# I Canonical Quantization

1. Nov 25, 2016

### klpskp

Suppose we have a classical system described by a Lagrangien $\mathscr{L}(x,t)$.
The same system can be described by the Lagrangien $\mathscr{L'}(x,t)=\mathscr{L}(x,t)+\frac{\mathrm{d}F(x,t)}{\mathrm{d}t}$. where $F(x,t)$ can be any function.

If we now quantize the system by calculating the Hamiltonian and promoting the canonical position and the canonical momentum to operators, we arrive at two different Hamiltonians describing two different quantum systems.

My question is: How do we choose the Lagrangien to arrive at the "right" quantum system?

2. Nov 25, 2016

### A. Neumaier

Quantiziation is not a unique procedure.

The quantum Lagrangian is chosen either arbitrarily (for a theoretical study) or such that it matches experiment. This decides whether the Lagrangian is correct.

3. Nov 28, 2016

### samalkhaiat

The correct equivalence relation is $$L'(x , \dot{x},t) = L(x,\dot{x},t) + \frac{d}{dt}F(x,t) .$$ The important fact here is $\frac{\partial F}{\partial \dot{x}}=0$, i.e., the primed and un-primed Lagrangians describe the same physical system if and only if the function $F$ dose not depend on the generalized velocity $\dot{x}$. This equivalence is true in classical mechanics as well as quantum mechanics where it modifies the wave-function by a space-time dependent phase, as expected given that $\Psi \sim \exp (\frac{i}{\hbar}\int L dt )$. To see this, rewrite the above relation as $$L' = L + \frac{\partial F}{\partial x} \dot{x} + \frac{\partial F}{\partial t} .$$ From this it follows that the primed generalized momentum $p' = \frac{\partial L'}{\partial \dot{x}}$ is related to un-primed one $p = \frac{\partial L}{\partial \dot{x}}$ by $$p' = p + \frac{\partial F}{\partial x} .$$ Compare this to the behaviour, under gauge transformation, of the generalized momentum of a charged particle in an electromagnetic field. Now, you can calculate the primed Hamiltonian from $$H'\left(x, p'\right) = p' \dot{x} - L' = H \left(x , p' - \frac{\partial F}{\partial x}\right) - \frac{\partial F}{\partial t} .$$ Using the above relations, you can easily establish the equivalence between the primed and un-primed Hamilton’s equations as well as the Poisson structure of classical mechanics. Indeed, the transformations $$x \to x , \ \ p \to p + \frac{\partial F}{\partial x} ,$$ and $$H \to H - \frac{\partial F}{\partial t} ,$$ are canonical “gauge” transformations.
Let us consider, as simple concrete example, the free particle $L = \frac{1}{2}m \dot{x}^{2}$. Under a Galilean boost, $\dot{x} \to \dot{x} + \beta$, the Lagrangian $L$ dose not remain invariant. As you can easily see, it changes by total derivative $$L \to L' = L + \frac{d}{dt}F(t,x;\beta) , \ \ \ \ \ (1)$$$$F \equiv m (x \cdot \beta + \frac{1}{2} \beta^{2}t).$$
In classical Mechanics, the system is not affected by the change (1), since the action integral, i.e., the equation of motion remains unchanged. The same is true in QM. Indeed, you can show (though not as easy as the CM case) that, under the Galilean boost, the free particle Schrödinger equation $$i\hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^{2}}{2m}\frac{\partial^{2}\Psi}{\partial x^{2}} ,$$ retains its form (in the primed system) if and only if the wave function transforms according to $$\Psi (t,x) \to \Psi'(t',x') = \exp \left(\frac{i}{\hbar}F(t,x;\beta) \right) \Psi (t,x) . \ \ \ \ \ \ (2)$$ This means that the transformation (1) of the classical system becomes a gauge transformation of the corresponding quantum system. The fact that the same function $F$ appears in (1) as well as (2) means that there is a common cause which can be shown to be the non-trivial cohomology of the Galilei group.

4. Nov 28, 2016

### vanhees71

If I remember my QT lectures right, this can be extended to canonical transformations too.

5. Nov 29, 2016

### samalkhaiat

Cohomology is a tricky subject, I might be able to add more advanced reatures to my previous post when I have some free time, may be over the weakend.

