VIPSolutions

VIPSolutions logo ✨ VIPSolutions

Home Dashboard Ask AI Archive

Alright Manish — here’s the full linear-variational solve, in your preferred format. (I also plotted (f_1,f_2) and put small tables of numbers for you above.) --- ### Step 1 — Given, objective, approach **Given:** Particle in a 1D infinite well (0\le x\le L), (V=0) inside; basis [ f_1(x)=\sqrt{\frac{105}{L^{7}}},x^{2}(L-x),\qquad f_2(x)=\sqrt{\frac{105}{L^{7}}},x(L-x)^{2} ] (Both vanish at (x=0,L) and are normalized: (S_{11}=S_{22}=1).) **Objective:** Use the linear variational method with the 2-function basis to get (i) (H_{ij}) and (S_{12}), (ii) the secular equation and the two approximate energies (W_1,W_2), (iii) percent errors vs exact (E_n), and (iv) normalized ((c_1,c_2)) for the **ground-state** variational wavefunction. **Approach:** Inside the box, (\hat H = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}). With (f_i(0)=f_i(L)=0), [ H_{ij}=\frac{\hbar^2}{2m}\int_0^L \frac{df_i}{dx}\frac{df_j}{dx},dx,\quad S_{ij}=\int_0^L f_i f_j,dx . ] Work in the dimensionless variable (y=x/L) so every energy comes out as a number times (\dfrac{\hbar^2}{2mL^2}). --- ### Step 2 — Overlap (S_{12}) Let (g_1(y)=y^2(1-y)), (g_2(y)=y(1-y)^2) so (f_i=\dfrac{\sqrt{105}}{\sqrt L},g_i(y)). Then [ S_{12}=105\int_{0}^{1} g_1 g_2,dy =105\int_0^1!\big[y^3-3y^4+3y^5-y^6\big]dy =\frac{3}{4}. ] **Explanation:** Pulled out constants; integrand is a polynomial → integrate termwise. --- ### Step 3 — Hamiltonian matrix (H_{ij}) With (g'*1=2y-3y^2), (g'*2=1-4y+3y^2), [ H*{ij}=\frac{\hbar^2}{2mL^2},105\int_0^1 g'*i g'*j,dy. ] Compute the three integrals: [ \int_0^1 g_1'^2dy=\frac{2}{15},\quad \int_0^1 g_2'^2dy=\frac{2}{15},\quad \int_0^1 g_1'g_2'dy=\frac{1}{30}. ] Hence [ H*{11}=H*{22}=14,\frac{\hbar^2}{2mL^2},\qquad H*{12}=H_{21}=\frac{7}{2},\frac{\hbar^2}{2mL^2}. ] **Explanation:** Same trick—polynomials integrate directly; the overall factor (105) gives the tidy integers. --- ### Step 4 — Secular equation and energies With (S_{11}=S_{22}=1), (S_{12}=3/4), define (\alpha=\dfrac{\hbar^2}{2mL^2}) and write the 2×2 determinant: [ \left| \begin{matrix} 14\alpha-W & \frac{7}{2}\alpha-\frac{3}{4}W[2pt] \frac{7}{2}\alpha-\frac{3}{4}W & 14\alpha-W \end{matrix}\right|=0. ] Solve: [ W_1=10,\alpha,\qquad W_2=42,\alpha. ] **Explanation:** Because the matrix is symmetric with equal diagonals, roots follow from (14\alpha-W=\pm\left(\frac{7}{2}\alpha-\frac{3}{4}W\right)). --- ### Step 5 — Compare with exact energies and percent errors Exact for an infinite well: (E_n=n^2\pi^2,\alpha). * (n=1): (E_1=\pi^2\alpha\approx 9.8696,\alpha) (\displaystyle %\text{ error}=\frac{W_1-E_1}{E_1}\times100\approx \mathbf{1.32%}) (upper bound) * (n=2): (E_2=4\pi^2\alpha\approx 39.4784,\alpha) (\displaystyle %\text{ error}=\frac{W_2-E_2}{E_2}\times100\approx \mathbf{6.39%}) (upper bound) **Explanation:** Variational energies are always ≥ exact—matches the overestimates. --- ### Step 6 — Ground-state coefficients ((c_1,c_2)) Solve ((H-WS)\vec c=0) at (W=W_1=10\alpha): [ \frac{c_1}{c_2} =-,\frac{H_{12}-W_1 S_{12}}{H_{11}-W_1} =-,\frac{\frac{7}{2}\alpha-10\alpha\cdot\frac{3}{4}}{14\alpha-10\alpha} =\frac{-4\alpha}{4\alpha}=1. ] Normalize with (c^\top S c = 1): [ c_1=c_2=k,\quad k^2(1+1+2S_{12})=1 ;\Rightarrow; k=\frac{1}{\sqrt{2+2\cdot(3/4)}}=\sqrt{\frac{2}{7}}. ] So the **variational ground-state wavefunction** is [ \boxed{;\phi_g(x)=\sqrt{\frac{2}{7}}\Big[f_1(x)+f_2(x)\Big];} ] with (W_1=10,\dfrac{\hbar^2}{2mL^2}). **Explanation:** The linear system gives only the ratio; normalization with the overlap yields the actual values. --- If you want, I can export the matrices/energies in LaTeX-ready form or generate a clean PDF page with the steps. just write this

