Let's solve **Exercise 4.12** **step-by-step** as per your request. --- ## **Given:** The Hamiltonian: \[ \hat{H} = \frac{1}{2} \hat{P}^2 + \frac{1}{2} \hat{X}^2 \] with \(\hat{P} = -i \frac{d}{dx}\). Wave function at time \( t=0 \): \[ \Psi(x,0) = \frac{1}{\sqrt{2}} \psi_0(x) + \frac{1}{\sqrt{8}} \psi_1(x) + \frac{1}{\sqrt{10}} \psi_2(x) \] The normalized eigenfunctions: \[ \begin{align*} \psi_0(x) &= \frac{1}{\sqrt{\sqrt{\pi}}} e^{-x^2/2} \\ \psi_1(x) &= \frac{\sqrt{2}}{\sqrt[4]{\pi}} x e^{-x^2/2} \\ \psi_2(x) &= \frac{1}{\sqrt{2\sqrt{\pi}}} (2x^2-1) e^{-x^2/2} \end{align*} \] --- ### **(a) Calculate \(\Delta x_n, \Delta p_n\) for \(n=0,1\)** #### **For Harmonic Oscillator Eigenstates** For the 1D quantum harmonic oscillator, the uncertainties in \(x\) and \(p\) for eigenstate \(\psi_n(x)\) are standard results: \[ \Delta x_n = \sqrt{ \langle x^2 \rangle_n - \langle x \rangle_n^2 } \] \[ \Delta p_n = \sqrt{ \langle p^2 \rangle_n - \langle p \rangle_n^2 } \] For the dimensionless oscillator: - \(\langle x \rangle_n = 0\) - \(\langle p \rangle_n = 0\) - \(\langle x^2 \rangle_n = n + \frac{1}{2}\) - \(\langle p^2 \rangle_n = n + \frac{1}{2}\) So, \[ \Delta x_n = \Delta p_n = \sqrt{ n + \frac{1}{2} } \] #### **Explicitly:** - For \(n = 0\): \[ \Delta x_0 = \Delta p_0 = \sqrt{ \frac{1}{2} } = \frac{1}{\sqrt{2}} \] - For \(n = 1\): \[ \Delta x_1 = \Delta p_1 = \sqrt{ 1 + \frac{1}{2} } = \sqrt{ \frac{3}{2} } = \frac{\sqrt{6}}{2} \] --- ### **(b) Calculate \( \hat{a}^\dagger \psi_0(x),\, \hat{a}\psi_0(x),\, \hat{a}^\dagger\psi_1(x),\, \hat{a}\psi_1(x),\, \hat{a}\psi_2(x) \) ** The annihilation and creation operators: \[ \hat{a} = \frac{1}{\sqrt{2}} \left( x + \frac{d}{dx} \right), \quad \hat{a}^\dagger = \frac{1}{\sqrt{2}} \left( x - \frac{d}{dx} \right) \] These act on harmonic oscillator eigenstates as: \[ \hat{a} \psi_n(x) = \sqrt{n} \psi_{n-1}(x) \] \[ \hat{a}^\dagger \psi_n(x) = \sqrt{n+1} \psi_{n+1}(x) \] #### **Now, let's compute:** - \( \hat{a}^\dagger \psi_0(x) = \sqrt{1} \psi_1(x) = \psi_1(x) \) - \( \hat{a} \psi_0(x) = 0 \) (since \( n = 0 \)) - \( \hat{a}^\dagger \psi_1(x) = \sqrt{2} \psi_2(x) \) - \( \hat{a} \psi_1(x) = \sqrt{1} \psi_0(x) = \psi_0(x) \) - \( \hat{a} \psi_2(x) = \sqrt{2} \psi_1(x) \) --- ## **Final Answers:** ### **(a)** \[ \boxed{ \begin{aligned} \Delta x_0 = \Delta p_0 &= \frac{1}{\sqrt{2}} \\ \Delta x_1 = \Delta p_1 &= \frac{\sqrt{6}}{2} \end{aligned} } \] ### **(b)** \[ \boxed{ \begin{aligned} \hat{a}^\dagger \psi_0(x) &= \psi_1(x) \\ \hat{a} \psi_0(x) &= 0 \\ \hat{a}^\dagger \psi_1(x) &= \sqrt{2} \psi_2(x) \\ \hat{a} \psi_1(x) &= \psi_0(x) \\ \hat{a} \psi_2(x) &= \sqrt{2} \psi_1(x) \\ \end{aligned} } \] If you need the explicit action in terms of \(x\), let me know!

In the harmonic oscillator, the uncertainties in position and momentum for eigenstates are given by \(\Delta x_n = \Delta p_n = \sqrt{n + \frac{1}{2}}\), reflecting minimal uncertainty states. The ladder operators \(\hat{a}\) and \(\hat{a}^\dagger\) act to lower or raise the quantum number \(n\) by one, satisfying \(\hat{a} \psi_n = \sqrt{n} \psi_{n-1}\) and \(\hat{a}^\dagger \psi_n = \sqrt{n+1} \psi_{n+1}\).