Thursday, February 5, 2015

Quaternionic Quantum Mechanics

 (part 1)

I will start a new series about the number system of quantum mechanics. Quantum mechanics can be expressed over real numbers, complex numbers, quaternions, and SL(2,C). I will simply follow the literature and try to present the interesting results which will help better understand the usual complex number formalism. 

The standard reference for quaternionic quantum mechanics is Adler's monograph: Quaternionic Quantum Mechanics and Quantum Fields

First, what is a quaternion? It is one of the 4 normed division number systems which consists of the elements of the form:

\(w = a+ ix + jy + kz\) with \( a,x,y,z \in R\) 
where \(i^2 = j^2 = k^2 = -1\) and \(ij=k, jk=i, ki=j\)

Quaternionic multiplication

Just like in complex quantum mechanics, a physical state is defined only up to a phase which here is a unit quaternion:

\(| \psi \rangle = \{|\psi \omega \rangle : |\omega| = 1\}\)

This works because the probability of the quantum transition between states \(\psi, \phi\) is given by the usual rule:

\(P = {|\langle \psi | \phi \rangle |}^2\)

Since unlike complex numbers quaternions are non-commutative (\( ij = -ji \ne ji\)) we have to be careful on the position of the numbers in the ket-bra notation. By convention we say:

\(|\psi \omega \rangle = |\psi \rangle \omega\)

and we will have the following linearity condition for an operator:

\(O (|\psi \rangle \omega) = (O|\psi\rangle) \omega\)

If \(1 \) is the identity operator, let us define:

\(1 = E_0\)
\(I = E_1 = i1\)
\(J=E_2 = j1\)
\(K=E_3 = k1\)

and an operator \(O\) has the decomposition: \(O = O_0 + I O_1 + J O_2 + K O_3\) where:

\(O_0 = 1/4 (O - IOI -JOJ -KOK)\)
\(O_1 = 1/4 (IO + OI -JOK + KOJ)\)
\(O_2 = 1/4 (JO + OJ -KOI + IOK)\)
\(O_3 = 1/4 (KO + OK -IOJ + JOI)\)

Similarly a quaternionic wavefunction can be decomposed as follows:

\(|\psi \rangle = |\psi_0 \rangle + I |\psi_1 \rangle + J |\psi_2 \rangle + K |\psi_3 \rangle\)


\(|\psi_0 \rangle = 1/4(|\psi\rangle - I |\psi\rangle i - J|\psi\rangle j -K|\psi\rangle k)\)
\(|\psi_1 \rangle = -1/4(I |\psi\rangle + |\psi\rangle i - J|\psi\rangle k + K|\psi\rangle j)\)
\(|\psi_2 \rangle = -1/4(J|\psi\rangle + |\psi\rangle j - K|\psi\rangle i + I|\psi\rangle k)\)
\(|\psi_3 \rangle = -1/4(K|\psi\rangle + |\psi\rangle k - I|\psi\rangle j + J|\psi\rangle i)\)

To be continued ...

No comments:

Post a Comment