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# 物理代写|核物理代考Nuclear Physics代写|One-particle states

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## 物理代写|核物理代考Nuclear Physics代写|One-particle states

The nucleon has isospin $1 / 2$. In other words, each of the operators $T_1, T_2$ and $T_3$ which are associated with this particle have eigenvalues $\pm 1 / 2$. The operator $T^2=T_1^2+T_2^2+T_3^2$ is proportional to the identity with eigenvalue $3 / 4$.

The states $|\mathrm{p}\rangle$ and $|\mathrm{n}\rangle$, are, by definition, the eigenstates of the particular operator $T_3$
$$T_3|\mathrm{p}\rangle=(1 / 2)|\mathrm{p}\rangle, \quad T_3|\mathrm{n}\rangle=(-1 / 2)|\mathrm{n}\rangle$$
In actual physics, the operator $T_3$ plays a special role since electric charge is related to $T_3$ by
$$Q=T_3+1 / 2$$
The action of $T_1$ and $T_2$ on these states, with $T_{ \pm}=T_1 \pm T_2$, can be written as
\begin{aligned} T_{+}|\mathrm{p}\rangle=0 & T_{-}|\mathrm{n}\rangle=0 \ T_1|\mathrm{p}\rangle=(1 / 2)|\mathrm{n}\rangle & T_1|\mathrm{n}\rangle=(1 / 2)|\mathrm{p}\rangle \ T_2|\mathrm{p}\rangle=(\mathrm{i} / 2)|\mathrm{n}\rangle & T_2|\mathrm{n}\rangle=(-\mathrm{i} / 2)|\mathrm{p}\rangle . \end{aligned}
An arbitrary nucleon state $|N\rangle$ is written
$$|N\rangle=\alpha|\mathrm{p}\rangle+\beta|\mathrm{n}\rangle \quad|\alpha|^2+|\beta|^2=1$$
We remark that all of this is an abstraction applicable only to a world without electromagnetism. A state such as
$$\frac{1}{\sqrt{2}}\left(\left|T_3=1 / 2\right\rangle+\left|T_3=-1 / 2\right\rangle\right),$$
which is oriented along the direction $T_2$ cannot be observed physically. Since it is a superposition of a proton and a neutron, it is both of charge 0 and 1 ; at the same time it creates and doesn’t create an electrostatic field. As such, it is a superposition of two macroscopically different states, an example of a “Schrödinger cat.”

## 物理代写|核物理代考Nuclear Physics代写|The generalized Pauli principle

The Pauli principle states that two identical fermions must be in an antisymmetric state. If the proton and the neutron were truly identical particles up to the projection of their isospin along the axis $T_3$, a state of several nucleons should be completely antisymmetric under the exchange of all variables, including isospin variables. If we forget about electromagnetic interactions, and assume exact invariance under rotations in isospin space, the Pauli principle is generalized by stating that an $A$-nucleon system is completely antisymmetric under the exchange of space, spin and isospin variables. This assumption does not rest on as firm a foundation as the normal Pauli principle and is only an approximation. However, we can expect that it is a good approximation, up to electromagnetic effects.

The generalized Pauli principle restricts the number of allowed quantum states for a system of nucleons. We shall see below how this determines the allowed states of the deuteron.

# 核物理代写

## 物理代写|核物理代考Nuclear Physics代写|One-particle states

$$T_3|\mathrm{p}\rangle=(1 / 2)|\mathrm{p}\rangle, \quad T_3|\mathrm{n}\rangle=(-1 / 2)|\mathrm{n}\rangle$$

$$Q=T_3+1 / 2$$
$T_1$和$T_2$对这些状态的作用，加上$T_{ \pm}=T_1 \pm T_2$，可以写成
\begin{aligned} T_{+}|\mathrm{p}\rangle=0 & T_{-}|\mathrm{n}\rangle=0 \ T_1|\mathrm{p}\rangle=(1 / 2)|\mathrm{n}\rangle & T_1|\mathrm{n}\rangle=(1 / 2)|\mathrm{p}\rangle \ T_2|\mathrm{p}\rangle=(\mathrm{i} / 2)|\mathrm{n}\rangle & T_2|\mathrm{n}\rangle=(-\mathrm{i} / 2)|\mathrm{p}\rangle . \end{aligned}

$$|N\rangle=\alpha|\mathrm{p}\rangle+\beta|\mathrm{n}\rangle \quad|\alpha|^2+|\beta|^2=1$$

$$\frac{1}{\sqrt{2}}\left(\left|T_3=1 / 2\right\rangle+\left|T_3=-1 / 2\right\rangle\right),$$

## 物理代写|核物理代考Nuclear Physics代写|The generalized Pauli principle

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## MATLAB代写

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