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I have found that (one?) definition of the trace norm is

$$\mid\mid A\mid\mid = \sqrt{A^*A} \tag{1}$$

but now I am reading this paper where (on page 4) it says

In particular, we will restrict the trace norm of the (driving) Hamiltonian $\| H_t \| < E_{\mathrm{max}}$.

Is the trace norm here defined as in Eq.(1)? I was looking around but couldn't find a definite answer.

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  • $\begingroup$ This doesn't sound very natural. Usually, one would bound the operator norm of a Hamiltonian. $\endgroup$ – Norbert Schuch Feb 3 at 15:29
  • $\begingroup$ @NorbertSchuch My understanding is that the trace norm is the sum of singular values while the operator norm is the maximum of singular values. Why exactly do we bound the Hamiltonian by the operator norm? I'm new to this field. $\endgroup$ – exp ikx Feb 3 at 17:05
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The trace norm $\|H\|$ of a matrix H is the sum of all the singular values of H. The singular values are the roots of the eigenvalues of $HH^\dagger$.

$\|H\| = Tr(\sqrt {HH^\dagger})$.

For a Hermitian matrix, like in our case Hamiltonian matrix, the singular values are the absolute value of eigenvalues. So the trace norm is just the sum of the absolute value of the eigenvalues of the Hamiltonian matrix at a particular time.

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