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the_variational_principle [2020/08/24 05:49] – [General Considerations] 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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 However, there are two major problems with this method.  First, as we increase $n$, the number of orthogonality conditions we have to impose increases, so it gets harder to choose an ansatz that satisfies all of the orthogonality conditions.  Second, and more seriously, when we want to know the energy $E_n$, we typically do not know the eigenstates $|E_1\rangle, |E_2\rangle, \cdots, |E_{n-1}\rangle$ exactly, but only the approximations to them $|\psi_1\rangle, |\psi_2\rangle, \cdots, |\psi_{n-1}\rangle$ that we have obtained from the variational method.  The errors in these states will accumulate as we calculate the energies of higher and higher excited states. However, there are two major problems with this method.  First, as we increase $n$, the number of orthogonality conditions we have to impose increases, so it gets harder to choose an ansatz that satisfies all of the orthogonality conditions.  Second, and more seriously, when we want to know the energy $E_n$, we typically do not know the eigenstates $|E_1\rangle, |E_2\rangle, \cdots, |E_{n-1}\rangle$ exactly, but only the approximations to them $|\psi_1\rangle, |\psi_2\rangle, \cdots, |\psi_{n-1}\rangle$ that we have obtained from the variational method.  The errors in these states will accumulate as we calculate the energies of higher and higher excited states.
  
-For this reason, the variational method is typically only used to estimate the ground state and perhaps the first few excited states, but after that, other approximation methods that we will learn about in PHYS 452 will be more useful..+For this reason, the variational method is typically only used to estimate the ground state and perhaps the first few excited states, but after that, other approximation methods that we will learn about in PHYS 452 will be more useful.
  
 {{:question-mark.png?nolink&50 |}} {{:question-mark.png?nolink&50 |}}
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   has   has
   \[\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}, \quad \mathrm{and} \quad \langle \hat{V} \rangle = \frac{m\omega^2}{8b}.\]   \[\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}, \quad \mathrm{and} \quad \langle \hat{V} \rangle = \frac{m\omega^2}{8b}.\]
 +  
 +  - For the same Hamiltonian, consider the ansatz $\psi(x) = \sqrt{b} e^{-b|x|}$.  Show that the potential energy term is
 +  \[\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}.\]