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| the_variational_principle [2020/08/24 05:49] – [General Considerations] admin | the_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, |
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| \[\langle \hat{T} \rangle = \frac{1}{2}m \omega^2 \int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* x^2 \psi(x).\] | \[\langle \hat{V} \rangle = \frac{1}{2}m \omega^2 \int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* x^2 \psi(x).\] |
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| 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 |
| For example, if the potential is $V(x) = k x^4$ for some positive constant $k$, then this potential looks similar to the Harmonic oscillator potential $V(x) = \frac{1}{2} m\omega x^2$ in the sense that they both have a minimum at $x=0$, they are both even functions, and they both curve upwards on either side of $x=0$. It would therefore make sense to choose the Gaussian ansatz $\psi(x) = Ae^{-bx^2}$ for the potential $V(x) = kx^4$ because we know that this is the form of the ground state of the harmonic oscillator. We also know that the ground state of any potential has no nodes and, if $V(x)$ is even, then the ground state is an even function. | For example, if the potential is $V(x) = k x^4$ for some positive constant $k$, then this potential looks similar to the Harmonic oscillator potential $V(x) = \frac{1}{2} m\omega x^2$ in the sense that they both have a minimum at $x=0$, they are both even functions, and they both curve upwards on either side of $x=0$. It would therefore make sense to choose the Gaussian ansatz $\psi(x) = Ae^{-bx^2}$ for the potential $V(x) = kx^4$ because we know that this is the form of the ground state of the harmonic oscillator. We also know that the ground state of any potential has no nodes and, if $V(x)$ is even, then the ground state is an even function. |
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| The other thing we can do is to make our ansatz depend on several (i.e. more than one) parameters $a,b,c,\cdots$, which are initially undetermined. We did this in the harmonic oscillator example with the parameter $b$. We can then calculate $\langle H \rangle = f(a,b,c,\cdots)$ as a function of those parameters and minimize $f(a,b,c,\cdots)$ over different choices of the parameters. The more parameters we use, the closer we will get to the true ground state energy, but the harder it will be to minimize $f$. In the extreme case where we leave $\psi(x)$ completely undetermined, we would have an infinite number of parameters and would get the exact ground state energy, but the minimization problem is then equivalent to solving $\hat{H}|\psi\rangle = E_{\mathrm{gs}}|\psi\rangle$ for the ground state, so it is of no help in simplifying the problem. The art of using the variational method is therefore to use both our knowledge of what the ground state looks like for similar Hamiltonians //and// a small number of undetermined parameters so that we are confident that we can get close to the true ground state with a minimization problem that is still tractable. | The other thing we can do is to make our ansatz depend on several (i.e. more than one) parameters $a,b,c,\cdots$, which are initially undetermined. We can then calculate $\langle H \rangle = f(a,b,c,\cdots)$ as a function of those parameters and minimize $f(a,b,c,\cdots)$ over different choices of the parameters. The more parameters we use, the closer we will get to the true ground state energy, but the harder it will be to minimize $f$. In the extreme case where we leave $\psi(x)$ completely undetermined, we would have an infinite number of parameters and would get the exact ground state energy, but the minimization problem is then equivalent to solving $\hat{H}|\psi\rangle = E_{\mathrm{gs}}|\psi\rangle$ for the ground state, so it is of no help in simplifying the problem. The art of using the variational method is therefore to use both our knowledge of what the ground state looks like for similar Hamiltonians //and// a small number of undetermined parameters so that we are confident that we can get close to the true ground state with a minimization problem that is still tractable. |
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| All this <del>guessing</del>ansatzing might seem not very rigorous, but this is actually how physical chemists calculate the ground states of atoms and molecules that are more complicated than the hydrogen atom. By introducing a large number of undetermined parameters and performing the minimizations numerically, they are able to achieve remarkable agreement with the ground state energies observed in the lab. | All this <del>guessing</del>ansatzing might seem not very rigorous, but this is actually how physical chemists calculate the ground states of atoms and molecules that are more complicated than the hydrogen atom. By introducing a large number of undetermined parameters and performing the minimizations numerically, they are able to achieve remarkable agreement with the ground state energies observed in the lab. |
| 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. |
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| 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. |
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| {{:question-mark.png?nolink&50 |}} | {{:question-mark.png?nolink&50 |}} |
| has | has |
| \[\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}, \quad \mathrm{and} \quad \langle \hat{V} \rangle = \frac{m\omega^2}{8b}.\] | \[\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}, \quad \mathrm{and} \quad \langle \hat{V} \rangle = \frac{m\omega^2}{8b}.\] |
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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 |
| | \[\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}.\] |
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