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If $H$ is a translationally invariant Hamiltonian, how can I convince myself that the correlation function (on the ground state $\left|G\right\rangle$) $\left\langle G|\psi(x)\psi(x’)|G\right\rangle$ depends only on $x-x’$.

How is the field operator $\psi(x)$ related to $\psi(0)$ thro’ the translation operator anyway? How is the translation operator defined in second quantization?

Qmechanic
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Ambrose Chau
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  • The unitary translation operator $U(x)$ is defined by the condition $U(x)\psi(0)U^\dagger(x)=\psi(x)$, and $U(x)U(x')=U(x+x')$, and since the ground state of a translation-invariant Hamiltonian tends to be translation invariant (ignoring the possibility of spontaneous symmetry breaking resulting in a crystalline ground state), this means $U(x)|G\rangle\propto |G\rangle$. These things can be used to prove that the correlation function depends only on $x-x'$. Or are you asking for an explicit expression for $U(x)$ in terms of the field operators? – Chiral Anomaly Jan 19 '19 at 16:23
  • I am not sure how to prove $U(x)\psi(0)U^\dagger(x)=\psi(x)$. And how to express $U(x)$ in terms of field operators too. I know in first quantization, $U(x)=\exp(-ix\hat{p}/\hbar)$. But in what way can we talk of $\hat{p}$ in second quantization? – Ambrose Chau Jan 19 '19 at 16:49
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    The relationship $U(x)\psi(0)\tilde U(x)=\psi(x)$ is not something to prove. It's what "translation operator" means. It's only a definition, just like in non-relativistic quantum mechanics ("first quantization"). The thing to prove is how the operator that has this property can be expressed in terms of the field operators ("second quantization" = quantum field theory). I'll write an answer about that... – Chiral Anomaly Jan 19 '19 at 16:52

2 Answers2

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Consider the simplest QFT, namely the free scalar field. The equation of motion (in the Heisenberg picture) is $$ (\partial_t^2-\nabla^2 +m^2)\phi(t,\mathbf{x})=0 \tag{1} $$ and the equal-time commutation relations are \begin{gather} [\phi(t,\mathbf{x}),\,\dot\phi(t,\mathbf{y})] =i\delta^3(\mathbf{x}-\mathbf{y}) \\ [\phi(t,\mathbf{x}),\,\phi(t,\mathbf{y})]=0 \hskip2cm [\dot\phi(t,\mathbf{x}),\,\dot\phi(t,\mathbf{y})]=0. \tag{2} \end{gather} We want to construct a unitary operator $U(\mathbf{x})$ that satisfies $$ U(\mathbf{x})\phi(t,\mathbf{0})U^\dagger(\mathbf{x})= \phi(t,\mathbf{x}), \tag{3} $$ which is the definition of the translation operator. The momentum operators $P_k$ are the hermitian generators of the translation group (by definition again), so $$ U(\mathbf{x})=\exp(i\mathbf{x}\cdot \mathbf{P}) \hskip2cm \mathbf{x}\cdot \mathbf{P}\equiv\sum_k x_k P_k. \tag{4} $$ in units where $\hbar=1$. Take the gradient of (3) with respect to $x_k$ to get $$ [P_k\,\phi(t,\mathbf{x})]=-i\nabla_k\phi(t,\mathbf{x}). \tag{5} $$ This is the defining property of the operators $P_k$. What we want, though, is an explicit expression for $P_k$ in terms of the field operators. In general, we can use Noether's theorem to get an expression for $P_k$ in terms of the field operators. Or, instead of going through Noether's theorem, we can write down an ansatz and then prove that it works. (The second approach is easier when we already know the answer, and since I do already know the answer, I'll use the second approach here.) The commutation relations (2) imply that the operator $$ P_k=\int d^3x\ \dot\phi(t,\mathbf{x})\nabla_k \phi(t,\mathbf{x}) \tag{6} $$ is hermitian and satisfies the condition (5), so this ansatz works. Equation (6) expresses the momentum operators in terms of the field operators, and then equation (4) gives the translation operators. If the field operators are written in terms of the usual creation/annihilation operators, then (6) becomes $$ P_k\propto \int d^3p\ p_k a^\dagger(\mathbf{p})a(\mathbf{p}). \tag{7} $$ The post

Derivation of total momentum operator QFT

writes out this last step a little more explicitly. Also see equation (2.21) in

which is one of the first sources I found in a quick search for "momentum operator in QFT."

The preceding equations explain how to express $U(\mathbf{x})$ in terms of the field operators. To address the original question about why the two-point correlation function is translation-invariant, we only need equations (3) and (4), together with the assumption that the ground state $|G\rangle$ is translation-invariant: $U(\mathbf{x})|G\rangle=|G\rangle$. This gives \begin{align} \langle G|\phi(\mathbf{x})\phi(\mathbf{x'})|G\rangle &= \langle G|U(\mathbf{x})\phi(\mathbf{0})U^\dagger(\mathbf{x}) U(\mathbf{x'})\phi(\mathbf{0})U^\dagger(\mathbf{x'})|G\rangle \\ &= \langle G|\phi(\mathbf{0}) U(\mathbf{x'}-\mathbf{x})\phi(\mathbf{0})|G\rangle, \end{align} which shows that the correlation function depends only on $\mathbf{x'}-\mathbf{x}$.

Chiral Anomaly
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The question is much more generic than you're leading on. To prove that any function $f(x,y)$ that is translationally invariant, i.e. a function for which $f(x, y) = f(x + a, y + a)$ for all translations $a$, reduces to a function of the difference $x - y$, just take $a = -y$ and you get $f(x, y) = f(x - y, 0)$, and then define $g(z) = f(z, 0)$. Then, $f(x, y) = g(x - y)$, which shows that the translationally-invariant function in $(x, y)$ reduces to a function of the difference $x - y$.

Lydia Marie Williamson
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