8
$\begingroup$

This question already has an answer here:

In the most general sense, the Time Dependent Schrödinger Equation (TDSE) reads $$\hat{H} \Psi = i \hbar ~{{\mathrm d} \over {\mathrm dt}}\Psi $$

Is it possible to get rid of the $i$ entirely? Does it need to be there?

Let me be clear on the type of answer I am looking for. I am aware that the spatial wave functions can be written entirely real (by a theorem, whose name i do not recall), and that complex constants can be chosen such that the the time dependent part cancels out.

What I am instead asking is, is it possible to get an equivalent TDSE that does not deal with imaginary numbers in any sense? Or is it impossible because the $i$ is required for Hermitian operators?

$\endgroup$

marked as duplicate by Qmechanic quantum-mechanics Nov 16 '16 at 7:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

17
$\begingroup$

You can easily eliminate all references to complex numbers in a rather trivial way, although doing so results in much less mathematically elegant expressions. For example, you could choose to work in the position basis and, instead of using a complex-valued wavefunction that assigns a complex number to every point in configuration space, you could use a wavefunction with two real components, i.e. that assigns an ordered pair of real numbers $\vec{\psi}(\{\vec{x}_i\})$ to every point in configuration space. Multiplication by $i$ (e.g. in the Schrodinger equation and in the canonical commutation relations) would be replaced by an application of the orthogonal matrix $$\hat{O} := \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right),$$ which rotates the "vector" $\vec{\psi}$ by 90 degrees counterclockwise. So the Schrodinger equation would become $\hat{H} \vec{\psi} = \hbar\, \hat{O}\, d\vec{\psi}/dt$ (in the position basis). But it would be much more complicated to do things like change basis.

In fact, this is exactly what all computational algorithms do "under the hood" - computers internally represent complex numbers as ordered pairs of real numbers, and all complex-number operations are converted into operations on pairs of reals.

Section 4 of this paper discusses some motivations for why we might expect that a theory like quantum mechanics might be most naturally expressed using complex numbers (which is very different from claiming that it can only be expressed using complex numbers). Basically, it would be nice if the field of numbers we use is algebraically closed. The square root of a complex number is always complex, but the square root of a real number isn't always real. However, the group $GL(2, \mathbb{R})$ of $2 \times 2$ real invertible matrices is algebraically closed, so that's just as good as the complex numbers in that sense. (Note for experts and nitpickers: I'm glossing over some subtleties here, like the fact that $GL(2, \mathbb{R})$ isn't closed under addition and therefore isn't a field, so strictly speaking you should be talking about closure under exponentiation rather than algebraic closure.)

$\endgroup$
  • $\begingroup$ Someone needs to create a binary data type for complex number... $\endgroup$ – The Great Duck Nov 16 '16 at 6:50
  • $\begingroup$ Some hard-core digital physics proponents might argue that "deep down," the laws of quantum mechanics do only use real numbers, and physicists' use of complex numbers are just a convenient mathematical shortcut. $\endgroup$ – tparker Nov 16 '16 at 6:59
  • $\begingroup$ to be fair though, if the computer can use complex numbers better than someone ought to make a complex number data type. $\endgroup$ – The Great Duck Nov 16 '16 at 15:06
0
$\begingroup$

I'm pretty sure Geometric Algebra does without the imaginary of complex numbers i for the Schrodinger Equation: https://en.wikipedia.org/wiki/Spacetime_algebra

Hestenes "Geometric Algebra" uses a Real valued Clifford Algebra

You still see i, I in Geometric Algebra equations but they are almost always Pseudoscalars

The Pseudoscalars are highest graded element of the Geometric Algebra of a given dimension, not the imaginary of complex numbers as I believe many assume without knowing the GA formalism

The Pseudoscalars do have negative square for 2D and 3D Euclidean and 4D Minkowski spaces giving isomorphisms with complex numbers in 2D, Quaternions in 3D

Pseudoscalar Geometric/Clifford product with lower grade elemnts gives the Dual

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.