Given a fermionic Hamiltonian in a matrix form, we can write it as a sum over Kronecker products of Pauli matrices using the Hilbert-Schmidt inner product. However if the same Hamiltonian is given in a operator form, we can use the Jordan-Wigner transformation to write it as a sum over Kronecker products of Pauli operators.

How can one show that both the methods will give the same result, or if that is not the case then how does one show that the two results are related in some way?

Any material that discusses this is also appreciated.


1 Answer 1


It's a good question, but the answer is that the Hilbert-Schmidt inner product and the Jordan-Wigner transformation are not the same, even for the special case of fermionic Hamiltonians.

First let us consider spin-5/2 particles (they are still fermions). The fermionic Hamiltonian in this case will be $6^n \times 6^n$ matrix for $n$ particles, so for one spin-5/2 particle we have a 6x6 matrix which cannot be decomposed as a Kronecker product of 2x2 Pauli matrices.

Now let me also mention that the Jordan-Wigner transformation is not the only way to convert a fermionic Hamiltonian into into a "Paulinomial" (polynomial of Pauli matrices). There's also, for example, the Bravyi-Kitaev transformation which will give a completely different Paulinomial compared to the Jordan-Wigner transformation, even when starting from the same fermionic Hamiltonian in operator form. Yet another example separate from the Jordan-Wigner and Bravyi-Kitaev bases is the parity basis (see Eqs 20-21 here).

  • $\begingroup$ That does clear up a few things. What I am understanding from this is that if the fermionic Hamiltonian matrix is of the size $2^n \times 2^n$ then the Hilbert-Schmidt inner product is one of the possible methods to express the Hamiltonian as a Paulinomial. Is that the correct interpretation? $\endgroup$
    – e-eight
    Jul 27, 2021 at 1:06
  • 1
    $\begingroup$ Correct! See this. So if you want to express a matrix in terms of a sum of products of operators, the Hilbert-Schmidt inner product can help you with that. You can think of a matrix as a representation of an operator after choosing a basis (remember in linear algebra, linear transformations have a matrix representation, which can be obtained by applying the transformation to the basis vectors). The fermionic operators can be expressed in the Jordan-Wigner basis, Bravyi-Kitaev basis, parity basis, etc. $\endgroup$ Jul 27, 2021 at 3:22

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