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3

This formula follows the usual heuristic discretization rules (here written in 1D): $$\tag{1} \text{discrete var.}\qquad i\in\{1, \ldots,N\}, ~~x_i=i\Delta ~~\longrightarrow~~x~\in~[0,L] \qquad \text{cont. var.},$$ $$\tag{2} \text{sum}\qquad \sum_{i=1}^N ~~\longrightarrow~~ \int_0^L \! \frac{dx}{\Delta} \qquad\text{integral},$$ $$\tag{3} ...


3

$X_i,Y_i,Z_i$ are three Pauli matrices acting on the $i$-th qubit where $i=1,2,3,4,5,6,7,8,9$ labels the qubit. In equation 4.1, the state is a superposition of tensor product of three states similar to $|000\rangle$. The latter is a state of three qubits, so if one takes the third power, it is a state of $3\times 3 = 9$ qubits. $X_1$ differs from $X_8$ by ...


3

For a free theory, say for one scalar field for simplicity, which gives a a linear differential equation for the field $\phi$, one can cast the hamiltonian $$ H=\frac{1}{2}\int d^3x \dot\phi^2+(\partial_i \phi)^2+ m^2\phi^2 $$ in this form (basically by taking a Fourier Transform) $$ H=\mathrm{const}+\int \frac{d^3 k}{(2\pi)^3} ...


2

For simplicity, let us talk about a scalar field $\phi : \mathbb{R}^4 \rightarrow \mathbb{R}$. The action for a free scalar field is $$S[\phi] = \frac{1}{2}\int_{\mathbb{R}^4} \partial_\mu\phi\partial^\mu\phi - m^2\phi^2$$ and its classical equations of motion is the Klein-Gordon equation $$ (\partial_\mu\partial^\mu + m^2) \phi = 0 $$ Now that looks ...


2

In a comment you write space time symmetries don't fit into the framework of the action since the action is a functional on the fields only not also on space time (space time here appears merely as a dummy variable This isn't quite right. A given spacetime transformation often induces a transformation on fields themselves, and in this way, spacetime ...


2

You can utilize the construction of the $Q$ space, as described in Reed and Simon vol.2, page 228-230. Oversimplifying, you can make the analogy $\lvert \phi\rangle \sim \lvert x\rangle$, but the associated momentum is not $\hat{p}$, but $\hat{\pi}$ (the canonical conjugate momentum of the field $\hat{\phi}$). With slightly more precision: the Fock space ...


1

They are very different. When you use a Higgs mechanism with a Yang-Mills action, symmetry breaking causes the gauge fields $A$ to gain mass. This is done in 4D. When you add a Chern-Simons term to a Yang-Mills action, you can see from the field equations that $\ast F$ becomes massive, not $A$. There is no symmetry breaking here. Also this is in 3D and ...


1

No $\hat\phi|0\rangle$ is not an eigenvector of $\hat\phi$. You can see this, for example, by writing out $\hat\phi$ in terms of creation and annihilation operators, then compare $\hat\phi|0\rangle$ against $\hat\phi^2|0\rangle$, and observe that one is not a scalar multiple of the other. So as you suspected, eq. 5 is not correct To obtain some analogy of ...


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Below that regime, we have the strongly coupled regime where perturbative approaches fail, due to the large value of coupling constant $\alpha_S$. The same is related to the QCD $\beta$ function via this relation. The behavior as a function of the energy scale looks roughly like this. Any perturbation expansion in this regime would give a divergent series, ...


1

Let's look to the expression for field with mass $m$ and spin $s$ (for massless case following statements exist in similar form): $$ \tag 1 \hat {\psi}_{a}(x) = \sum_{\sigma = -s}^{s}\int \frac{d^{3}\mathbf p}{\sqrt{(2 \pi )^{3} 2E_{\mathbf p}}}\left( u^{\sigma}_{a}(\mathbf p )e^{-ipx}\hat{a}_{\sigma}(\mathbf p ) + v^{\sigma}_{a}(\mathbf p ...



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