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--- the_variational_principle [2020/08/18 07:00] – [5.i.4 Dealing with A Discontinuous Ansatz] admin
+++ the_variational_principle [2020/08/24 07:02] (current) – [In Class Activities] admin
@@ Line 1: / Line 1: @@
 ====== 5.i.1 Introduction ======
-This section discusses two related concepts: the //variational principle// and the //variational method// (also called the //Rayleigh-Ritz method//).  The variational principle shows that solving the Time Independent Schrödinger Equation (TISE) is equivalent to the stationarity of the energy functional $E[\psi]$ in the calculus of variations.  From classical mechanics, you should be familiar with the idea that dynamical laws can be formulated as variational principles.  In that case, the trajectories of a classical system are those for which the variation of the action $\delta S$ is zero.  The //variational method// is a method of approximating the energy spectrum and stationary states, particularly the ground state, that is based on the variational principle.
+This section discusses two related concepts: the //variational principle// and the //variational method// (also called the //Rayleigh-Ritz method//).  The variational principle shows that solving the Time Independent Schrödinger Equation (TISE) is equivalent to the stationarity of the energy functional $E[\psi]$ in the calculus of variations.  From classical mechanics, you should be familiar with the idea that dynamical laws can be formulated as variational principles.  In that case, the trajectories of a classical system are those for which the variation of the action $\delta S$ is zero.  The //variational method// is a method of approximating the energy spectrum and stationary states of a Hamiltonian $\hat{H}$, particularly the ground state, that is based on the variational principle.
 There are many potentials for which it is difficult or impossible to calculate the energy spectrum and stationary states exactly.  For this reason, we either have to use numerical methods or employ suitable approximation schemes to solve most physically realistic potentials.  Several approximation methods are covered in PHYS 452, but the variational method, which we discuss here, is perhaps the simplest.  A better name for it would be "guessing", but physicists do not like to admit that they are just guessing in their research papers, hence the more sophisticated sounding name.
-We also often use the word //ansatz// instead of "guess".  When quantum mechanics was being developed in the 1920's and 1930's, the language that most physics papers were published in was German, so it was natural to use the German word "ansatz".  It has several different meanings and is not precisely translatable into English, but it has the same sense as words like "attempt", "first approach", or "beginnings".  As the center of world physics moved westwards around the time of World War II, and the lingua franca of physics became English, and it proved more appealing to retain the word "ansatz" than replace it by the accurate, but less impressive sounding, word "guess".
+We also often use the word //ansatz// instead of "guess".  When quantum mechanics was being developed in the 1920's and 1930's, the language that most physics papers were published in was German, so it was natural to use the German word "ansatz".  It has several different meanings and is not precisely translatable into English, but it has the same sense as words like "attempt", "first approach", or "beginnings".  As the center of world physics moved westwards around the time of World War II, and the lingua franca of physics became English, it proved more appealing to retain the word "ansatz" than replace it by the accurate, but less impressive sounding, word "guess".
 ====== 5.i.2 The Variational Principle ======
@@ Line 119: / Line 119: @@
 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
@@ Line 143: / Line 143: @@
 Let's think about how to use the variational method more generally.  If we have a Hamiltonian $\hat{H}$ for which it is too difficult to solve for the eigenstates directly, then can still get an upper bound on the ground state energy by simply guessing any quantum state we like and calculating the expectation value of its energy.  Unfortunately, this is an //uncontrolled// approximation, meaning that we do not have a way of estimating how far above the ground state energy the expectation value is.  However, we can often use what we know about the potential to make an intelligent <del>guess</del>ansatz.
-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 ground state of the Harmonic oscillator as an ansatz for the potential $V(x) = k x^4$.  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.
-The other thing we can do is to make our ansatz depend on some 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.
 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.
@@ Line 183: / Line 183: @@
 The first step is to normalize the wavefunction
-\[\int_{-\infty}^{+\infty} \mathrm{d}x\, |\psi(x)|^2 = A^2 \int_{-\infty}^{+\infty} \mathrm{d}x\, e^{-2b|x|} = A^2 \left [ \int_{-\infty}^{0} \mathrm{d}x\, e^{2bx} + \int_{}^{+\infty} \mathrm{d}x\, e^{-2bx} = \frac{A^2}{b},\right ]\]
+\[\int_{-\infty}^{+\infty} \mathrm{d}x\, |\psi(x)|^2 = A^2 \int_{-\infty}^{+\infty} \mathrm{d}x\, e^{-2b|x|} = A^2 \left [ \int_{-\infty}^{0} \mathrm{d}x\, e^{2bx} + \int_{0}^{+\infty} \mathrm{d}x\, e^{-2bx} = \frac{A^2}{b},\right ]\]
 so the normalization constant is $A = \sqrt{b}$ and $\psi(x) = \sqrt{b} e^{-b|x|}$.