6. Dec 14, 2016

### samalkhaiat

Sorry, for late reply: busy at new job and also busy doing X-mass shopping with my family.
Okay, for obvious reasons, this is not going to be a detailed study about group cohomology, central extensions or projective representations. Rather, we will try to learn something about these concepts from equations (1) and (2) in #3. Basically, we will use (1) to learn one piece of information then jump to (2) and learn another piece before introducing the poor-man version of the underline mathematics.
I will start by setting up my notations for the Galilei group $G$. The action of $G$ on space and time $(\vec{x},t)$ is given by $$\vec{x} \to R \vec{x} + \vec{\beta} t + \vec{a} ,$$$$t \to t + \tau ,$$ where $R \in SO(3)$. General group element $g$ is, therefore, parametrized by ten parameters $g = (\tau , \vec{a} , \vec{\beta} , R)$. The above transformations allow us to deduce the group law $g_{1} g_{2} = g_{12} \in G$ to be of the form $$\tau_{12} = \tau_{1} + \tau_{2} ,$$ $$\vec{a}_{12} = \vec{a}_{1} + R_{1}\vec{a}_{2} + \vec{\beta}_{1}\tau_{2} ,$$$$\vec{\beta}_{12} = \vec{\beta}_{1} + R_{1}\vec{\beta}_{2} ,$$ $$R_{12} = R_{1}R_{2} .$$ The identity element is $e = (0 , \vec{0} , \vec{0} , 1)$, and the inverse element is given by $$g^{-1} = \left( - \tau , - R^{-1}( \vec{a} - \vec{\beta} \tau ), - R^{-1} \vec{\beta} , R^{-1} \right).$$ We will mostly ignore rotations by setting $R = 1$. When no confusion arises, I will put $x = (t , \vec{x})$ and write the coordinate transformations as $x’ = gx$. The quantum operators corresponding to the conjugate pair $(\vec{x}, \vec{p})$ will be denoted by $(\vec{X},\vec{P})$. The group $G$ has three abelian subgroups: time translations $g(\tau) = (\tau , \vec{0} , \vec{0} , 1)$, space translations $g(\vec{a}) = (0 , \vec{a} , \vec{0} , 1)$ and boosts $g(\vec{\beta}) = (0 , \vec{0} , \vec{\beta} , 1)$. Notice, in particular, that translations and boosts are commuting subgroups: $g(\vec{a}) g(\vec{\beta}) = g(\vec{\beta}) g(\vec{a})$. This, however, will no longer be true in Galilean-covariant physical theories, i.e., non-relativistic physics. Indeed, we will see that the Lie algebra of the generators does not follow from the Lie algebra of the group $G$, there will be an extension characterized by the mass.
Recall that under a finite Galilean boost $g = (0 , \vec{0} , -\vec{\beta} , 1)$, the free particle Lagrangian $L = \frac{1}{2} m \dot{x}^{2}$ changes by a total time derivative $$L’ - L = - \frac{d}{dt} \omega_{1}(t,\vec{x};\vec{\beta}) ,$$ where $$\omega_{1} (x ; \vec{\beta}) \equiv m \left( \vec{x} \cdot \vec{\beta} - \frac{1}{2} \beta^{2} t \right) . \ \ \ \ \ \ \ \ \ (1)$$ Actually, $\omega_{1}$ has a name in mathematics: A quantity depending on the variable $x$ and $n$ group elements, $\omega_{n}(x ; g_{1}, \cdots , g_{n})$, is called n-cochain, and when certain combination of n-cochains vanish, we then call it n-cocycle. We will come to that later but not bother ourselves too much with the names. Okay, infinitesimally (i.e., $\beta^{2} \approx 0$) we have $$\delta_{(\beta)} x_{i} = - \beta_{i}t , \ \ \ \delta_{(\beta)}L = - \frac{d}{dt}(m \vec{\beta} \cdot \vec{x}) . \ \ \ \ \ \ (2)$$ Now, on actual trajectories, any variation of the Lagrangian is given by total derivative $$\delta L = \frac{d}{dt} (\vec{p} \cdot \delta \vec{x}) . \ \ \ \ \ \ \ \ \ \ \ (3)$$ From (2) and (3) we obtain the (conserved) infinitesimal boost generator $$\mathcal{C} = \vec{\beta}\cdot \vec{C} = \vec{\beta} \cdot (m \vec{x} - t \vec{p}) .