# What is the physical meaning of equivalence of 1st and 2nd quantization formalism?

Ref (Superstring theory (Green, Schwarz, Witten))

Take an $n$ dimensional euclidean space-time $x_0,x_1...x_{n -1}$, a relativist real scalar field, with a propagator $G_E(x,y)$.

The propagator could be written : $G_E(x,y) = <x|\Box^{-1}|y>$ (with $<x|y> = \delta (x - y))$

We could write $<x|\Box^{-1}|y> = \int_0^{+\infty} ~d\tau~ <x|e^{- \tau \Box}|y>$

Now, we can consider a $n+1$ dimensional space-time, considering non-relativist particles. The hamiltonian is then $H_{n+1} = \frac{p^2}{2m} = \frac{\Box}{2m}$

By taking $m = \frac{1}{2}$, we get :

$<x|\Box^{-1}|y> = \int_0^{+\infty} ~d\tau~ <x|e^{- \tau H_{n+1}}|y>$

Now, we can use standard Quantum mechanics path integrals, that is :

$<x|e^{- \tau H_{n+1}}|y> = \int_{x(0) = x}^{x(\tau) = y} Dx(u) e^{-\frac{1}{4} \int du ~\dot x^2(u)}$,

where $x(u)$ represents a path from $x$ to $y$, done in a "proper time" $\tau$.

So, finally :

$G_E(x,y) = \int_0^{+\infty} ~d\tau~ \int_{x(0) = x}^{x(\tau) = y} Dx(u) e^{-\frac{1}{4} \int du ~\dot x^2(u)}$

Now, the question is :

What is the philosophy of these equivalence between 1st and 2nd quantization formalism ?

• Isn't it just that the object $\langle vac|\phi(x)\phi(y)|vac\rangle$ in the field theory language (2nd quantization) is proportional to $\langle x|e^{-\tau H}|y\rangle$ in single-particle QM (1st quantization), which follows from the fact that the operator $\phi(x)$ acting on the vacuum state creates a single particle at $x$? – higgsss Jul 12 '13 at 8:34
• More precisely, the euclidean propagator $G_E(x,y)$ is the euclidean version of $<0|T(\Phi(x)\Phi(y))|0>$. The problem is that, in First Quantization formalism, there is no explicit creation or annihilation operator . – Trimok Jul 12 '13 at 8:42
• Apparently I haven't read your question carefully. Sorry about that. Now it looks to me that it is a mathematical coincidence that the same differential operator (n-dimensional Laplacian) is involved in the two different problems. – higgsss Jul 12 '13 at 9:39
• Even it this is not maybe directly related, note that, when you treat a relativist particle in the light-cone gauge, The expression of the energy looks like a non relativistic expression, see for instance this reference See for instance formulae (11.1.12) and (11.3.10) – Trimok Jul 12 '13 at 10:02