# Why is $i$ used in the equations for quantum mechanics? [duplicate]

Coming from someone who knows a tiny bit about the subject but who really wants to learn. I know it's the square root of -1 but I would like some insight as to why it's used at all.

• May 28, 2017 at 17:27
• I wanted to know about i in particular, not complex numbers in general, also not necessarily the Schrodinger equation May 28, 2017 at 17:34
• If you want to know about $i$, then you also want to know about complex numbers; they're kind of inseparable. Otherwise, I'm not quite sure what you're asking. May 28, 2017 at 17:39

As @AccidentalFourierTransform pointed out, the use of complex numbers isn't strictly necessary to describe quantum mechanics. It's simply the first (and in a sense, least complicated) mathematical structure people learn that is able to describe quantum mechanics properly.

For an easy example of quantum mechanics entirely using real numbers, we simply note that the map $f$ from complex numbers to invertible 2x2 matrices given by

$f(a+bi)=\begin{bmatrix}a&-b\\b&a\end{bmatrix}$

is an isomorphism. This means we can effectively replace all of the complex numbers in quantum mechanics with these matrices, which contain entirely real entries.

• How would you implement Dirac quantization using that isomorphism? It seems really difficult to interpret $\hat{x}$ as the position operator and $\hat{p}$ as the momentum operator if $\hat{x}$ and $\hat{p}$ have the matrix structure necessary to enforce $[\hat{x},\hat{p}]=\left( \begin{array}{cc}0 & -1 \\ 1 & 0 \end{array}\right)$.
– WAH
May 28, 2017 at 18:31
• The isomorphism can be used in the same way in that situation, too. Granted, it's nasty, but I never claimed that this interpretation made things easier. There's a reason we use complex numbers, after all. May 28, 2017 at 18:36
• I appreciate your reply, but I don't see any more clearly how in practice I'd think about or use $\hat{x}$ as the position operator. Maybe it's just not worth thinking about in detail because there's no reason not to use $i$. But I guess I'm disappointed because I've said in the past that QM is the only place where the use of complex numbers is not merely helpful but necessary.
– WAH
May 28, 2017 at 18:42
• Complex numbers are certainly helpful, but not by any means necessary. The reason that we use complex numbers is that they are a simple representation of an underlying group structure (like $SO(3)$ or $SU(2)$, or Lie and Clifford algebras for more complicated/relativistic situations). As long as you respect that group structure, you can employ any representation you want. May 28, 2017 at 18:54
• @WAH you would split $\hat x,\hat p$ into its real and imaginary parts, so that they become "column vectors", $\hat x\to\begin{pmatrix} \hat x_1\\ \hat x_2\end{pmatrix}$ and similarly for $\hat p$. In this case, the commutation relations become $[\hat x_i,\hat p_j]=i\hbar \omega_{ij}$, where $\omega=\begin{pmatrix}0&-1\\ 1&0\end{pmatrix}$. May 28, 2017 at 19:09

Quantum mechanical models use solutions of wave equations. These are second degree differential equations which accept solutions using the complex number formalism.

This is not particular to quantum mechanics. Electromagnetic radiation is described by solutions of the maxwell equations and are also formulated with the complex numbers formalism.

It is the quantum mechanics postulates that define the difference between classical wave solutions and quantum mechanical, not the i.

Whereas for classical equations the variables in the solutions are measurable in the laboratory , as for example electric and magnetic fields, in quantum mechanics it is the complex conjugate squared of the complex wavefunctions that have a connection with measurable quantities, describing probability distributions for the observables.