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# 物理代写|量子力学代写Quantum mechanics代考|PHYS4141 Young diagrams

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## 物理代写|量子力学代写Quantum mechanics代考|Young diagrams

A Young diagram consists of an array of boxes (or some other symbol) arranged in one or more left-justified rows, with each row being at least as long as the row beneath. The correspondence between a diagram and a multiplet label is: The top row juts out $\alpha$ boxes to the right past the end of the second row, the second row juts out $\beta$ boxes to the right past the end of the third row, etc. A diagram in $\mathrm{SU}(n)$ has at most $n$ rows. There can be any number of “completed” columns of $n$ boxes buttressing the left of a diagram; these don’t affect the label. Thus in $\mathrm{SU}(3)$ the diagrams
represent the multiplets $(1,0),(0,1),(0,0),(1,1)$, and $(3,0)$. In any $\mathrm{SU}(n)$, the quark multiplet is represented by a single box, the antiquark multiplet by a column of $(n-1)$ boxes, and a singlet by a completed column of $n$ boxes.

## 物理代写|量子力学代写Quantum mechanics代考|Coupling multiplets together

The following recipe tells how to find the multiplets that occur in coupling two multiplets together. To couple together more than two multiplets, first couple two, then couple a third with each of the multiplets obtained from the first two, etc.

First a definition: A sequence of the letters $a, b, c, \ldots$ is admissible if at any point in the sequence at least as many $a$ ‘s have occurred as $b$ ‘s, at least as many $b$ ‘s have occurred as $c$ ‘s, etc. Thus $a b c d$ and $a a b c b$ are admissible sequences and $a b b$ and $a c b$ are not. Now the recipe:
(a) Draw the Young diagrams for the two multiplets, but in one of the diagrams replace the boxes in the first row with a’s, the boxes in the second row with b’s, etc. Thus, to couple two SU(3) octets (such as the $\pi$-meson octet and the baryon octet), we start with $\square$ and a a $b$ . The unlettered diagram forms the upper left-hand corner of all the enlarged diagrams constructed below.
(b) Add the a’s from the lettered diagram to the right-hand ends of the rows of the unlettered diagram to form all possible legitimate Young diagrams that have no more than one $a$ per column. In general, there will be several distinct diagrams, and all the $a$ ‘s appear in each diagram. At this stage, for the coupling of the two SU(3) octets, we have:
(c) Use the b’s to further enlarge the diagrams already obtained, subject to the same rules. Then throw away any diagram in which the full sequence of letters formed by reading right to left in the first row, then the second row, etc., is not admissible.
(d) Proceed as in (c) with the $c$ ‘s (if any), etc.
The final result of the coupling of the two $\mathrm{SU}(3)$ octets is:
Here only the diagrams with admissible sequences of $a$ ‘s and $b$ ‘s and with fewer than four rows (since $n=3$ ) have been kept. In terms of multiplet labels, the above may be written
$$(1,1) \otimes(1,1)=(2,2) \oplus(3,0) \oplus(0,3) \oplus(1,1) \oplus(1,1) \oplus(0,0) .$$

## 物理代写|量子力学代写|量子力学代考|将多子耦合在一起

(a) 画出这两个多子的杨氏图，但在其中一个图中用a替换第一行的方框，用b替换第二行的方框，等等。因此，为了耦合两个SU(3)八重体（如$pi$介子八重体和重子八重体），我们从$/square$和a a $b$开始。这个无字图构成了下面构建的所有放大图的左上角。
(b) 将有字图中的a添加到无字图各行的右端，形成所有可能的合法杨氏图，每列不超过一个$a$。一般来说，会有几个不同的图，所有的$a$都出现在每个图中。在这个阶段，对于两个sU(3)八边形的耦合，我们有。
(c) 使用b来进一步扩大已经得到的图，并遵守同样的规则。然后扔掉任何图，在其中通过从右到左读第一行，然后读第二行等形成的完整字母序列是不被允许的。
(d) 按(c)的方法处理$c$’s（如果有的话），等等。

$$(1,1) \otimes(1,1)=(2,2) \oplus(3,0) \oplus(0,3) \oplus(1,1) \oplus(0,0) 。$$

## MATLAB代写

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