$$ For the quantum boost operator, we will write $$C_{i} = m X_{i} - t P_{i} . \ \ \ \ \ \ \ \ (4)$$ Using the commutation relations $[X_{i} , P_{j}] = i\hbar \delta_{ij}$ (or the classical Poisson brackets) we can easily obtain the following commutation relation: $[C_{i}, C_{j}] = 0$ which confirms the abelian nature of boosts; $[C_{i} , H] = [C_{i}, P^{2}/2m] = iP_{i}$ confirming that $C_{i}$ is constant of motion; $[C_{i}, L_{j} ] = [C_{i} , (\vec{X} \times \vec{P})_{j} ] = i\hbar \epsilon_{ijk}C_{k}$ showing that $C_{i}$ is a 3-vector; and more importantly $[C_{i}, P_{j}] = m i\hbar \delta_{ij}$ saying that we are now dealing with the (centrally) extended Galilei group $\bar{G}_{m}$ by $\mathbb{R}$. $\bar{G}_{m}$ is an 11-parameter Lie group $\bar{g} = (\tau , \vec{a} , \vec{\beta} , R , \alpha)$ with composition law given by that of the original Galilei group $G$ plus $$\alpha_{12} = \alpha_{1} + \alpha_{2} + \omega_{2}(g_{1} , g_{2}) ,$$ where, as we will explain later, $$\omega_{2}(g_{1},g_{2}) = m(\vec{\beta}_{1} \cdot R_{1}\vec{a}_{2} - \frac{1}{2}\beta_{1}^{2}\tau_{2} ),$$ is the non-trivial 2-cocycle which defines the extension.
In QM, we are naturally interested in the unitary representations of symmetry groups on Hilbert space $\mathscr{H}$ of the states, i.e., in unitary operators $U(g)$ that implement transformations $x \to gx$ of the coordinates on the states and wave functions $$\Psi^{’} (x) = \langle x | \Psi^{’} \rangle = \langle x | U(g) | \Psi \rangle .$$ So for our boosts $g = (0 , \vec{0} , -\vec{\beta} , 1)$, the unitary operator, which effects the finite transformation on the wave functions, is given by $$U(\vec{\beta}) = e^{\frac{i}{\hbar} \mathcal{C}} = e^{\frac{i}{\hbar} \ \vec{\beta} \cdot (m\vec{X} - t \vec{P})} . \ \ \ \ (5)$$ For ease of notations I will from now on drop the arrows from the scalar products of vectors, for example, the expression $a \cdot P$ will mean $\vec{a} \cdot \vec{P}$. So, instead of (5) I will simply write $$U(\vec{\beta}) = e^{\frac{i}{\hbar} \ \beta \cdot (m X - t P)} . \ \ \ \ (5')$$
Applying the identity $$e^{A + B} = e^{-\frac{1}{2}[A,B]} e^{A}e^{B} ,$$ to the RHS of Eq(5'), we get $$U(\vec{\beta}) = e^{- \frac{i}{2\hbar} m t \beta^{2}} \ e^{\frac{i}{\hbar} m \beta \cdot X} \ e^{-\frac{i}{\hbar} t \beta \cdot P} . \ \ \ \ \ (5'')$$ Now, using the eigen-value equation $\langle \vec{x}|e^{\frac{i}{\hbar}m \beta \cdot X} = \langle \vec{x}|e^{\frac{i}{\hbar}m \beta \cdot x}$, and the definition of the translation operator $$e^{-\frac{i}{\hbar} a \cdot P} |\vec{x}\rangle = |\vec{x} + \vec{a}\rangle \ \Rightarrow \ \langle \vec{x}| e^{\frac{i}{\hbar} a \cdot P} = \langle \vec{x} - \vec{a}|,$$ we can easily evaluate the action of $U(\beta)$ on wave functions $\Psi (t , \vec{x})$ from Eq(5'')
