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the_variational_principle [2020/08/24 05:56] – [5.i.4 Finding Excited States] adminthe_variational_principle [2020/08/24 07:02] (current) – [In Class Activities] admin
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 and for the potential term we have, and for the potential term we have,
  
-\[\langle \hat{T} \rangle = \frac{1}{2}m \omega^2 \int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* x^2 \psi(x).\]+\[\langle \hat{V} \rangle = \frac{1}{2}m \omega^2 \int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* x^2 \psi(x).\]
  
 In an in class activity, you will show that these integrals evaluate to $\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}$ and $\langle \hat{V} \rangle = \frac{m\omega^2}{8b}$, so In an in class activity, you will show that these integrals evaluate to $\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}$ and $\langle \hat{V} \rangle = \frac{m\omega^2}{8b}$, so
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   - For the same Hamiltonian, consider the ansatz $\psi(x) = \sqrt{b} e^{-b|x|}$.  Show that the potential energy term is   - For the same Hamiltonian, consider the ansatz $\psi(x) = \sqrt{b} e^{-b|x|}$.  Show that the potential energy term is
-  \[\langle \hat{V} \rangle = \frac{m\omega^2}{4b}.\]+  \[\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}.\]