Friday, February 27, 2015

Real and quaternionic quantum mechanics


Wrapping up the discussion about quaternionic quantum mechanics, in this theory the inner product \(<f|g> = \int d^3 x \bar{f}(x,t) g(x,t)\) decomposes into a complex inner product and a symplectic inner product:

\(<f|g> = {<f|g>}_{C} + {<f|g>}_{S}\)

Here "f bar" means complex conjugation with respect to all imaginary elements: i, j, k.

If we consider quaternionic wavefunction \(f\) decomposition: \(f = f_{\alpha} + j f_{\beta}\) where \(f_{\alpha} = f_0 + i f_1\) and \(f_{\beta} = f_2 - i f_3\) then:

\({<f|g>}_{C} = \int d^3 x {f}^{*}_{\alpha}(x,t) {g}_{\alpha}(x,t) + {f}^{*}_{\beta}(x,t) {g}_{\beta}(x,t) \)

and

\({<f|g>}_{S} = \int d^3 x {f}_{\alpha}(x,t) {g}_{\beta}(x,t) - {f}_{\beta}(x,t) {g}_{\alpha}(x,t) \)

Therefore in quaternionic quantum mechanics the probabilities are different then in complex quantum mechanics because it includes the symplectic part:

\({|<f|g>|}^2\ = {|{<f|g>}_{C}|}^2 + {|{<f|g>}_{S}|}^2\)


Now real quantum mechanics is defined over the real numbers and we can compare it with complex quantum mechanics. If we consider the complex quantum mechanics decomposition of the wavefunction:

\( f = f_0+ i f_1\)

the complex inner product is:

\(<f|g> = {<f|g>}_{C} + {<f|g>}_{I}\)

with

\({<f|g>}_{R} = \int d^3 x {f}_{0}(x,t) {g}_{0}(x,t) + {f}_{1}(x,t) {g}_{1}(x,t) \)
and
\({<f|g>}_{I} = \int d^3 x {f}_{0}(x,t) {g}_{1}(x,t) - {f}_{1}(x,t) {g}_{0}(x,t) \)

Then 

\({|<f|g>|}^2\ = {|{<f|g>}_{R}|}^2 + {|{<f|g>}_{I}|}^2\)

and real quantum mechanics has only this term: \({|{<f|g>}_{R}|}^2\)

Now for the main problem of real quantum mechanics: in real quantum mechanics there are no energy eigenstates. Because of this we need to embed real quantum mechanics in complex quantum mechanics!!!

Quaternionic quantum mechanics does not have such problems, but in quaternionic quantum mechanics a De Finetti theorem does not hold. The root cause of it is the fact that quaternionic quantum mechanics does not respect the tensor product. 

In conclusion, both real and quaternionic quantum mechanics have very limited usefulness in describing nature and the most general way to express quantum mechanics is over complex numbers. 

Now the million dollar question: are the quantum mechanics number systems exhausted by the reals, complex numbers, or quaternions? The answer is no and there is at least one more: a direct sum of two SL(2,C). This will lead to Dirac's equation of the electron. Please stay tuned.

Friday, February 20, 2015

Is Quaternionic Quantum Mechanics Detectable?


Let us start by explaining why quaternionic quantum mechanics may be interesting for describing nature despite its problem with the tensor product. 

There are several arguments put forward, but I think only two of them have merit. The first obvious usage can be in justifying the \(SU(2)\) gauge group of the Standard Model which can be achieved in a similar fashion of how \(U(1)\) (the gauge group of electromagnetism) arises when one considers local symmetry. 

The second potential usage stems from the peculiar behavior of quaternionic quantum mechanics: the ground energy level is uniquely determined and the zero point energy cannot be freely shifted like in complex quantum mechanics. The obvious application of this is in solving the puzzle of the cosmological constant which by naive field theory arguments should be \({10}^{120}\) times larger than the observed value.

A pure state in quaternionic quantum mechanics is defined only up to a quaternionic phase (a unit quaternion) and one may ask if such phases which are different than complex wavefunction phases are experimentally detectable. In fact Asher Peres

Asher Peres


proposed such an experiment: pass a neutron beam through slabs of dissimilar materials and search for the non-commutativity of the phase shift when the slabs are reversed. The experiment was performed and the answer was negative. The theoretical clarification came later on from Adler's analysis which showed that the S-matrix quaternionic scattering is in fact indistinguishable from the usual complex quantum mechanics scattering.

This result is not surprising given the previous post: quaternionic quantum mechanics can be understood as a constrained complex quantum mechanics and there the square roof of negative one is represents a map between observables (hermitean operators) and generators of continuous symmetries (anti-hermitean operators). This map is also known as "dynamic correspondence".

