# Heisenberg equation of motion — why is $\vec{\sigma}_H=\vec{\sigma}$?

Trying to obtaining the Heisenberg EOM ( "for $$\vec{\sigma}$$" ) for the following Hamiltonian

$$H = - \mu \vec{B}\cdot\vec{\sigma}$$ where the magnetic field $$\vec{B}$$ is generic for now, $$\vec{B}=\{B_x,B_y,B_z\}$$, and $$\vec{\sigma}$$ is the Pauli vector (of Pauli matrices).

$$\frac{d}{dt} A(t)_H = \frac{i}{\hbar} [H,A(t)_H] + (\frac{\partial A_H}{\partial t})_H$$

The (well my professor's, and so I would like to mimic her style as she's brilliant) solutions to the problem we got after trying this for HW last week makes a step that I am confused about:

Why is the operator $$\vec{\sigma}$$ not transformed to the Heisenberg picture? Ought we to first send

$$\vec{\sigma} \rightarrow \vec{\sigma}_H=e^{\frac{it}{\hbar}H}\vec{\sigma}e^{\frac{-it}{\hbar}H}$$

My only thought is that my professor already did so! See the following work from a similar problem I found on the web. The Hamiltonian in this work's case is more specific,

$$H=-B \sigma_x$$

In this case, using Mathematica enables us to show that for the case of $$\sigma_x$$,

$$(\sigma_x)_H = e^{\frac{it}{\hbar}H} \sigma_x e^{\frac{-it}{\hbar}H} = \sigma_x$$

The following code verifies it:

MatrixForm[Simplify[MatrixExp[i t/h PauliMatrix[1]].PauliMatrix[1].MatrixExp[-i t/h PauliMatrix[1] ]]]


Now... if I define my generic Hamiltonian:

where I use the Pauli vector now for the more generic Hamiltonian instead of one of the matrices like the example above... the same formula for $$(\sigma_x)_H$$ is .... disturbing.

Can someone point out my mistake (computational or conceptual) and point me to how I can show $$\vec{\sigma}_H=\vec{\sigma}$$

Edit: [Incorrect!]

I was thinking about the following but it is still a complete (or/nor coded correctly) idea

Edit based on @ZeroTheHero's Baker–Campbell–Hausdorff formula insight

using this form of the BCH formula (eq 3 here)

$$e^A B e^{-A} = B + [A,B] + 1/2[A,[A,B]] + ...$$

shouldn't I be able to say

$$\begin{eqnarray*} (\vec{\sigma})_H &=& e^{\frac{it}{\hbar}H} \vec{\sigma} e^{\frac{-it}{\hbar}H} \\ &=& \vec{\sigma} + [H,\vec{\sigma}] + ... \\ &=& \vec{\sigma} -\mu [\vec{B} \cdot \vec{\sigma},\vec{\sigma}] ... \\ &=& \vec{\sigma}- \mu \vec{B} [\vec{\sigma},\vec{\sigma}] + ... \\ &=& \vec{\sigma}+ 0 \\ &=& \vec{\sigma} \end{eqnarray*}$$

This seems correct to me, but if so, the result $$[H,\vec{\sigma}]=0$$ reduced the Heisenberg EOM to

$$\frac{d}{dt} \vec{\sigma} = 0$$

which I know to be false... :/

My thoughts at the moment is that my $$[H,\vec{\sigma}]$$ is wrong based on my professor's calculation above where she does NOT factor out the constant $$\vec{B}$$. I was motivated to factor out based on this Stack answer.

• Please don’t post scans of math. – G. Smith Sep 11 '20 at 22:24
• – Cosmas Zachos Sep 12 '20 at 11:30

Clearly $$\exp(\hat A)\hat A\exp(-\hat A)=\hat A$$ by Baker-Campbell-Hausdorff. Here, set $$\hat A=\hat H=-it\mu\vec B\cdot \vec \sigma/\hbar$$.
Note that \begin{align} \exp(-i t\mu \vec B\cdot \vec \sigma/\hbar)\ne \exp(-it B_x\sigma_x)\exp(-it B_y\sigma_y)\exp(-it B_z\sigma_z) \end{align} or any other kind of factorization suggested by your snippet of Mma code since in general $$\exp(A+B)\ne \exp(A)\exp(B)$$. The correct implementation would be to compute $$-it\mu\vec B\cdot\vec \sigma/\hbar$$ as a single matrix and then exponential this single matrix.
• @LopeyTall added just a bit to make it more clear. Seems like your error is that you are writing the exponential as a product of three parts, one for each components, but $\exp(A+B)\ne \exp(A)\exp(B)$ unless $A,B$ commute which is NOT the case for Pauli’s. – ZeroTheHero Sep 12 '20 at 16:37
• Hmm, I am a bit confused by the new material in your answer. I understand your comment, and I've altered the way I'm approaching the problem based on your CBH suggestion, but I'm not sure I ought to be showing $\hat{H}_H=\hat{H}$ (please correct me if I'm wrong!) I was using your CBH suggestion to try to show the following: $(\vec{\sigma})_H = e^{\frac{it}{\hbar}H} \vec{\sigma} e^{\frac{-it}{\hbar}H} = \vec{\sigma}$ then by the form of CBH appearing here with $e^A O e^-{A} = O + [A,O] + ...$ And so I was focusing on $[H,\vec{\sigma}]$ which I show is zero.... but then the EOM = triv, no? – Lopey Tall Sep 12 '20 at 18:58