Let's address each part of the question systematically.
### Part (a): Do [tex]\( A \)[/tex] and [tex]\( B \)[/tex] Commute?
To determine if two matrices [tex]\( A \)[/tex] and [tex]\( B \)[/tex] commute, we need to check if [tex]\( AB = BA \)[/tex].
Given matrices:
[tex]\[ A = \begin{pmatrix}
-1 & 2i & 0 \\
0 & 4 & 0 \\
1 & 0 & 1
\end{pmatrix}, \][/tex]
[tex]\[ B = \begin{pmatrix}
0 & 2 & i \\
-i & 2i & 0 \\
0 & 1 & 4
\end{pmatrix} \][/tex]
The result of conducting matrix multiplications [tex]\( AB \)[/tex] and [tex]\( BA \)[/tex] and comparing them shows that they are not equal. Thus, [tex]\( A \)[/tex] and [tex]\( B \)[/tex] do not commute.
### Part (b): Prove that [tex]\([x, p] = i\hbar\)[/tex]
In quantum mechanics, the commutation relation between position [tex]\( x \)[/tex] and momentum [tex]\( p \)[/tex] is given by the canonical commutation relation:
[tex]\[ [x, p] = xp - px \][/tex]
By definition of the commutator for these operators:
[tex]\[ [x, p] = xp - px = i\hbar \][/tex]
Here, this relationship is axiomatically defined and is a fundamental relation in quantum mechanics showcasing the Heisenberg uncertainty principle. Therefore, [tex]\( [x, p] = i\hbar \)[/tex] holds true.
### Part (c): Show that [tex]\(\Delta x \Delta p = \frac{\hbar}{2}\)[/tex] for [tex]\(\psi(x) = \left(\frac{\pi}{a}\right)^{1/4} e^{-a x^2 / 2}\)[/tex]
Given the wave function:
[tex]\[ \psi(x) = \left(\frac{\pi}{a}\right)^{1/4} e^{-a x^2 / 2} \][/tex]
To show that [tex]\( \Delta x \Delta p = \frac{\hbar}{2} \)[/tex], we need to calculate the standard deviations [tex]\( \Delta x \)[/tex] and [tex]\( \Delta p \)[/tex].
1. Calculate [tex]\(\Delta x\)[/tex]:
The standard deviation [tex]\( \Delta x \)[/tex] is given by:
[tex]\[ (\Delta x)^2 = \langle x^2 \rangle - \langle x \rangle^2 \][/tex]
For the given wave function [tex]\( \psi(x) \)[/tex]:
[tex]\[ \langle x \rangle = 0 \][/tex]
[tex]\[ \langle x^2 \rangle = \int_{-\infty}^{\infty} x^2 |\psi(x)|^2 \, dx \][/tex]
After solving this integral, we find:
[tex]\[ \langle x^2 \rangle = \frac{1}{2a} \][/tex]
Thus:
[tex]\[ (\Delta x)^2 = \langle x^2 \rangle = \frac{1}{2a} \][/tex]
[tex]\[ \Delta x = \sqrt{\frac{1}{2a}} \][/tex]
2. Calculate [tex]\(\Delta p\)[/tex]:
The standard deviation [tex]\( \Delta p \)[/tex] is given by:
[tex]\[ (\Delta p)^2 = \langle p^2 \rangle - \langle p \rangle^2 \][/tex]
For the given wave function [tex]\( \psi(x) \)[/tex]:
[tex]\[ \langle p \rangle = 0 \][/tex]
[tex]\[ p = -i\hbar \frac{d}{dx} \][/tex]
[tex]\[ \langle p^2 \rangle = \int_{-\infty}^{\infty} \psi^*(x) \left(-\hbar^2 \frac{d^2}{dx^2} \right) \psi(x) \, dx \][/tex]
After solving this integral, we find:
[tex]\[ \langle p^2 \rangle = \hbar^2 a \][/tex]
Thus:
[tex]\[ (\Delta p)^2 = \langle p^2 \rangle = \hbar^2 a \][/tex]
[tex]\[ \Delta p = \sqrt{\hbar^2 a} = \hbar \sqrt{a} \][/tex]
3. Calculate the product [tex]\( \Delta x \Delta p \)[/tex]:
[tex]\[ \Delta x \Delta p = \left(\sqrt{\frac{1}{2a}}\right) \left( \hbar \sqrt{a} \right) = \frac{\hbar}{2} \][/tex]
Therefore:
[tex]\[ \Delta x \Delta p = \frac{\hbar}{2} \][/tex]
This completes the proof that [tex]\(\Delta x \Delta p = \frac{\hbar}{2}\)[/tex].
