Suppose we have a time independent hamiltonian $$H = \hbar g (\sigma_x + \sigma_y + \sigma_z)$$

I know that the unitary operator is as follows:

$$U(t) = exp(-iHt/{\hbar})$$ Sinnce the pauli spin operators in H pairwise anticommute we know that

$$exp(-itg(\sigma_x + \sigma_y + \sigma_z)) \neq exp(-itg\sigma_x) exp(-itg\sigma_y) exp(-itg\sigma_z)$$

My question is how do we use this information to deduce the time-evolved state vector for the particle $|\psi(t)\rangle = U(t)|\psi(0)\rangle$.

Suppose we know that $|\psi(0)\rangle = |\uparrow\rangle$

We can state that $$|\psi(t)\rangle = exp(-itg(\sigma_x + \sigma_y + \sigma_z))|\uparrow\rangle$$

I also know that for an operator A and scalar $\alpha$ we have the identity $$exp(iA\alpha) = cos(\alpha ) + isin(\alpha)A$$

Generally for these sort of problems the operators commute so i would try to solve something along the lines of $$ |\psi(t)\rangle = exp(-itg\sigma_x) exp(-itg\sigma_y) exp(-itg\sigma_z) |\uparrow \rangle$$ and then invoke the operator A relation. However since the pauli operators pairwise anti commute I cannot do this and am unsure how to solve simplify such as, $$|\psi(t)\rangle = exp(-itg(\sigma_x + \sigma_y + \sigma_z))|\uparrow\rangle$$

and so any helo would be appreciated:

  • $\begingroup$ Did you try finding the eigenbasis of the Hamiltonian first? Once you know the eigenbasis of the Hamiltonian, you can express the propagator simply as $\sum_n e^{-iE_nt}\vert n\rangle\langle n\vert$ and the state at time $t$ will be simply $\sum_n e^{-iE_nt}\vert n\rangle\langle n\vert\psi(0)\rangle$. $\endgroup$
    – user87745
    Apr 12, 2021 at 21:48
  • $\begingroup$ I would also mention that the Hamiltonian being given as a summation of operators that don't commute isn't anything exotic. Almost all our examples in QM courses involve Hamiltonians of the type $\hat{p}^2 + V(\hat{x})$ and we know that $[\hat{p},\hat{x}]\neq 0$ ;-) $\endgroup$
    – user87745
    Apr 12, 2021 at 21:54
  • $\begingroup$ Thats very true, but normally I am used to working out stuff such as $H |\psi \rangle = E|\psi \rangle$ solving things that are of the form $exp(-iHt /\hbar)|\psi \rangle$ is a bit more difficult for me at the present moment. $\endgroup$
    – DJA
    Apr 12, 2021 at 21:59
  • 1
    $\begingroup$ Yes, but the point of my comment is that you only need to solve the eigenvalue problem for the Hamiltonian. The $H\vert\psi\rangle = E\vert\psi\rangle$ that you mentioned. Once that is done, it is trivial. You don't need to work with the exponentiated operators because it takes the form of exponentiated scalars in the eigenbasis of the Hamiltonian. $\endgroup$
    – user87745
    Apr 12, 2021 at 22:02
  • $\begingroup$ Still dont see how to solve the problem above though. Sorry that I cant quite see what to do! $\endgroup$
    – DJA
    Apr 12, 2021 at 22:27

1 Answer 1


There is a very useful identity for exponentials of Pauli matrices (see https://math.stackexchange.com/questions/3236998/exponential-of-pauli-matrices/3237834 for a proof):

$\begin{eqnarray} e^{i\theta \hat{\bf n}\cdot \sigma} = \cos \theta I + i(\hat{\bf n}\cdot \sigma) \sin \theta \end{eqnarray}$

For your Hamiltonian $\theta = -tg$ and $\hat{\bf n}\cdot \sigma = \sigma_x + \sigma_y + \sigma_z$, which gives:

$U(t) = \cos(tg)I - i\sin(tg)(\sigma_x + \sigma_y + \sigma_z)$

Note that $U(t)$ can also be conveniently represented as a $2x2$ matrix that acts on vectors representing the quantum state:

$U(t) = \begin{bmatrix}\cos(tg) - i\sin(tg)& (-1-i)\sin(tg) \\ (1-i)\sin(tg) & \cos(tg) + i\sin(tg)\end{bmatrix}$

So, with your example state,

$|\psi(t)\rangle = U(t)|\uparrow\rangle = (\cos(tg) - i\sin(tg))|\uparrow\rangle + (1-i)\sin(tg) |\downarrow\rangle$

You can alternatively follow the method from the above comments, where you take the matrix $H$, diagonalize this matrix to find the eigenvectors and eigenvalues, represent your initial state in this eigenbasis, and then time evolve the state using the eigenvalues. Both methods are equivalent, and the best method often depends on the application.

  • $\begingroup$ from the matrix U(t) how did you read off the final line? $\endgroup$
    – DJA
    Apr 13, 2021 at 2:54
  • $\begingroup$ First, in vector notation, $|\uparrow\rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $|\downarrow\rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$. This is true because of how we define $\sigma_z$ and that $\sigma_z|\uparrow\rangle = |\uparrow\rangle$ and $\sigma_z|\downarrow\rangle = - |\downarrow\rangle$. Now we can just directly multiply the state in vector notation by the matrix $U$, and finally rewrite in bra-ket notation. $\endgroup$ Apr 13, 2021 at 3:57
  • $\begingroup$ Sorry this was a while back but I also noticed that the new state is not normalised. Why is this and surely this is a problem? $\endgroup$
    – DJA
    Apr 30, 2021 at 9:53
  • 1
    $\begingroup$ Yeah, you are right, my apologies. There should be a factor of $1/\sqrt(3)$, which comes from the fact that $\hat{n}$ must be a unit vector. You should get a properly normalized state if you define $\hat{n}$ correctly with that extra factor. $\endgroup$ May 2, 2021 at 20:00

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