$$\langle \vec{x}|\Psi^{’}(t)\rangle = \langle \vec{x} | U(\beta)|\Psi (t)\rangle = e^{\frac{i}{\hbar} m ( \beta \cdot x - \frac{1}{2} \beta^{2} t )} \langle \vec{x} - \vec{\beta}t | \Psi (t)\rangle , \ \ \ \ (6)$$ or $$\Psi^{’} (t , \vec{x}) = U(\beta) \Psi (t , \vec{x}) = e^{\frac{i}{\hbar} \omega_{1} (t , \vec{x} ; \vec{\beta})} \Psi (t , \vec{x} - \vec{\beta}t) . \ \ \ \ \ (6’)$$ Thus, the finite transformation of wave function contains the phase (or “1-cochain”) $\omega_{1}(x ; \beta) = m(\beta \cdot x - \frac{1}{2} \beta^{2}t)$, that we have already seen in the finite change of the Lagrangian $L$ under boosts. Notice that the above transformation law agrees with the fact (which was stressed by Weyl during the early days of QM) that physical (pure) states are described rays $\mathscr{R}$, not by vectors of a Hilbert space $\mathscr{H}$: $\{\mathscr{R}\} = \mathscr{H} / \sim$, where $\sim$ is the equivalence relation which identifies vectors $|\Psi \rangle$ and $|\Psi^{’} \rangle$ of $\mathscr{H}$ which differ in a phase.
Of course, boosts from an abelian group, we have actually shown that $[\beta_{1}\cdot C , \beta_{2} \cdot C] = 0$. Thus, the composition law for two successive boosts is simply given by $$U(\vec{\beta}_{2})U(\vec{\beta}_{1}) = U(\vec{\beta}_{1} + \vec{\beta}_{2} ) \equiv U(\vec{\beta}_{12}) .$$ From this it follows that $$\langle \vec{x}|U(\vec{\beta}_{2})U(\vec{\beta}_{1})|\Psi (t)\rangle = \langle \vec{x}|U(\vec{\beta}_{12})|\Psi (t)\rangle . \ \ \ \ \ \ \ \ \ \ \ (7)$$ Let us now use the transformation law (6’) to evaluate both sides of (7). We have already calculated the RHS, it is $$\langle \vec{x}|U(\vec{\beta}_{12})|\Psi (t)\rangle = e^{\frac{i}{\hbar}\omega_{1}(x ; \vec{\beta}_{12})} \langle \vec{x} - \vec{\beta}_{12}t | \Psi (t)\rangle ,$$ or
$$\Psi (t , \vec{x} - \vec{\beta}_{12}t ) = e^{- \frac{i}{\hbar}\omega_{1}(x ; \vec{\beta}_{12})} \langle \vec{x}|U(\vec{\beta}_{12})|\Psi (t)\rangle . \ \ \ \ \ \ \ \ \ (8)$$ On the LHS of (7), we do the followings
\begin{align*} \langle \vec{x}|U(\vec{\beta}_{2})U(\vec{\beta}_{1})|\Psi (t)\rangle &= \int d^{3}y \ \langle \vec{x}|U(\vec{\beta}_{2}) | \vec{y}\rangle \langle \vec{y}|U(\vec{\beta}_{1})|\Psi (t)\rangle \\ &= \int d^{3}y \ \langle \vec{x}|U(\vec{\beta}_{2}) | \vec{y}\rangle e^{\frac{i}{\hbar}\omega_{1} (t , \vec{y} ; \vec{\beta}_{1})} \Psi (t , \vec{y} - \vec{\beta}_{1} t ) \\ &= \int d^{3}y \ e^{-\frac{i}{2\hbar} m t \beta_{2}^{2}} \ \langle \vec{x}| e^{\frac{i}{\hbar} m \beta_{2} \cdot X} \ e^{- \frac{i}{\hbar} t \beta_{2} \cdot P} | \vec{y} \rangle e^{\frac{i}{\hbar}\omega_{1} ( y ; \vec{\beta}_{1})} \Psi (t , \vec{y} - \vec{\beta}_{1} t ) \\ &= e^{\frac{i}{\hbar}\omega_{1}(x ; \vec{\beta}_{2})} \ \int d^{3}y \ \langle \vec{x}| e^{- \frac{i}{\hbar} t \beta_{2} \cdot P} | \vec{y} \rangle \ e^{\frac{i}{\hbar}\omega_{1} ( y ; \vec{\beta}_{1})} \Psi (t , \vec{y} - \vec{\beta}_{1} t ) \\ &= e^{\frac{i}{\hbar}\omega_{1}(x ; \vec{\beta}_{2})} \ \int d^{3}y \ \delta^{3}(x - y - \beta_{2}t ) e^{\frac{i}{\hbar}\omega_{1} ( y ; \vec{\beta}_{1})} \Psi (t , \vec{y} - \vec{\beta}_{1} t ) \\ &= e^{\frac{i}{\hbar}\left[ \omega_{1}(x ; \beta_{2}) + \omega_{1}(x - \beta_{2}t ; \beta_{1}) \right]} \ \Psi (t , \vec{x} - \vec{\beta}_{12} t ) . \end{align*}
Substituting Eq(8) in the last equality, we obtain
