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As is well known, two examples of basis invariant functions are the trace and determinant functions. These functions can therefore be thought of as a property of a linear operator, and not just a property of a matrix.

For a (linear) operator $\hat{A}$, in terms of a bracket the trace is defined as: $$\operatorname{tr}(\hat{A}) = \sum_i \langle u_i|\hat{A}|u_i \rangle,$$ where $\{|u_i \rangle\}$ is any orthonormal basis set.

If the operator $\hat{A}$ is represented by some square matrix $A$, then finding $\det(A)$, the determinant of the matrix is not difficult.

My question is, can the determinant of an operator be expressed in terms of kets, bras, or brackets in some way as is done with the trace?

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    $\begingroup$ $\det{A}=\sum_{\sigma\in \text{Sym}}\text{Sign}(\sigma)\Pi_i \langle u_i|A|u_{\sigma(i)}\rangle$ $\endgroup$
    – Rescy_
    Commented Dec 18, 2022 at 11:24
  • $\begingroup$ @Rescy - I'm guessing $\sigma$ here is the permutation symbol. Would you mind, as an answer, explicitly working through as an example the $2 \times 2$ case? Here I assume $\sigma = (12) \in S_2$, but I am not sure how one goes about determining $\operatorname{sgn}(\sigma)$. $\endgroup$
    – omegadot
    Commented Dec 18, 2022 at 12:14
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    $\begingroup$ Usually this is taught in first year linear algebra in the UK, so if you search Part IA Cambridge/Oxford Linear algebra notes you should be able to find some discussions on the Internet. Just to give a concise answer: the signature of a permutation is determined by whether it is even or odd number of products or transpositions when you write a permutation as a product of transpositions. It can be proved that although you may write the same permutation in different ways as products of transpositions, they are all even/odd and this unique property can be assigned a number $\pm 1$. $\endgroup$
    – Rescy_
    Commented Dec 18, 2022 at 14:14
  • $\begingroup$ In your example, $\text{Sign}((1~2))=-1$ because the number of transposition is 1 which is odd. $\endgroup$
    – Rescy_
    Commented Dec 18, 2022 at 14:21
  • $\begingroup$ In your 2×2 example (only), you just have det A =((trA)² - tr(A²) )/2 . $\endgroup$ Commented Dec 18, 2022 at 17:31

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See determinant of an $n\times n$ matrix and convert to bra/ket language.

$$ \text{det}\left(\hat{A}\right) = \sum_{i_1, \ldots, i_n}\epsilon_{i_1,\ldots,i_n}\langle u_1|\hat{A}|u_{i_1}\rangle \ldots \langle u_{n}|\hat{A}|u_{i_n}\rangle = \sum_{\sigma \in S_{n}}\left(\text{sgn}(\sigma) \langle u_1|\hat{A}|u_{i_1}\rangle \ldots \langle u_{n}|\hat{A}|u_{i_n}\rangle \right) $$

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The determinant $\det(A)$ of an $n\times n$ matrix can be expressed in terms of $\mathrm{tr}(A^k)$, where $k=1,\dots,N$ -- see e.g. Wikipedia for a closed form. This immediately gives an expression of the type you want, using that $\mathrm{tr}(A^k) = \sum \langle u_i\vert A^k \vert u_i\rangle$.

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My question is, can the determinant of an operator be expressed in terms of kets, bras, or brackets in some way as is done with the trace?

Given an operator A, you have already learned that: $$\operatorname{tr}(\hat{A}) = \sum_i \langle u_i|\hat{A}|u_i \rangle,$$

Now let some other operator $\hat B$ be given by: $$ \hat B = e^{\hat A} $$

The determinant of $B$ is: $$ \operatorname{det}(B) = e^{\operatorname{tr}(A)} = e^{\sum_i \langle u_i|\hat{A}|u_i \rangle} = \Pi_i e^{\langle u_i|\hat{A}|u_i \rangle} $$

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  • $\begingroup$ This does not explain how, given $B$, you can compute its determinant. $\endgroup$ Commented Dec 18, 2022 at 18:18
  • $\begingroup$ Yes, it does. Given B, take $A = \log(B)$. This is based on the identify $\log \det (M) = Tr \log (M)$. See: math.stackexchange.com/questions/1487773/… $\endgroup$
    – hft
    Commented Dec 18, 2022 at 19:35
  • $\begingroup$ Taking the log involves diagonalizing the matrix. Once you have the eigenvalues, you can compute the determinant directly. $\endgroup$ Commented Dec 18, 2022 at 20:15
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    $\begingroup$ No, the last expression here does not require diagonalizing the matrix! $\endgroup$ Commented Dec 18, 2022 at 20:36
  • $\begingroup$ @CosmasZachos I'm not sure which expression you are referring to -- my comment was about finding an A such that $e^A=B$ (which is what the answer suggests to do), and don't see an expression for that. Can you be more specific? $\endgroup$ Commented Jan 11, 2023 at 12:06

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