-The potential energy term is straightforward to calculate, and is left as an in class activity.  It is $\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}$.  The interesting part is the kinetic energy term.
+The potential energy term is straightforward to calculate, and is left as an in class activity.  It is $\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}$.  The interesting part is the kinetic energy term.  Let's first see what the difficulty is by doing it the naive way.
+For $x>0$, the first derivative is
+\[\frac{\mathrm{d}\psi}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x} \left ( \sqrt{b} e^{-bx} \right ) = -b^{3/2} e^{-bx},\]
+and for $x < 0$, it is
+\[\frac{\mathrm{d}\psi}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x} \left ( \sqrt{b} e^{bx} \right ) = b^{3/2}e^{bx}.\]
+Similarly, for $x>0$, the second derivative is
+\[\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = b^{5/2} e^{-bx},\]
+and for $x<0$, it is
+\[\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = b^{5/2} e^{bx}.\]
+Therefore, a naive computation of the expectation value of the kinetic energy yields
+\[\langle \hat{T} \rangle = - \frac{\hbar^2}{2m} \int_{-\infty}^{+\infty} \mathrm{d}x\, \psi^*(x) \frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = -\frac{\hbar^2}{2m} \left ( \int_{-\infty}^0 \mathrm{d}x\, b^3 e^{2bx} + \int_0^{+\infty} \mathrm{d}x\, b^3 e^{-2bx}\right ).\]
+By substituting $x'=-x$ and swapping the order of the limits, one can see that the first integral is the same as the second.  Therefore, we get
+\[\langle \hat{T} \rangle = -\frac{\hbar^2 b^3}{m} \int_{0}^{+\infty} \mathrm{d}x \, e^{-2bx} = + \frac{\hbar^2 b^2}{2m} \left [ e^{-2bx}\right ]_{0}^{+\infty} = -\frac{\hbar^2 b^2}{2m}.\]
+This result cannot possibly be right, because it says that the average kinetic energy is //negative//.
+The problem is that there is a jump discontinuity in the first derivative at $x=0$.  We can rewrite the first derivative in the following way:
+\[\frac{\mathrm{d}\psi}{\mathrm{d}x} = -\mathrm{sgn}(x) b^{3/2} e^{-b|x|},\]
+where $\mathrm{sgn}(x)$ is the sign function
+\[\mathrm{sgn}(x) = \begin{cases} -1, & x<0, \\ +1, & x\geq 0. \end{cases}\]
+We have that
+\[\frac{\mathrm{d} |x|}{\mathrm{d}x} = \mathrm{sgn}(x),\]
+and we previously learned that the derviative of the Heaviside step function
+\[\theta(x) = \begin{cases} 0, & x<0, \\ +1, & x\geq 0, \end{cases}\]
+is a Dirac delta function $\frac{\mathrm{d}\theta}{\mathrm{d}x} = \delta(x)$.  Since $\mathrm{sgn}(x) = 2\theta(x) - 1$ we have $\frac{\mathrm{d}\mathrm{sgn}(x)}{\mathrm{d}x} = 2\delta(x)$.  Therefore, we should compute the second derivative of the wavefunction as follows.
+\[\frac{\mathrm{d}^2\psi}{\mathrm{d}x^2} = \frac{\mathrm{d}}{\mathrm{d}x} \left ( -\mathrm{sgn}(x) b^{3/2} e^{-b|x|} \right ) = -2 b^{3/2}\delta(x) e^{- b|x|} + -b^{3/2} \mathrm{sgn}(x) \left ( -b \mathrm{sgn}(x) \right ) e^{-b |x|} = -2 b^{3/2}\delta(x) e^{-b|x|} + b^{5/2} e^{-b |x|},\]
+where we have used the product and chain rule for the derivatives and the fact that $\mathrm{sgn}(x)\mathrm{sgn}(x) = 1$.
+The second term is just what we used for the second derivative in the naive calculation, and so will contribute $-\frac{\hbar^2 b^2}{2m}$ to the average kinetic energy.  The first term, with the delta function, is due to the discontinuity in the first derivative, and will contribute an extra term
+\[\frac{\hbar^2 b^2}{m}\int_{-\infty}^{+\infty} \mathrm{d}x\,\delta(x) e^{-2b |x|} = \frac{\hbar^2 b^2}{m}.\]
+Adding the contributions from both terms gives
+\[\langle \hat{T} \rangle = \frac{\hbar^2 b^2}{m} - \frac{\hbar^2 b^2}{2m} = \frac{\hbar^2 b^2}{2m},\]
+which is the correct value and is (thankfully) positive.