$$\langle \vec{x}|U(\vec{\beta}_{2})U(\vec{\beta}_{1})|\Psi \rangle = e^{\frac{i}{\hbar}\left[ \omega_{1}(x ; \beta_{2}) + \omega_{1}(x - \beta_{2}t ; \beta_{1}) - \omega_{1}(x ; \beta_{12})\right]} \langle \vec{x}|U(\vec{\beta}_{12})|\Psi \rangle .$$
Comparing this result with Eq(7), we obtain the following algebraic condition on the 1-cochain $\omega_{1}(x;\beta)$
$$\Delta \omega_{1} \equiv \omega_{1}(x - \beta_{2}t ; \beta_{1}) + \omega_{1}(x ; \beta_{2}) - \omega_{1}(x ; \beta_{12}) = 0 \ \mbox{mod} (2 \pi \hbar \mathbb{Z}) . \ \ \ \ (9)$$ When the 1-cochain $\omega_{1}(x;\beta)$ satisfies (9), it is called 1-cocycle. And $\Delta$ is the so-called co-boundary operator (I will say few more things about them later). Of course it is trivial thing to check that our phase function $\omega_{1}(x;\beta) = m \beta \cdot x - \frac{1}{2} m \beta^{2}t$ does satisfy the 1-cocycle condition (9), after all, our derivation of the condition (9) was based on the explicit form of $\omega_{1}(x;\beta)$. However, notice that the 1-cocycle $\omega_{1}(x;\beta)$ can identically be written as $$\omega_{1}(t, \vec{x} ; \beta) = m \beta \cdot x - \frac{1}{2} m \beta^{2}t \equiv -\frac{m}{2t}(\vec{x} - \vec{\beta}t)^{2} + \frac{m}{2t} (\vec{x})^{2} .$$ If we now introduce the function (i.e., 0-cochain) $\alpha_{0}(t,\vec{x}) = -\frac{m}{2t} (\vec{x})^{2}$, we obtain $$\omega_{1}(t, \vec{x} ; \beta) = \alpha_{0}(t , \vec{x} - \vec{\beta}t) - \alpha_{0}(t , \vec{x}) \equiv \Delta \alpha_{0} . \ \ \ \ (10)$$ In general, a n-cocycle $\omega_{n}$ is called trivial (i.e., it can be removed) if it can be written as $\Delta$ of a (n-1)-cochain. Thus, (10) means that our boosts 1-cocycle $\omega_{1}$ is a trivial one, i.e., we can removed it by adjusting the phase of the wave function. Indeed, using (10) we can rewrite the transformation law (6’) in the form $$\left(e^{\frac{i}{\hbar}\alpha_{0}(x)}U(\beta) e^{-\frac{i}{\hbar}\alpha_{0}(x)}\right) \left(e^{\frac{i}{\hbar}\alpha_{0}(x)} \Psi (x) \right) = e^{\frac{i}{\hbar}\alpha_{0}(x - \beta t)} \Psi (x - \beta t) .$$ Thus, by defining new wave function $$\Phi (t , \vec{x}) = e^{\frac{i}{\hbar}\alpha_{0}(t , \vec{x})} \Psi (t, \vec{x}) , \ \ \ \ \ \ \ \ \ (11)$$ and a new unitary operator $$V(\beta) = e^{\frac{i}{\hbar}\alpha_{0}(x)}U(\beta) e^{-\frac{i}{\hbar}\alpha_{0}(x)} , \ \ \ \ (12)$$ we find the following simple transformation and composition laws $$\Phi^{’} (t , \vec{x}) = V(\beta) \Phi (t , \vec{x}) = \Phi (t , \vec{x} - \vec{\beta}t) ,$$ $$V(\beta_{1}) V(\beta_{2}) = V(\beta_{1} + \beta_{2}) \equiv V(\beta_{12}) .$$ The function $\alpha_{0}(t,\vec{x})$ also removes $\omega_{1}$ from our original Lagrangian $L = \frac{1}{2}m \dot{x}^{2}$. Indeed, we know that adding $\frac{d}{dt}\alpha_{0}(t , \vec{x})$ to $L$ produces a new equivalent Lagrangian $\hat{L}$ except that now $\hat{L}$ is invariant under boosts: $$\hat{L} = L + \frac{d\alpha_{0}}{dt} \ \Rightarrow \ \Delta \hat{L} = \Delta L + \frac{d}{dt} \Delta \alpha_{0} .