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Sunday, May 17, 2015

Exploring Quantum Physics - Week 7 Question 7

Question:

Compute the root mean square of electron velocity in ground state quantized motion under the Earth's constant magnetic field.

Solution:

The problem seems daunting at first. But experience tell me once again this is most likely just number substituting exercise, so let's try expanding $ \hat{a}^\dagger \hat{a} $ and see what is going on there.

$ \begin{eqnarray*} & & \hat{a}^\dagger \hat{a} \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1} - i\hat{v_2})(\hat{v_1} + i\hat{v_2}) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i[\hat{v_1}, \hat{v_2}]) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 + i(i\frac{\hbar\omega}{m})) \\ &=& \frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) \\ \end{eqnarray*} $

So we can expand $ H_{2D} $ as follow:

$ \begin{eqnarray*} & & H_{2D} \\ &=& \hbar\omega(\hat{a}^\dagger \hat{a} + \frac{1}{2}) \\ &=& \hbar\omega(\frac{m}{2\hbar\omega}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{1}{2}) \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2 - \frac{\hbar\omega}{m}) + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) - \frac{\hbar\omega}{2} + \frac{\hbar\omega}{2} \\ &=& \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) \\ \end{eqnarray*} $

Now we will equate the ground state energies

$ \begin{eqnarray*} \epsilon_0 | \psi_0 \rangle &=& H_{2D} | \psi_0 \rangle \\ \langle \psi_0 | \epsilon_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 \langle \psi_0 | \psi_0 \rangle &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \epsilon_0 &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | H_{2D} | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \langle \psi_0 | \frac{m}{2}(\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{2} &=& \frac{m}{2}\langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \frac{h\omega}{m} &=& \langle \psi_0 | (\hat{v_1}^2 + \hat{v_2}^2) | \psi_0 \rangle \\ \end{eqnarray*} $

Last but not least, the actual value root mean square value is 32.15.

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