+While this method of working out the correct delta function contribution in the second derivative is perfectly fine, and gives the correct answer, it can be simpler to employ the alternative method that only involves the first derivative.  In this case, we get:
+\[ \langle \hat{T} \rangle = \frac{\hbar^2}{2m} \int_{-\infty}^{+\infty} \mathrm{d}x\, \left | \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |^2 = \frac{\hbar^2}{2m} \left ( \int_{-\infty}^0 \mathrm{d}x\, \left | b^{3/2}e^{bx} \right |^2 + \int_0^{\infty} \mathrm{d}x\, \left | -b^{3/2}e^{-bx}\right |^2 \right ) = \frac{\hbar^2}{2m} \left ( \int_{-\infty}^0 \mathrm{d}x\,  b^{3}e^{2bx} + \int_0^{\infty} \mathrm{d}x\, b^{3}e^{-2bx} \right ).\]
+In the first integral, if we substitute $-x$ for $x$ and swap the limits, it turns out to be the same as the second integral.  Therefore.
+\[ \langle \hat{T} \rangle = \frac{\hbar^2}{m} \int_0^{\infty} \mathrm{d}x\, b^3 e^{-2bx} = -\frac{\hbar^2 b^2}{2m} \left [ e^{-2bx}\right ]_0^{\infty} = \frac{\hbar^2 b^2}{2m},\]
+which agrees with the answer we had before.  Use whichever of the two methods you feel most comfortable with, but just be aware that if there is a discontinuity in the first derivative then you need to be careful about how you calculate the expectation value of the kinetic energy.
+At this stage, we have the expected value of the energy
+\[\langle \hat{H} \rangle = \langle \hat{T} \rangle + \langle \hat{V} \rangle = \frac{\hbar^2 b^2}{2m} + \frac{m\omega^2}{4b^2},\]
+so we can optimize over $b$ to obtain an upper bound on the ground state energy.  We have
+\[\frac{\mathrm{d} \langle \hat{H} \rangle}{\mathrm{d} b} = \frac{\hbar^2 b}{m} - \frac{m\omega^2}{2b^3}.\]
+Setting this equal to zero gives
+\[b = \frac{1}{2^4} \sqrt{\frac{m\omega}{\hbar}}.\]
+Substituting this into the expectation value gives
+\[\langle \hat{H} \rangle = \frac{\hbar^2 m \omega}{2\sqrt{2} \hbar m} + \frac{m\omega^2 \sqrt{2}\hbar}{4 m \omega} = \frac{\sqrt{2}}{2}\hbar \omega.\]
+The true ground state energy of the harmonic oscillator is $\frac{1}{2} \hbar \omega$.  It should not be a surprise that the upper bound obtained from our ansatz is a bit larger than this, since we know that the ground state is gaussian.
 ====== 5.i.4 Finding Excited States ======
-{{:question-mark.png?nolink&50 |}}====== In Class Activities ======
+Since all of the energy eigenstates are stationary points under variations of the expectation value of the Hamiltonian we can, in principle, find the excited states by the same method.  If you know the exact ground state $|E_1\rangle$ then you can choose an ansatz $|\psi_2\rangle$ that is orthogonal to the ground state $\langle \psi_2 | E_1 \rangle = 0$.  Then, we know that $|\psi_2 \rangle$ is a superposition of excited states $|\psi_2 \rangle = \sum_{j\geq 2}^n c_j |E_j \rangle$, with no ground state component in the superposition.  By the same argument we gave for the ground state, we will have $\langle \hat{H} \rangle \geq E_2$, so by varying the parameters of the ansatz we can obtain a bound on the energy of the first excited state.
+We can proceed for the higher excited states analagously.  If we want an upper bound on $E_n$ and we know the states $|E_1\rangle, |E_2\rangle, \cdots, |E_{n-1}\rangle$ then we can choose an ansatz $|\psi_n\rangle$ that is orthogonal to all of them, and then we will have $\langle \hat{H} \rangle \geq E_n$.
+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.
+{{:question-mark.png?nolink&50 |}}
+====== In Class Activities ======
   - For the harmonic oscillator Hamiltonian
@@ Line 201: / Line 294: @@
   has
   \[\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}, \quad \mathrm{and} \quad \langle \hat{V} \rangle = \frac{m\omega^2}{8b}.\]
+  - For the same Hamiltonian, consider the ansatz $\psi(x) = \sqrt{b} e^{-b|x|}$.  Show that the potential energy term is
+  \[\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}.\]