$$ But, we know $\Delta L = - \frac{d}{dt}\omega_{1}$, then (10) leads to $\Delta \hat{L} = \frac{d}{dt}(-\omega_{1} + \omega_{1} ) = 0$. Okay, let’s finish the talk about 1-cocycle by showing that the phase-freedom in QM corresponds to the freedom of adding a total time-derivative to the Lagrangian. In other words, we would like to prove that a phase change in a wave function corresponds to a canonical transformation which changes the Lagrangian by a total time-derivative. In the Schrodinger equation for $\Psi (t,q)$, we make the substitution (11) $$i\hbar \frac{\partial}{\partial t}\left( e^{-\frac{i}{\hbar}\alpha_{0}(t , Q)} \Phi (t , q)\right) = \frac{P^{2}}{2m}\left( e^{-\frac{i}{\hbar}\alpha_{0}(t , Q)} \Phi (t , q)\right) .$$ Notice that I replaced the variable $q$ by the operator $Q$, this is possible because in the coordinate representation we can use $f(Q) \Phi (q) = f(q) \Phi (q)$. So, by doing the trivial algebra, we find
\begin{align*} i\hbar \frac{\partial \Phi}{\partial t} &= \frac{1}{2m} \left( e^{\frac{i}{\hbar}\alpha_{0}(t,Q)} \ P \ e^{-\frac{i}{\hbar}\alpha_{0}(t,Q)} \right)^{2} \Phi - \frac{\partial \alpha_{0}}{\partial t} \Phi \\ &= \frac{1}{2m} \left( P + \frac{i}{\hbar}[\alpha_{0}(t,Q) , P] \right)^{2} \Phi - \frac{\partial \alpha_{0}}{\partial t} \Phi \\ &= \left[ \frac{1}{2m} \left( P - \frac{\partial \alpha_{0}}{\partial q} \right)^{2} - \frac{\partial \alpha_{0}}{\partial t} \right] \Phi . \end{align*}
So, the Hamiltonian relevant to $\Phi$ is $$\hat{H} = \frac{1}{2m} \left( P - \frac{\partial \alpha_{0}}{\partial q} \right)^{2} - \frac{\partial \alpha_{0}}{\partial t} ,$$ and the corresponding Lagrangian is $$\hat{L} = \frac{1}{2}\dot{q}^{2} + \frac{d}{dt}\alpha_{0}(t,q) = L + \frac{d}{dt}\alpha_{0} .$$ The important thing to notice is the following: by removing the trivial 1-cocycle $\omega_{1}$, the resulting pair $( \Phi , \hat{L})$ are scalars with respect to boosts only, i.e., new phase and new time-derivative will reappear in $( \Phi , \hat{L})$ when we consider Galilean transformations other than boosts, or compositions of boosts with translations. This is connected with the fact that $G$ has intrinsic projective representations.
I am sorry because I have consumed all the free time I had. I will talk about the appearance of 2-cocycle (projective representations) and 3-cocycle in QM sometime soon.

Last edited: Dec 16, 2016
7. Dec 15, 2016

### vanhees71

Great posting (put it to the Insights section for reference!). Do you have a citation for where and when Weyl stressed the important fact that pure states are represented by rays rather than vectors in Hilbert space (I guess in his book on group theory in QT which appeared in German in 1931?).

8. Dec 16, 2016

### samalkhaiat

Thank you.
Well, that means that I have to re-type the whole thing again? That will be a waste of my time, for I prefer to produce new post with different material.
Yes, that one as well as "The classical group", Princeton Univ. Press (1946).

9. Dec 17, 2016

### A. Neumaier

No. If you edit your post you can copy the source text and paste it into the editor area for the Insight article!

10. Dec 18, 2016

### dextercioby

Amazing posting by sam, again. It's like Bargmann, Levy-Leblond and Voisin writing at the same time. :)

11. Dec 18, 2016

### samalkhaiat

Thanks Dexter. I will be heading toward those great people soon. I hope, I will be able to represent their work well, and do them justice :)