Before comparing quaternionic quantum mechanics with real and complex quantum mechanics there is one last result of relative importance. Wigner's theorem states that in complex quantum mechanics symmetries can be represented by a unitary or anti-unitary transformation. Emch, Piron, Uhlhorn and Bargman generalized this to the quaternionic quantum mechanics case and here there are only quaternionic-unitary transformations.


Thursday, February 12, 2015

Quaternionic Quantum Mechanics

(part 2)



Continuing the quaternionic discussion, let us see when one might encounter it. Here I will follow the discussion in "Geometry of Quantum States" by Ingemar Bengtsson and Karol Zyczkowski. The idea is that of a time reversal.


We start in the usual complex quantum mechanics and here by Wigner's theorem we have that every symmetry is represented by a unitary or anti-unitary transformation. So suppose we have a time reversal operator \(\Theta \). When we reverse time, from the time evolution equation in the Heisenberg picture this is equivalent with \(i\) changing signs, and therefore \(\Theta\) must be an anti-unitary transformation. 

If the system is invariant under time reversal, we have:

\(\langle \Psi| \Phi \rangle = \langle \Theta\Psi| \Theta\Phi \rangle\)

and this means that there are two options for \(\Theta\): \(\Theta^2 = \pm 1\).

Now the discussion depends on the angular momentum of the system. For fermions \(\Theta^2 = - 1\). If we cannot tell the direction of time by any measurement, the observables commute with \(\Theta\):

\([O, \Theta] = 0\)

and this defines a superselection rule. Then one can define:

\( i, \Theta, i\Theta\ = i, j, k\)

the quaternionic imaginary elements and one arrives at the quaternionic projective space. Here is how quaternionic quantum mechanics can arise.

However, because in this case one talks about superselection rules, composing two quaternionic systems breaks the superselection constraint and there are problems defining the tensor product. We'll talk about this next time.

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:

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

where:

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

To be continued ...

Friday, January 30, 2015

I've been had by Mr. Bender's lectures


Bamboozled, duped, hoodwinked, well, you get the idea. I drank his cool aid on (lack of) mathematical rigor and I bought his idea of summing infinite series like

\( 1+ 2+ 3+ \cdots = -1/12\)

What I did wrong was swallowing hook, line, and sinker his postulates:

  • Rule 1: Summation property:
    • \(S(a_0 + a_1 + a_2 + \cdots )= a_0 + S(a_1 + a_2 + \cdots )\)
  • Rule 2: Linearity:
    • \(S(\sum (\alpha a_n + \beta b_n) )= \alpha S(\sum (a_n)) + \beta S(\sum (b_n))\)


    Why? Because in addition to assuming a unique result, they are inconsistent. If:

    \(1+2+3+ \cdots = -1/12 = A\)

    Then for example consider this:
    \(A-A = (1+2+3+ \cdots ) - (1+2+3+ \cdots )\) 
    which by rule 1:
    \( 0 = (1+2+3+ \cdots ) - (0+1+2+3+ \cdots ) \) 
    such that
    \(0 = 1+1+1+1+ \cdots\) 

    Aha! This now implies that
    \(0=1 + (1+1+1+\cdots ) = 1+0\) and so \(0 = 1\)!!!!

    The two rules work for alternating sums, but when the sign of the sum terms is the same the two rules are clearly inconsistent. 

    But does this mean that \(1+2+3+ \cdots \) is not \(-1/12\) ? Not at all. The result is still valid due to deeper reasons: analytic continuation of Riemann zeta function.

    It is not easy to find why things like this work in math, but in general physics intuition is a very good clue that there must be a solid and rigorous foundation. It is just that physicists' focus is on solving the practical problems and not on the deeper mathematical theory. One may say that a physicist to a mathematician is like an engineer to a physicist :) This is not that bad though: the engineers make more money than physicists, and physicists make more money than mathematicians. 

    I still regard Mr. Bender's lectures as outstanding, but I should have trusted my mathematical intuition more and not disregarded the alarm bells of mathematical rigor. The inconsistent argument above is due to David Joyce and I ran across it on Quora where the -1/12 result is discussed often.

    Friday, January 23, 2015

    Quantum Groups and curved space-time


    Today I will present the last post on Hopf algebras and I will talk about quantum groups (which are a special kind of Hopf algebras) and their amazing application to curves space-time. Alfredo Iorio pointed me his paper (http://arxiv.org/pdf/hep-th/0104162v1.pdf) which I studied in detail.

