Physics Asked by T. Arthur on May 8, 2021

I’ve once read somewhere that the following two Hamiltonians each describing a two-level system (TLS) are equivalent to each other and can thus be used in describing the same TLS.

$$mathcal{H}_1=Delta sigma_x + epsilon mathop{I} quad;quad mathcal{H}_2=Delta sigma_x + epsilon sigma_z.$$

$sigma_x$ and $ sigma_z$ are Pauli spin matrices and $mathop{I} $ is the 2×2 identity matrix.

Personnaly, I think that they describe two completely different TLS because $sigma_z neq mathop{I}$, such as the eigenvalues of the two Hamiltonians.

Please can you clarify the situation to me ? i.e. Am I right or wrong with more possible explanations ?

[**Revised and corrected**]

Hamiltonians $mathcal H_2$ and $mathcal H_1$ are equivalent provided there exists a unitary transformation $U$ taking $mathcal H_1$ into $mathcal H_2$, that is, $$ {mathcal H}_2 = U{mathcal H}_1 U^dagger $$ As pointed out in a comment, this means that the two Hamiltonians must share the same spectrum, although they will generally have different eigenvectors. Whenever this is the case, anything calculated using ${mathcal H}_1$, also holds for ${mathcal H}_2$, up to a unitary transformation of eigenvectors.

Technically, the identity of spectra provides one quick check that the two hamiltonians are equivalent, which in turn implies the existence of the unitary $U$. However, the spectra are understood to be identical up to arbitrary energy offsets, which do not contribute to the dynamics, therefore the relevant factor is the eigenvalue gap. That is, two hamiltonians are exactly equivalent if and only they correspond to identical energy gaps.

In the case of $mathcal H_1$, the term $epsilon I$ is just an energy offset, meaning that we are looking in fact at the equivalence of ${tilde{mathcal H}}_1 = Deltasigma_x$ and ${mathcal H}_2 = Deltasigma_x + epsilonsigma_z$. But the gap of ${tilde{mathcal H}}_1$ is simply $2Delta$, while that of $mathcal H_2$ is $2sqrt{epsilon^2 + Delta^2} neq 2Delta$. So formally, $mathcal H_2$ is only equivalent to $mathcal H_1$ to first order in $epsilon$, when $2sqrt{epsilon^2 + Delta^2} approx 2Delta + {mathcal O}(epsilon^2)$. In this case the unitary $U$ can be found easily if we rearrange ${mathcal H}_2$ as
$$
{mathcal H}_2 = Deltasigma_x + epsilonsigma_z = sqrt{epsilon^2 + Delta^2}left[ frac{Delta}{sqrt{epsilon^2 + Delta^2}}sigma_x + frac{epsilon}{sqrt{epsilon^2 + Delta^2}}sigma_zright] approx Delta (sigma_xcostheta + sigma_zsintheta )
$$
where $costheta = frac{Delta}{sqrt{epsilon^2 + Delta^2}}$, $sintheta = frac{epsilon}{sqrt{epsilon^2 + Delta^2}}$. In other words, $mathcal H_2$ lies along direction $(costheta, 0, sintheta)$ in the $x,z$-plane of the Bloch sphere. Then the transformation from $tilde{mathcal H}_1$ or $sigma_x$ to $mathcal H_2$ or $(sigma_xcostheta + sigma_zsintheta )$ is just a rotation around the $y$-axis, generated by
$$
U = e^{-ifrac{theta}{2} sigma_y} = I cosfrac{theta}{2} - isigma_ysinfrac{theta}{2}
$$

since indeed, we have
$$
Usigma_x U^dagger = left[I cosfrac{theta}{2} - isigma_ysinfrac{theta}{2}right] sigma_x left[I cosfrac{theta}{2} + isigma_ysinfrac{theta}{2}right] = sigma_xcostheta + sigma_zsintheta
$$
$$
;
$$
Alternatively, even when $2sqrt{epsilon^2 + Delta^2} neq 2Delta + {mathcal O}(epsilon^2)$, one may use the reverse $y$-rotation to bring $mathcal H_2$ to the simpler form
$$
tilde{mathcal H}_2 = sqrt{epsilon^2 + Delta^2} sigma_x
$$
The difference between $mathcal H_1$ and $mathcal H_2$ is then reduced to different time scales of the dynamics.
$$
;
$$
**Note on relation to the** *rotating frame approximation*:

Hamiltonian $mathcal H_2$ is often seen in the context of textbook examples of the "*rotating frame approximation*", where it arises as a result of a transformation to a frame rotating about the $z$-axis, not about the $y$-axis as here. See sec.5.4 of these lecture notes, for instance. The reason is that the starting point of the transformation is not Hamiltonian $mathcal H_1$, but a more complicated (and time-dependent) form, with an unperturbed term in $sigma_z$ and perturbation terms in $sigma_x$ and $sigma_y$. In the rotating frame the unperturbed $sigma_z$ part remains the same and the perturbation reduces to a $sigma_x$ term after discarding rapidly oscillating contributions. The equivalence questioned in the OP entails an additional transformation.

Answered by udrv on May 8, 2021

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