}e^{-i\frac{E_n}{\hbar}t}= E_n|\psi_n(0)\rangle e^{-i\frac{E_n}{\hbar}t} = E_n|\psi_n(t)\rangle$$

As a less trivial example, one can formally solve Schrodinger's equation for an arbitrary state to find

$$|\psi(t)\rangle = |\psi(0)\rangle e^{-\frac{i}{\hbar}\hat H t}$$

Which again can be formally checked:

$$\hat H |\psi(t)\rangle=i\hbar \frac{\partial}{\partial t}|\psi(t)\rangle = i\hbar |\psi(0)\rangle \frac{\partial}{\partial t}e^{-\frac{i}{\hbar}\hat H t} = \hat H |\psi(0)\rangle e^{-\frac{i}{\hbar}\hat H t} = \hat H |\psi(t)\rangle$$

Now, one might wonder how to interpret an operator valued exponent. However, a simple Taylor expansion yields

$$e^{-\frac{i}{\hbar}\hat H t} = 1 - i\frac{t}{\hbar} \hat H - \frac{t^2}{2!\,\hbar^2}\hat H^2 + i\frac{t^3}{3!\,\hbar^3} \hat H^3 + ...$$

So this operator evolves the state $|\psi(0)\rangle$ to the state $|\psi(t)\rangle$.

Answer 3

One reason is because harmonic oscillation is so common in nature. An oscillation is called harmonic if the force is proportional to how far the system is from equilibrium. E.g. a mass attached to a spring in a uniform gravitational field. If you lift the mass a bit, gravity takes over and there's a force in the down direction. If you pull the mass down past the equilibrium point, the force from the spring wants to pull it up. If you let it go, a harmonic oscillation starts.

If you solve the equations for harmonic oscillators you find they are sine (or cosine) functions. A lot of the maths becomes much easier if you use the Euler equation connecting sin and cos,

$$e^{i\phi}=\cos\phi + i\sin\phi.$$

Answer 4

As a specific case of @ACuriousMind s answer, the $e^{i\phi}$ and $e^{i\phi}$ forms (where $\phi$ is something real) appear as solutions to the differential equation form $$\frac{d^2}{dy^2}f(y)=-\phi^2f(y),$$ where $y$ is some independent variable (not necessarily spatial). This differential equation shows up quite often in physics, including when the Schrodinger wave equation is separated. It is definitely the form of the time differential equation, and sometimes the spatial equation (e.g. infinite square well in rectangular coordinates).

This $e^{i\phi}$ also relates to sinusoidal functions by Euler's equation$$e^{i\phi}=\cos\phi + i\sin\phi.$$

EDIT: An interesting side note is that the solutions of the quantum mechanical harmonic oscillator do not explicitly include any $e^{i\phi}$ forms in the spatial solution. The oscillatory nature of the solution is contained in a finite-order polynomial. This is in contrast to the classical harmonic oscillator solution. (There is a Gaussian decay term $e^{-\alpha x^2}$ in the QMHO solution.)