    In second quantization he starts with the Weyl-Heisenberg algebra \((a, a^{\dagger}, N, c)\):

    \([a, a^{\dagger}] = 2c\)
    \([N, a] = −a \)
    \([N, a^{\dagger}] = a^{\dagger} \)
    \([c, \cdot] = 0\)

    and then the Hopf algebra coproduct is introduced:

    \(\Delta a = a\otimes I + I \otimes a\)

    So far nothing special but then the coproduct can be deformed forming a quantum group:

    \(\Delta a_q = a_q \otimes q^c + q^{-c}\otimes a_q\)

    where \(q\) is the deformation parameter related to a geometric-like series:

    \({[x]}_q = \frac{q^x - q^{-x}}{q - q^-1}\)

    As a nice math puzzle, what is the series which gives rise to \({[n]}_q\)? It is obviously related to \( 1+q+q^2+\cdots + q^n\) but it is not quite that.

    So what is this to do with anything? 

    The q-deformation o reproduces the typical structures of a quantized field in a space–time with horizon!!! 



    Quantum group deformations can be induced by gravitational fields. For technical details please read Iorio's paper from above.

    I am still studying quantum groups and I do not want to venture stating more because my intuition in this area is work in progress, but clearly they have a real physical interpretation.

    Friday, January 16, 2015

    My first physics paper


    I was looking through a pile of old papers and I discovered a copy of my very first physics paper which I published in my 4th year in college - quite a long time ago. I remember I was attending a standard electromagnetism class and the teacher said: to obtain the invariants of the electromagnetic filed, compute the characteristic polynomial:

    \( det(F -I_4 x) = 0\)

    where \( F\) is electromagnetic field tensor. The coefficients are the invariants.

    I went home and double checked the math and surprise: the equation implied:

    \( x^4 -  (B^2 +E^2 ) x^2 + {(E\cdot B)}^2 = 0 \) 

    So the next lecture I confronted the professor and showed that you get \(B^2 + E^2\) instead of \(B^2 - E^2\) and his statement was wrong. To my surprise he said that he knew it was wrong, but it was close to the answer and it must have a kernel of truth but he did not know how to fix it. Now in college I never liked functional analysis (and only recently I developed the right intuition in the area) but I was always good at algebra and I could come up with the answer to any algebra problem at first sight. So I said to the teacher: I know how to fix it: just make the electric field imaginary. The professor than said: OK, write it up and if you can do it we'll write a paper. 

    This was not that easy -it took me an afternoon- but I worked it at home starting backwards. Suppose you have a \(4 \times 4\) diagonal matrix \(J\) with the diagonal elements \((1,1,1,i)\) Then if you multiply the electromagnetic tensor \(F\) with \(J\) on the left and on the right: \(J F J\) then you are in business with \( det(J F J -I_4 x) = 0\). But why does this work? What is happening is a transition from a pseudo-orthogonal group to an orthogonal group. Here is how:

    The electromagnetic tensor changes under a Lorentz boost \( \Lambda \) like this:

    \( F^{'} = \Lambda F {\Lambda}^t\)

    where 

    \({\Lambda}^t Y \Lambda = Y\)

    with \(Y \) the diagonal matrix \((1,1,1,-1)\)

    Now if we sandwich \( F \) with \( J\) we get:

    \(J F^{'} J = J \Lambda F {\Lambda}^t J = (J \Lambda J^{-1}) (J F J) {(J \Lambda J^{-1})}^t \)

    But what about \(\Lambda\) ? Here is the fireworks: \(Y = J J\) 

    So let us compute \({(J \Lambda J^{-1})}^t (J \Lambda J^{-1}) \):

    \({(J \Lambda J^{-1})}^t J \Lambda J^{-1} = J^{-1} {\Lambda}^t J J \Lambda J^{-1} = J^{-1} {\Lambda}^t Y \Lambda J^{-1} = \)

    \( = J^{-1} Y J^{-1} = J^{-1} J J J^{-1} = 1\)

    Bingo: we manage to get an orthogonal transformation from a pseudo-orthogonal one. This is reminiscent of Dirac's trick because we basically take the square root of \(Y\) and we get two \(J \) matrix instead. This is how people used to write imaginary \(ict \) time and preserve the usual orthogonal rotations. And for orthogonal group there is an elementary linear algebra theorem which states that the only invariants of a similarity transformation (which is how the electromagnetic tensor changes under a Lorentz transformation in the new complexified orthogonal transformation) are the coefficients of the characteristic polynomial. 

    So lo and behold, my first physics paper appeared in the Romanian Journal of Physics, Volume 38, Number 9, pages 873-875 in 1993. The teacher's name was Andrei Ludu and he was also doing seminars on quantum groups. At that time I had no clue on Lie algebras, let alone on quantum groups and I could not find any motivation or intuition to a bunch of very long and way too abstract things. 

    But guess what? Quantum groups are actually Hopf algebras and they have very interesting physics applications. I'll talk about it next time.