# General Ehrenfest Theorem applied to N-qubit system operator

Please advise if the following short calculation of the derivative of the expectation value of an all spin Pauli-$$\hat{Y}$$ operator (acting on a $$N$$-qubit system) is consistent:

The general Ehrenfest theorem states $$\frac{d }{d t}\langle \hat{S}(t) \rangle = \bigg\langle \frac{\partial \hat{S}(t)}{\partial t} \bigg\rangle + \frac{1}{i \hbar}\langle [\hat{S}(t),\hat{H}]\rangle~~~~~~~~(*),$$ for some operator $$\hat{S}$$. Consider the expectation value of the all Pauli-spin $$Y$$ measurement operator $$\hat{S}= \underbrace{\hat{Y}\otimes \cdot \cdot \cdot \otimes \hat{Y}}_{N}$$ (in the Heisenberg picture) $$\big\langle\hat{S}(t)\big\rangle = \big\langle e^{i \frac{t}{2} \hat{X_{1}}}\hat{Y_{1}}e^{-i \frac{t}{2} \hat{X_{1}}}\otimes\cdot\cdot\cdot\otimes e^{i \frac{t}{2} \hat{X_{N}}}\hat{Y_{N}}e^{-i \frac{t}{2} \hat{X_{N}}} \big\rangle$$ acting on an $$N$$-qubit system, where the Hamiltonian is given by $$\hat{H} := \sum_{l=1}^{N}\hat{X}_{l}$$. Would I be correct in stating that this implies that the expectation value terms $$(*)$$ are given by \begin{align}\frac{\partial}{\partial t}[e^{i \frac{t}{2} \hat{X_{1}}}\hat{Y_{1}}e^{-i \frac{t}{2} \hat{X_{1}}}\otimes\cdot\cdot\cdot\otimes e^{i \frac{t}{2} \hat{X_{N}}}\hat{Y_{N}}e^{-i \frac{t}{2} \hat{X_{N}}}] = \frac{\partial}{\partial t}[e^{i t \hat{X_{1}}}\hat{Y}_{1}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y}_{N}]&\\=\underbrace{[i\hat{X}e^{i t \hat{X_{1}}}\hat{Y}_{1}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y}_{N}]+\cdot\cdot\cdot +[e^{i t \hat{X_{1}}}\hat{Y}_{1}\otimes\cdot\cdot\cdot\otimes i\hat{X}e^{i t \hat{X_{N}}}\hat{Y}_{N}] }_{N}\end{align}

using the anti-commutation of Pauli $$\hat{X}$$ and $$\hat{Y}$$ operators and then the product rule of differentiation of a tensor product. For the second term we get \begin{align}[\hat{S}(t), \hat{H}]= [e^{i t \hat{X_{1}}}\hat{Y_{1}}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y_{N}},\sum_{l}\hat{X}_{l}] = \underbrace{[e^{i t \hat{X_{1}}}\hat{Y}_{1}\hat{X_1}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y_{N}}]+\cdot\cdot\cdot +[e^{i t \hat{X_{1}}}\hat{Y_{1}}\otimes\cdot\cdot\cdot\otimes e^{i t \hat{X_{N}}}\hat{Y}_{N}\hat{X_N}] }_{N}. \end{align}

Thanks for any assistance.

The expression you have for $$\langle S(t)\rangle$$ on the top is already the time dependence - in this case there's no need to calculate anything beyond that. If, however, you want to see how the calculation turns out when doing it explicitly, it is:

First of all, because there are no interactions, it's much easier to look at a single qubit at a time and in the end take the tensor product. So let's do that: $$\frac{d}{dt}\langle Y\rangle = \bigg\langle \frac{\partial Y(t)}{\partial t} \bigg\rangle -i\langle [Y,H]\rangle$$ Now, $$Y$$ has no explicit time dependence, so its partial derivative w.r.t time is zero. Secondly, $$-i[Y(t),H]=-i[Y(t),X]=-2Z(t)$$ so (no need for the averaging anymore, we're basically working in the Heisenberg picture): $$\frac{dY(t)}{dt}=-2Z(t)$$ Similarly, $$\frac{dZ(t)}{dt} =2Y(t).$$ Since $$Y(0)=Y$$ and $$Z(0)=Z$$, the solution is: $$Y(t)=Y\cos 2t-Z\sin2 t$$ And the solution for the full many qubit system is the tensor product of this. From it, you just need to calculate the expectation value.

• Made a mistake by considering $\hat{S}$ as having time dependence. Commented Feb 14, 2022 at 16:35
• @JohnDoe yeah, up to the factor of 1/2 that I'm not sure why it's there. You can also derive $Y(t) = e^{iHt}Y e^{-iHt}$ by $t$ directly, observing the chain rule for derivatives and the commutation relations, and you'll that you get $-i[Y(t),H]$.