Question:

Alright Manish — here’s the full linear-variational solve, in your preferred format. (I also plotted (f_1,f_2) and put small tables of numbers for you above.) --- ### Step 1 — Given, objective, approach **Given:** Particle in a 1D infinite well (0\le x\le L), (V=0) inside; basis [ f_1(x)=\sqrt{\frac{105}{L^{7}}},x^{2}(L-x),\qquad f_2(x)=\sqrt{\frac{105}{L^{7}}},x(L-x)^{2} ] (Both vanish at (x=0,L) and are normalized: (S_{11}=S_{22}=1).) **Objective:** Use the linear variational method with the 2-function basis to get (i) (H_{ij}) and (S_{12}), (ii) the secular equation and the two approximate energies (W_1,W_2), (iii) percent errors vs exact (E_n), and (iv) normalized ((c_1,c_2)) for the **ground-state** variational wavefunction. **Approach:** Inside the box, (\hat H = -\dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}). With (f_i(0)=f_i(L)=0), [ H_{ij}=\frac{\hbar^2}{2m}\int_0^L \frac{df_i}{dx}\frac{df_j}{dx},dx,\quad S_{ij}=\int_0^L f_i f_j,dx . ] Work in the dimensionless variable (y=x/L) so every energy comes out as a number times (\dfrac{\hbar^2}{2mL^2}). --- ### Step 2 — Overlap (S_{12}) Let (g_1(y)=y^2(1-y)), (g_2(y)=y(1-y)^2) so (f_i=\dfrac{\sqrt{105}}{\sqrt L},g_i(y)). Then [ S_{12}=105\int_{0}^{1} g_1 g_2,dy =105\int_0^1!\big[y^3-3y^4+3y^5-y^6\big]dy =\frac{3}{4}. ] **Explanation:** Pulled out constants; integrand is a polynomial → integrate termwise. --- ### Step 3 — Hamiltonian matrix (H_{ij}) With (g'*1=2y-3y^2), (g'*2=1-4y+3y^2), [ H*{ij}=\frac{\hbar^2}{2mL^2},105\int_0^1 g'*i g'*j,dy. ] Compute the three integrals: [ \int_0^1 g_1'^2dy=\frac{2}{15},\quad \int_0^1 g_2'^2dy=\frac{2}{15},\quad \int_0^1 g_1'g_2'dy=\frac{1}{30}. ] Hence [ H*{11}=H*{22}=14,\frac{\hbar^2}{2mL^2},\qquad H*{12}=H_{21}=\frac{7}{2},\frac{\hbar^2}{2mL^2}. ] **Explanation:** Same trick—polynomials integrate directly; the overall factor (105) gives the tidy integers. --- ### Step 4 — Secular equation and energies With (S_{11}=S_{22}=1), (S_{12}=3/4), define (\alpha=\dfrac{\hbar^2}{2mL^2}) and write the 2×2 determinant: [ \left| \begin{matrix} 14\alpha-W & \frac{7}{2}\alpha-\frac{3}{4}W[2pt] \frac{7}{2}\alpha-\frac{3}{4}W & 14\alpha-W \end{matrix}\right|=0. ] Solve: [ W_1=10,\alpha,\qquad W_2=42,\alpha. ] **Explanation:** Because the matrix is symmetric with equal diagonals, roots follow from (14\alpha-W=\pm\left(\frac{7}{2}\alpha-\frac{3}{4}W\right)). --- ### Step 5 — Compare with exact energies and percent errors Exact for an infinite well: (E_n=n^2\pi^2,\alpha). * (n=1): (E_1=\pi^2\alpha\approx 9.8696,\alpha) (\displaystyle %\text{ error}=\frac{W_1-E_1}{E_1}\times100\approx \mathbf{1.32%}) (upper bound) * (n=2): (E_2=4\pi^2\alpha\approx 39.4784,\alpha) (\displaystyle %\text{ error}=\frac{W_2-E_2}{E_2}\times100\approx \mathbf{6.39%}) (upper bound) **Explanation:** Variational energies are always ≥ exact—matches the overestimates. --- ### Step 6 — Ground-state coefficients ((c_1,c_2)) Solve ((H-WS)\vec c=0) at (W=W_1=10\alpha): [ \frac{c_1}{c_2} =-,\frac{H_{12}-W_1 S_{12}}{H_{11}-W_1} =-,\frac{\frac{7}{2}\alpha-10\alpha\cdot\frac{3}{4}}{14\alpha-10\alpha} =\frac{-4\alpha}{4\alpha}=1. ] Normalize with (c^\top S c = 1): [ c_1=c_2=k,\quad k^2(1+1+2S_{12})=1 ;\Rightarrow; k=\frac{1}{\sqrt{2+2\cdot(3/4)}}=\sqrt{\frac{2}{7}}. ] So the **variational ground-state wavefunction** is [ \boxed{;\phi_g(x)=\sqrt{\frac{2}{7}}\Big[f_1(x)+f_2(x)\Big];} ] with (W_1=10,\dfrac{\hbar^2}{2mL^2}). **Explanation:** The linear system gives only the ratio; normalization with the overlap yields the actual values. --- If you want, I can export the matrices/energies in LaTeX-ready form or generate a clean PDF page with the steps. just write this

