物理代写|弦论代写string theory代考|PHY-897S

物理代写|弦论代写string theory代考|Non-oriented Strings

All solutions we have obtained for open and closed strings can be built from solutions which are either even or odd with respect to world-sheet parity
$$\Omega: \sigma^1 \mapsto \pi-\sigma^1 \cong-\sigma^1 \bmod \pi .$$
As indicated, $\Omega$ can be interpreted as a reflection up to boundary conditions of the spatial world-sheet coordinate $\sigma^1$. Since $\Omega^2=\mathbb{1}$, the eigenvalues of this transformation are $\pm 1$, and we refer to the corresponding eigenstates as even and odd, respectively. For closed strings world-sheet parity interchanges left- and rightmoving excitations, whereas for open strings the mode number decides which excitations are even and odd:
$$\Omega: \begin{cases}\alpha_m^\mu \leftrightarrow \tilde{\alpha}_m^\mu, & \text { (closed), } \ \alpha_n^\mu \mapsto(-1)^n \alpha_n^\mu, & \text { (Neumann), } \ \alpha_n^\mu \mapsto(-1)^{n+1} \alpha_n^\mu, & \text { (Dirichlet). }\end{cases}$$
Non-oriented strings theories are defined by restricting to configurations which are even. Since in such a theory all configurations are invariant under world-sheet parity, one looses the information about the orientation of the string. In other words, one cannot distinguish between the ends of the string and the two sides of the worldsheet. Non-oriented closed strings are symmetric under exchange of left- and rightmoving waves, while for non-oriented open strings, depending on the boundary conditions, either the odd- or the even-numbered modes cannot be excited. While this reduces the number of independent excitations, it increases the number of worldsheet topologies, because we now have to admit non-orientable surfaces, such as the Moebius strip (for non-oriented open strings) and the Klein bottle (for non-oriented closed strings).

物理代写|弦论代写string theory代考|Quantised Relativistic Particles

The usual heuristic approach to quantisation is to promote the canonical coordinates and canonical momenta of a classical theory to self-adjoint operators acting on a separable Hilbert space $\mathcal{H}$, and to impose the canoncial commutation relations. The canonical commutation relations can be motivated through replacing the Poisson brackets ${\cdot, \cdot}$ of the classical theory by quantum commutators $[\cdot, \cdot]$, using the formal substitution rule ${\cdot, \cdot} \rightarrow-i[\cdot, \cdot]$. We will not evaluate the Poisson brackets of the classical theory, but directly postulate the canonical commutation relations.

In the case of a free non-relativistic particle with Cartesian coordinates $x^i$ and momenta $p^j$, the canonical commutation relations are
$$\left[x^i, p^j\right]=i \delta^{i j},$$
where we have set $\hbar=1$, and where the unit operator on the Hilbert space $\mathcal{H}$ is understood on the right-hand side. We will procede formally and ignore the technical complications caused by the fact that $x^i, p^j$ are unbounded operators on an infinite dimensional Hilbert space.
For a relativistic particle the natural generalisation of $(3.1)$ is
$$\left[x^\mu, p^v\right]=i \eta^{\mu v} .$$
However, we know that the components of the relativistic momentum are subject to the mass shell condition $p^2+m^2=0$. One option is to solve this constraint in the classical theory, and then to quantise the theory using only gauge-inequivalent quantities. A specific version of this procedure is the so-called light cone quantisation, which will be discussed later (see Chapter 10). Any such scheme has the disadvantage that Lorentz invariance is no longer manifest. Here we will follow the complementary, covariant approach, where canonical commutation relations are imposed on Lorentz covariant quantities, while the constraint $p^2+m^2=0$ is imposed afterwards and selects a subspace of physical states.

We start by constructing a representation space $\mathcal{F}$ for the commutations relations (3.2), which we call the Fock space. This is done by postulating the existence of a distinguished state, the vacuum or ground state $|0\rangle$, which is translation invariant:
$$p^v|0\rangle=0 .$$
The Fock space $\mathcal{F}$ is generated by applying operators built out of the canonically conjugate operator $x^\mu$. We assume that $p^v$ has a complete set of eigenstates, so that $\mathcal{F}$ is spanned by momentum eigenstates $|k\rangle$,
$$p^v|k\rangle=k^v|k\rangle .$$

物理代写|弦论代写string theory代考|Non-oriented Strings

$$\Omega: \sigma^1 \mapsto \pi-\sigma^1 \cong-\sigma^1 \bmod \pi .$$

$\Omega:\left{\alpha_m^\mu \leftrightarrow \tilde{\alpha}_m^\mu, \quad\right.$ (closed), $\alpha_n^\mu \mapsto(-1)^n \alpha_n^\mu, \quad$ (Neumann), $\alpha_n^\mu \mapsto(-1)^{n+1} \alpha_n^\mu, \quad$ (Dirichlet).

物理代写|弦论代写string theory代考|Quantised Relativistic Particles

$$\left[x^i, p^j\right]=i \delta^{i j},$$

$$\left[x^\mu, p^v\right]=i \eta^{\mu v} .$$

$$p^v|0\rangle=0 .$$

$$p^v|k\rangle=k^v|k\rangle .$$

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