2

During my masters my prof. mentioned that the Youngs' tableau method can be used to calculate the number independent components of any arbitrary rank symmetric tensor in arbitrary dimensions. I wasn't much familiar with the concepts of Youngs tableau method back then. Now I am reading those methods and can't understand how they can be used to calculate the number of independent components of a symmetric tensor. Can anyone explain this concept or suggest a book or article to understand this in details?

Update after the comment: Suppose we have a completely symmetric rank 3 tensor in 4 dimensions, can you show how many independent components it would require. I know about the formulae to obtain this number using combinatorics. But I want to do it using Youngs’ tableau.

Qmechanic
  • 201,751
Sagar K. Biswal
  • 377
  • 3
  • 13
  • 1
    Related : Symmetry in terms of matrices. – Frobenius Dec 04 '23 at 12:56
  • 1
    Thank you for sharing. I read your answer but I am still struggling to understand it completely. I have updated my question, can you write a detailed answer please. – Sagar K. Biswal Dec 05 '23 at 20:05
  • My answer in above link doesn't answer your question using Youngs’ tableau. – Frobenius Dec 05 '23 at 20:10
  • From your answer I got the idea that, in Youngs tableau each box can be treated as an index of a tensor and the number of quantum states can be treated as the no. of values each index can take. Using this I got answer 20 for my above question. Is it correct? – Sagar K. Biswal Dec 05 '23 at 20:15
  • 1
    Exactly for the number 20 : The independent components of a completely symmetric rank $:p=3:$ tensor in $:d=4:$ dimensions is \begin{equation} N!\left(p,d\right)=\binom{p!+!d!-!1}{d!-!1} \tag{N-04}\label{N-04} \end{equation} that is \begin{equation} N!\left(3,4\right)=\binom{3!+!4!-!1}{4!-!1}=\binom{6}{3}=20 \tag{N-04'}\label{N-04'}
    \end{equation} We say that the tensor is of dimension $:N=20$.     – Frobenius Dec 05 '23 at 20:45

1 Answers1

3

The independent components of a tensor whose symmetry type is given by a given Young diagram each correspond to a semistandard tableau with that same shape. And the number of semistandard tableaux for a given Young diagram can be found via a hook length formula: $$ N = \prod_{(i,j) \in \lambda} \frac{ k - i + j}{h_\lambda(i,j)} $$ where (using the English conventions for Young diagrams)

  • $k$ is the range of entries allowed in the tableau (in your case, the dimension of the underlying vector space)
  • $i$ and $j$ are the index of the row and column, respectively (numbered starting from the top left, beginning with 1)
  • $h_\lambda(i,j)$ is the hook length associated with the cell $(i,j)$; it is found by counting all the cells that are directly below the cell $(i,j)$ and all the cells that are directly to the right of $(i,j)$, and adding 1 for the cell itself. In other words, if you draw a line that comes from below the diagram, hits $(i,j)$, and then "hooks" to the right and exits the diagram, $h_\lambda(i,j)$ is the number of cells you passed through.

So, for a general $k$ and a Young diagram consisting of three cells in a single row, this becomes $$ N = \underbrace{\left( \frac{k - 1 + 1}{3} \right)}_{(i,j) = (1,1)} \underbrace{\left( \frac{k - 1 + 2}{2} \right)}_{(i,j) = (1,2)} \underbrace{\left( \frac{k - 1 + 3}{1} \right)}_{(i,j) = (1,3)} = \frac{k(k+1)(k+2)}{6} $$ which gives $N = 20$ when $k = 4$, as you found.

As another example, for the mixed-symmetry Young diagram with three cells, we have $$ N = \underbrace{\left( \frac{k - 1 + 1}{3} \right)}_{(i,j) = (1,1)} \underbrace{\left( \frac{k - 1 + 2}{1} \right)}_{(i,j) = (1,2)} \underbrace{\left( \frac{k - 2 + 1}{1} \right)}_{(i,j) = (2,1)} = \frac{k(k+1)(k-1)}{3}. $$ (This also happens to yield $N = 20$ when $k = 4$, but this is just a coincidence; for general $k$, the dimensionality of this representation is different.)

Michael Seifert
  • 48,772