This is AI generated Response.
Asked by: manish
Created at: 2025-11-05 16:34:41

Final Answer:

Rendered Raw Text
Did you like this answer?
# Linear Variational Solution for Particle in 1D Infinite Well ## Step 1 — Problem Setup **Given:** - Particle in a 1D infinite well, \( \le x \le L \), \( V(x) = \) inside. - Basis functions: \[ f_1(x) = \sqrt{\frac{105}{L^7}}\, x^2(L-x),\qquad f_2(x) = \sqrt{\frac{105}{L^7}}\, x(L-x)^2 \] Both vanish at \( x=,L \); normalized so \( S_{11}=S_{22}=1 \). **Objective:** Use the two-function basis to obtain: 1. \( H_{ij} \) and \( S_{12} \) 2. Secular equation & approximate energies (\( W_1, W_2 \)) 3. Percent errors vs exact \( E_n \) 4. Normalized coefficients (\( c_1, c_2 \)) for the ground state variational wavefunction **Approach:** - Hamiltonian: \( \hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} \) - \( H_{ij} = \frac{\hbar^2}{2m} \int_^L f_i'(x) f_j'(x) dx \) - \( S_{ij} = \int_^L f_i(x) f_j(x) dx \) - Use \( y = x/L \) (dimensionless); energies in units of \( \frac{\hbar^2}{2mL^2} \) --- ## Step 2 — Overlap \( S_{12} \) Let \( g_1(y) = y^2(1-y) \), \( g_2(y) = y(1-y)^2 \), so \( f_i = \sqrt{105/L} \cdot g_i(y) \): \[ S_{12} = 105 \int_^1 g_1(y) g_2(y) dy = 105 \int_^1 [y^3 - 3y^4 + 3y^5 - y^6]\,dy = \frac{3}{4} \] **Explanation:** Expanded product, integrate termwise (all polynomials). --- ## Step 3 — Hamiltonian Matrix \( H_{ij} \) Derivatives: - \( g_1'(y) = 2y - 3y^2 \) - \( g_2'(y) = 1 - 4y + 3y^2 \) \[ H_{ij} = \frac{\hbar^2}{2mL^2} \cdot 105 \int_^1 g_i'(y) g_j'(y)\,dy \] Compute: - \( \int_^1 [g_1'(y)]^2 dy = \frac{2}{15} \) - \( \int_^1 [g_2'(y)]^2 dy = \frac{2}{15} \) - \( \int_^1 g_1'(y) g_2'(y) dy = \frac{1}{30} \) So, \[ H_{11} = H_{22} = 14 \frac{\hbar^2}{2mL^2},\qquad H_{12} = H_{21} = \frac{7}{2} \frac{\hbar^2}{2mL^2} \] --- ## Step 4 — Secular Equation and Energies Given \( S_{11} = S_{22} = 1 \), \( S_{12} = 3/4 \), let \( \alpha = \frac{\hbar^2}{2mL^2} \): \[ \left| \begin{matrix} 14\alpha - W & \frac{7}{2}\alpha - \frac{3}{4}W \\ \frac{7}{2}\alpha - \frac{3}{4}W & 14\alpha - W \end{matrix} \right| = \] This yields: \[ W_1 = 10\alpha, \qquad