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In the book Quantum Information, Computation, and Communication by Jaksch and Jones, the authors claim (on page 117) that in first-order perturbation theory, we can simply neglect the off-diagonal components of the perturbation (in the basis of the unperturbed Hamiltonian) to get an approximate Hamiltonian.

The approximate Hamiltonian will have the correct first-order energy eigenvalues, but incorrect eigenstates. This can have an impact on experiments in that there will be a mixing of states to first order in the perturbation, so how is it allowed?


To be specific, the authors consider a Hamiltonian for two interacting nuclear spins $$H=\frac{1}{2}\omega_1\sigma_{1z}+\frac{1}{2}\omega_2\sigma_{2z}+\frac{1}{4}\omega_{J_{12}}\boldsymbol{\sigma}_1\cdot\boldsymbol{\sigma}_2$$ where the last term is considered as a perturbation. The approximation is then made that the off-diagonal terms of the Heisenberg-like perturbation can be neglected such that it becomes Ising-like, giving $$H\approx\tilde{H}=\frac{1}{2}\omega_1\sigma_{1z}+\frac{1}{2}\omega_2\sigma_{2z}+\frac{1}{4}\omega_{J_{12}}\sigma_{1z}\sigma_{2z}.$$ It is said for example here that this corresponds to first-order perturbation theory.

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The second Hamiltonian you wrote is already diagonal, $$ \tilde H= \mathrm{diag} \Bigl ( ( \omega_1 +\omega_2)/2 + \omega/4, ~ ( \omega_1 -\omega_2)/2 - \omega/4,\\ -( \omega_1 -\omega_2)/2 - \omega/4, ~ -( \omega_1 +\omega_2)/2 + \omega/4 \Bigr ), $$ just like the unperturbed one, and the perturbation, so the eigenstates are the four diagonal basis vectors.

The discarded piece of the original H is just $\tfrac{1}{2}\omega \sigma_x$ mixing the 2nd and 3rd unperturbed states, off diagonal, so it does not affect the first order energies. The eigenstate first order shifts only affect the second order energy shifts.

  • To first order, the two hamiltonians, $H$ and $\tilde H$ have the same eigenvalues, $UHU^\dagger= \tilde H$, so they only differ by a change of basis, an unphysical entity, which cannot be probed by any and all experiments.

Your paradigm is flawed. The situation you are dealing with here, where the discarded terms live in the 2nd & 3rd eigenvector subspace are, with the obvious correspondences, $\epsilon= \omega/2(\omega_1-\omega_2)$, $$ H=\begin{pmatrix} 1-\epsilon &2\epsilon\\ 2\epsilon& -1-\epsilon \end{pmatrix} ~~~\leadsto ~~~\tilde H= \begin{pmatrix} 1-\epsilon & 0\\ 0& -1-\epsilon \end{pmatrix}. $$ To lowest order in $\epsilon$, they have the same eigenvalues, $1-\epsilon$ and $-1-\epsilon$, but slightly different eigenvectors, $(1, \epsilon)^T$ and $(-\epsilon,1)^T$ for H, and $(1,0)^T$ and $(0,1)^T$ for $\tilde H$.

To this order in $\epsilon$, then, the two systems are equivalent, since, given the eigenvalues, no "preferred" basis is defined: you cannot describe to a Martian which is which; and any evolution, experiment, etc..., cannot tell you you are working with H or $\tilde H$. The (infinitesimal) basis rotation U connecting the two hamiltonians/bases is unphysical, to this order. Diagonalizing H yields $\tilde H$, $$ \begin{pmatrix}1&\epsilon\\-\epsilon & 1 \end{pmatrix} H \begin{pmatrix}1& -\epsilon\\\epsilon & 1 \end{pmatrix} = \tilde H. $$

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  • $\begingroup$ I understand that the Hamiltonians have the same eigenvalues, but they correspond to different eigenstates! Why do we only care that the eigenvalues are the same? $\endgroup$ Mar 5, 2023 at 23:57
  • $\begingroup$ @Ghorbalchov you’re confusing things. To first order, it is the diagonal part of the perturbation that remains, in addition to an unperturbed Hamiltonian already diagonal. Surely then the original eigenstates are also eigenstates of new but still diagonal Hamiltian $H_0$+ “diagonal part of the perturbation”. $\endgroup$ Mar 6, 2023 at 0:09
  • $\begingroup$ The first corrections to the eigenstates appear when you compute the second order corrections to the energy. $\endgroup$ Mar 6, 2023 at 0:14
  • $\begingroup$ In an experiment, if there is degeneracy, the eigenstate cannot be uniquely characterized by the (common) eigenvalue. So, to observe its "movement", you need something else to break the degeneracy. Evolution will not change it. $\endgroup$ Mar 6, 2023 at 0:16
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    $\begingroup$ Flawed paradigm . You, in fact, should have $k\propto \epsilon$ and work strictly to first order in $\epsilon$... $\endgroup$ Mar 6, 2023 at 1:34
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Unfortunately I think Cosmas has not understood what I am asking. Of course, his mathematics is correct, being ultimately just linear algebra, but I do not agree with the physical interpretation of it.

The eigenbasis of the unperturbed Hamiltonian is determined by the fixed external apparatus, in this case a magnetic field. Similarly, we will be able to apply alternating magnetic fields perpendicular to this, and readout using a fixed detection coil, etc. This will decide our 'preferred' basis, and we will ultimately have to perform all experiments in this basis.

In order to diagonalize the perturbed Hamiltonian, we have to make a change of basis, and this essentially tilts the quantization axis relative to the external apparatus. Cosmas says the basis change is unphysical and cannot be probed by experiments, but I think he is imagining a hypothetical situation in which the apparatus gets 'tilted' accordingly, which would not be done in practice.

In this regard, I think what the authors write is a bit misleading. It should be $UHU^\dagger \approx \tilde H$ (where $\approx$ means up to first order, as Cosmas has shown) whereas they write $H\approx\tilde{H}$, implying the same basis should be used for both Hamiltonians. Alternatively, they should write something like $$H\approx\frac{1}{2}\omega_1\sigma_{1z}'+\frac{1}{2}\omega_2\sigma_{2z}'+\frac{1}{4}\omega_{J_{12}}\sigma_{1z}'\sigma_{2z}'$$ where the prime indicates the transformed basis.

If, for example, a perpendicular magnetic field was applied to enact some pulse, it would be fixed by the external apparatus, and hence be proportional to $\sigma_x$, not to $\sigma_x'$. The effect of pulses not being applied in the primed basis is that there will be some mixing of states to first order in the perturbation, and the result will deviate from the ideal pulse sequence, in which the pulses are also carried out in the primed basis. When we ultimately perform readout, the results will differ from the ideal case in that there will be a small proportion of the wrong result mixed in.

The point I think the authors are trying to make is that this can be neglected if the mixing is small, that is if $$\left|\frac{\omega_{J_{12}}}{\omega_1-\omega_2}\right|\ll 1.$$ We then carry out our quantum computing as normal, assuming that the tilt between the primed and unprimed basis is negligible, with the understanding that our final results will be accurate only up to some correction, first order in the perturbation.

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