# Spin of a particle in uniform magnetic field along arbitrary direction

Suppose initially, at time $t=0$, a spin 1/2 particle is in the spin up state along some arbitrary direction in the XY plane, and given by $a(\hat{x} + \hat{y})$.

Also, at time $t=0$, a uniform magnetic field is applied in some arbitrary direction in the XZ plane, and given by $\vec{B} = b(\hat{x}+\hat{z})$.

How would I compute the probability of the particle being in the spin down state along the z-direction after a time $t=T$?

• Your specifications for the magnetic field are inconsistent. – Emilio Pisanty Mar 19 '18 at 2:02
• My apologies. I updated the magnetic field equation. I should also add that $a$ and $b$ are constants. – MomoTheSir Mar 19 '18 at 2:59

I think you just have to solve the spin-evolution part of the corresponding Pauli-Schroedinger equation. It goes like this: The behavior of a half spin particle is encoded by a 2D complex state vector function $$|\, \psi(t) \,\rangle =\left(\begin{matrix} \psi_1(t)\\ \psi_2(t)\end{matrix}\right)$$ that depends on time $t$. Let $$\vec{\sigma} = \sigma_x \, \hat{x} + \sigma_x \, \hat{y} + \sigma_x \, \hat{z}$$ be the vector whose coefficients are the three Pauli matrices $$\sigma_x = \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right), \,\, \sigma_y = \left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix}\right), \,\, \sigma_z = \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right)$$ If the homogeneous constant magnetic field is $$\vec{B} = b \, \hat{x} + b \, \hat{z}$$ then one has to form the dot-product-like operator $$\vec{B} \circ \vec{\sigma} = b \, \sigma_x + 0 \, \sigma_y + b \, \sigma_z = b \, \sigma_x + b \, \sigma_z = b \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) + b \left(\begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix}\right) = \left(\begin{matrix} b & b \\ b & -b \end{matrix}\right)$$ Then, the Pauli-Schroedinger equation is $$i \hbar\, \frac{\partial}{\partial t} \, | \, \psi \, \rangle = -\, \frac{q\,\hbar}{2m} \, \big( \vec{B}\circ\vec{\sigma}\big) \,| \, \psi \, \rangle$$ $$\frac{\partial}{\partial t} \, | \, \psi \, \rangle = \frac{i\, q}{2m} \, \big( \vec{B}\circ\vec{\sigma}\big) \,| \, \psi \, \rangle$$ which written ''explicitly" is $$i \hbar \, \frac{\partial}{\partial t} \left(\begin{matrix} \psi_1\\ \psi_2\end{matrix}\right) = -\, \frac{q\,\hbar}{2m} \,\left(\begin{matrix} b & b \\ b & -b \end{matrix}\right) \left(\begin{matrix} \psi_1\\ \psi_2\end{matrix}\right)$$ $$\frac{\partial}{\partial t} \left(\begin{matrix} \psi_1\\ \psi_2\end{matrix}\right) = \, \frac{i\, q}{2m} \,\left(\begin{matrix} b & b \\ b & -b \end{matrix}\right) \left(\begin{matrix} \psi_1\\ \psi_2\end{matrix}\right)$$ When you solve this linear system of differential equations you obtain the evolution operator $$U(t) = \exp\left(i\,\frac{q\, t}{2m} \, \big( \vec{B}\circ\vec{\sigma}\big)\right)$$ The initial state, the one that is spin up along $a(\hat{x}+\hat{y})$, can be found for example as the eigenvector: $$a\big(\sigma_x+\sigma_y\big)\,|\, \psi_0 \,\rangle = |\, \psi_0 \,\rangle$$ Since $$a\big(\sigma_x+\sigma_y\big) = a \left(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix}\right) + a \left(\begin{matrix} 0 & -i \\ i & 0 \end{matrix}\right) = a \left(\begin{matrix} 0 & 1-i \\ 1+i & 0 \end{matrix}\right)$$ the latter eigenproblem is $$a \left(\begin{matrix} 0 & 1-i \\ 1+i & 0 \end{matrix}\right) \left(\begin{matrix} \psi_{01}\\ \psi_{02}\end{matrix}\right)= \left(\begin{matrix} \psi_{01}\\ \psi_{02}\end{matrix}\right)$$ choosing a normalized eigenvector $$\langle \, \psi_0 \, | \, \psi_0 \, \rangle = |\psi_{01}|^2 + |\psi_{02}|^2 = 1$$ The evolution of this state with time is given by $$|\, \psi(t) \, \rangle = U(t) \, |\, \psi_0 \, \rangle = \exp\left(i\,\frac{q\, t}{2m} \, \big( \vec{B}\circ\vec{\sigma}\big)\right) \, |\, \psi_0 \, \rangle$$ To find the probability of the particle to be in spin down state along the $z$-direction you need the down eigenstate which is $$|\, \text{z-down} \, \rangle = \left(\begin{matrix} 0\\ 1\end{matrix}\right)$$ and the quantum phase $$\langle \, \text{z-down} \,|\, U(T)\, |\, \psi_0\, \rangle = \langle \, \text{z-down} \,|\, \exp\left(i\,\frac{q\, T}{2m} \, \big( \vec{B}\circ\vec{\sigma}\big)\right) \, |\, \psi_0\, \rangle$$ Finally the probabilyt you seek is $$P(T) = \big|\, \langle \, \text{z-down} \,|\, U(T)\, |\, \psi_0\, \rangle \, \big|^2$$