W_2 = 42\alpha \] --- ## Step 5 — Comparison to Exact Energies Exact levels for infinite well: \( E_n = n^2 \pi^2 \alpha \) | \( n \) | \( E_n/\alpha \) | \( W_n/\alpha \) | % Error | |---------|------------------|------------------|---------| | 1 | \( \pi^2 \approx 9.8696 \) | 10 | \( +1.32\% \) | | 2 | \( 4\pi^2 \approx 39.4784 \) | 42 | \( +6.39\% \) | \[ \%\,\text{error} = \frac{W_n - E_n}{E_n} \times 100 \] --- ## Step 6 — Ground-State Coefficients (\( c_1, c_2 \)) Solve \( (H - WS)\vec{c} = \) at \( W = W_1 = 10\alpha \): \[ \frac{c_1}{c_2} = -\frac{H_{12} - W_1 S_{12}}{H_{11} - W_1} = -\frac{\frac{7}{2}\alpha - 10\alpha \cdot \frac{3}{4}}{14\alpha - 10\alpha} = \frac{-4\alpha}{4\alpha} = 1 \] Normalize with \( c^\top S c = 1 \): \[ c_1 = c_2 = k,\qquad k^2 (1 + 1 + 2 \cdot S_{12}) = 1 \implies k = \frac{1}{\sqrt{2 + 2 \cdot 3/4}} = \sqrt{\frac{2}{7}} \] **Final ground-state variational wavefunction:** \[ \boxed{ \phi_g(x) = \sqrt{\frac{2}{7}}\left[f_1(x) + f_2(x)\right] } \] with \( W_1 = 10 \frac{\hbar^2}{2mL^2} \). --- ## Summary Table | Quantity | Value | |------------------|---------------------------------------| | \( S_{11}, S_{22} \) | 1 | | \( S_{12} \) | \( \frac{3}{4} \) | | \( H_{11}, H_{22} \) | \( 14\, \frac{\hbar^2}{2mL^2} \) | | \( H_{12} \) | \( \frac{7}{2}\, \frac{\hbar^2}{2mL^2} \) | | \( W_1 \) | \( 10\, \frac{\hbar^2}{2mL^2} \) | | \( W_2 \) | \( 42\, \frac{\hbar^2}{2mL^2} \) | | \( (c_1, c_2) \) | \( \left(\sqrt{\frac{2}{7}},\,\sqrt{\frac{2}{7}}\right) \) | --- If needed: The above matrices and results can be exported in LaTeX or as a formatted PDF.

📢 Instructions:

  • Ask anything relatd to above question only.
  • For new Question goto :
    Ask AI

Related Answers

Study Help, Fast Answers

VIPSolutions focuses on clear, step‑by‑step explanations so you can learn quickly. Whether you need an AI‑generated walkthrough or a short hint to get unstuck, each solution is organized for fast reading and easy review later.

Search similar questions, compare approaches, and bookmark the best answers for revision. Our goal is simple: quick, reliable study help that feels natural—not noisy.