Electron spin in magnetic field

You are trying to get the probability of finding the electron with spin down at time t, but your calculation is giving you the probability of finding the electron with spin up at time t. To get the correct probability, you need to take the inner product of the state vector at time t with the spin down state vector, which would give you a probability of 0.5.
  • #1
fyzikapan
12
0

Homework Statement


An electron is an eigenstate of sz at time t = 0 (spin up). It is in a magnetic field [tex]\vec B = (B \sin \theta, 0,B\cos\theta)[/tex]. Find the probability of finding the electron with spin down at time t.


Homework Equations


[tex]
U(t) = \exp \left( -i \mathcal{H} t/\hbar\right)\\\\

P(t) = \left| \left \langle \downarrow | \chi(t) \right \rangle \right|
[/tex]


The Attempt at a Solution


[tex]
\begin{align*}
U(t) &= \exp \left( -i \mathcal{H} t/\hbar\right)\\
&= \exp \left( -i \frac{et}{mc\hbar} \vec S \cdot \vec B\right)\\
&= \exp \left( -i \frac{\omega_0 t}{2} \left(\sigma_x \sin \theta + \sigma_z \cos \theta\right)\right)\\
&= \exp \left(-i \frac{\omega_0 t}{2} \left[ \begin{array}{cc}\cos \theta & \sin \theta \\
\sin \theta & \cos \theta \end{array}\right]\right)\\
&= \left( \begin{array}{cc}
\exp (- i \frac{\omega_0 t}{2} \cos \theta) & \exp (- i \frac{\omega_0 t}{2} \sin \theta)\\
\exp (- i \frac{\omega_0 t}{2} \sin \theta) & \exp ( i \frac{\omega_0 t}{2} \cos \theta)
\end{array}\right)
\end{align*}[/tex]

Then at time t,
[tex]
\begin{align*}
\chi(t) &= U(t) \left| \chi(0) \right \rangle\\
&= \left( \begin{array}{cc}
\exp (- i \frac{\omega_0 t}{2} \cos \theta) & \exp (- i \frac{\omega_0 t}{2} \sin \theta)\\
\exp (- i \frac{\omega_0 t}{2} \sin \theta) & \exp ( i \frac{\omega_0 t}{2} \cos \theta)
\end{array}\right)\left( \begin{array}{c}1\\0\end{array}\right)\\
&= \left( \begin{array}{cc}
\exp (- i \frac{\omega_0 t}{2} \cos \theta)\\
\exp (- i \frac{\omega_0 t}{2} \sin \theta)\end{array}\right)
\end{align*}
[/tex]

But this gives [tex]P(t) = \left| (\begin{array}{cc} 0 & 1 \end{array})\left( \begin{array}{cc}
\exp (- i \frac{\omega_0 t}{2} \cos \theta)\\
\exp (- i \frac{\omega_0 t}{2} \sin \theta)\end{array}\right)\right|^2 = 1[/tex], which is obviously wrong.

I've looked over my math and don't see any obvious mistakes, so I'm not sure what I'm doing wrong.



The Attempt at a Solution

 
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  • #2
First, you have used for sigma_z the identity matrix. It should be 1 and -1 on the diagonal.
Also, you seemed to assume that the exponential of a matrix is the matrix of the exponentials of the elements of the initial matrix. This is not true in general (works only for a diagonal matrix)
 

1. What is electron spin in a magnetic field?

Electron spin refers to the intrinsic angular momentum of an electron. When placed in a magnetic field, the electron will experience a torque that causes it to align its spin with the magnetic field.

2. How does electron spin affect the behavior of atoms in a magnetic field?

The orientation of electron spin in a magnetic field can affect the energy levels of atoms, leading to changes in their electronic and magnetic properties. This phenomenon is known as the Zeeman effect.

3. What is the role of electron spin in MRI technology?

In MRI (magnetic resonance imaging), the magnetic field causes the protons in the body's tissues to align their spin. When exposed to radio waves, these protons emit a signal that can be used to create detailed images of the body's internal structures.

4. Can electron spin be manipulated and controlled?

Yes, electron spin can be manipulated and controlled through the use of external magnetic fields. This is the basis for technologies such as magnetic storage devices and spintronics, which utilize the spin of electrons to store and transmit information.

5. How is electron spin related to quantum mechanics?

The concept of electron spin was first introduced in quantum mechanics to explain certain phenomena, such as the splitting of spectral lines in the presence of a magnetic field. Electron spin is a crucial aspect of quantum mechanics and is essential for understanding the behavior of atoms and particles on a microscopic level.

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