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, it proved more appealing to retain the word “ansatz” than replace it by the accurate, but less impressive sounding, word “guess”.

The variational method works due to a theorem in the calculus of variations called the variational principle or the Ritz theorem. Calculus of variations should be familiar from your mechanics courses, but in any case it is only needed for this subsection, so you can skip to the next section if you do not understand what is going on. In the context of quantum mechanics, the expectation value of the energy \[ E[ \psi] = \langle \hat{H} \rangle = \langle \psi | \hat{H} | \psi \rangle ,\] is a functional of the state vector $|\psi\rangle$. Note here we are assuming that the state vector is normalized, which will be important later. The wavefunction is a function of $x$, and the energy expectation is a function of this function, and hence a functional. As we vary the wavefunction $\psi(x)$, $E[\psi]$ varies. The Ritz theorem says that the energy eigenstates are the stationary points of this variation.

More precisely, let $|\delta \psi \rangle$ be a small change in the state vector and let \[\delta E [\psi] = E[\psi + \delta \psi ] - E[\psi].\] The energy eigenstates are precisely those states such that \[\delta E [\psi ] = 0,\] for all variations $|\delta \psi \rangle$ to first order in $|\delta \psi \rangle$.

Note that, when we vary $|\psi \rangle$, we want to impose that the variations preserve the normalization of the state, so we first need to add a Lagrange multiplier to the energy functional in order to enforce this condition.

\[E[\psi] = \langle \psi | \hat{H} |\psi \rangle - \lambda (\langle \psi | \psi \rangle - 1),\] where $\lambda$ is the Lagrange multiplier.

We now apply the calculus of variations, using the product rule, which gives

\[\delta E[\psi] = \langle \delta \psi | \hat{H} | \psi \rangle + \langle \psi | \hat{H} | \delta \psi \rangle - \lambda (\langle \delta \psi | \psi \rangle + \langle \psi | \delta \psi \rangle).\]

Rearranging this gives

\[\delta E[\psi] = \langle \delta \psi | \left ( \hat{H} |\psi \rangle - \lambda |\psi \rangle \right ) + \left ( \langle \psi |\hat{H} - \lambda \langle \psi | \right ) | \delta \psi \rangle.\]

Note that $\langle \delta \psi |$ and $|\delta \psi \rangle$ are independent variations, just as, for a complex number $z$, $\mathrm{d}z^*$ and $\mathrm{d}z$ are independent, so setting $\delta E[\psi] = 0$ yields the two equations

\[\hat{H} |\psi \rangle - \lambda |\psi \rangle = 0,\] \[\langle \psi |\hat{H} - \lambda \langle \psi | = 0.\]

Since $\hat{H}$ is Hermitian, these equations are equivalent, and rearranging gives

\[\hat{H}|\psi\rangle = \lambda |\psi \rangle.\]

This equation says that the functional $E[\psi]$ is stationary precisely when $|\psi \rangle$ is an eigenstate of the Hamiltonian $\hat{H}$, and the Lagrange multiplier turns out to be the energy eigenvalue. But this is exactly the same thing that the TISE says, so solving the TISE is equivalent to solving the variational problem $\delta E[\psi] = 0$.

What does this mean in practice. Well, suppose our quantum system has a Hilbert space $\mathcal{H}$, and suppose that all the unit vectors $|\psi\rangle$ in $\mathcal{H}$ can be described in terms of a set of real parameters $a_1, a_2, \cdots$. For a two dimensional Hilbert space, this is easy to do, as (up to an irrelevant global phase) an arbitrary unit vector can be written as

\[|\psi \rangle = \cos \theta |\phi_1 \rangle + e^{i\phi} \sin \theta |\phi_2 \rangle, \]

where $|\phi_1 \rangle, |\phi_2 \rangle$ form an orthonormal basis. We will need a larger number of parameters as the dimension of the Hilbert space increases and, in fact, even the Hilbert space of a quantum particle in one spatial dimension is infinite dimensional, so we would need an infinite number of parameters for most of the systems we care about.

We can then compute the energy functional $E[\psi]$ as a function of these parameters, i.e.

\[E[\psi] = \langle \psi(a_1,a_2,\cdots) |\hat{H} |\psi(a_1,a_2,\cdots) \rangle = f(a_1,a_2,\cdots),\] where $f(a_1,a_2,\cdots)$ is just notation for the function of the parameters $a_1,a_2,\cdots$ so defined.

Since we know that solving the TISE is equivalent to solving $\delta E[\psi] = 0$, we can find the energy eigenstates by solving the set of equations:

\[\frac{\mathrm{d}f}{\mathrm{d}a_1} = 0, \quad \frac{\mathrm{d}f}{\mathrm{d}a_2} = 0, \quad \cdots \]

Generally, finding an appropriate parameterization and solving all of these equations is much more convoluted than just solving the TISE directly. However, some numerical methods for solving the TISE work like this and there is an example of how to do this for a system with a two dimensional Hilbert space on Homework 5.

The method is only really practical if there are a relatively small number of parameters but, as noted above, we are very often dealing with infinite dimensional Hilbert spaces which require an infinite number of parameters. We can get around this by restricting attention to a subset of vectors (our ansatz) that can be specified by a small number of parameters. Since we are no longer considering all possible vectors, the variation is no longer guaranteed to give us the exact energy eigenstates, but if we choose a good ansatz then they will be good approximations. In particular, we will always get an upper bound on the ground state energy. This is the basis of the variational method discussed in the next section.

Consider a Hamiltonian $\hat{H}$ and suppose we want to find its ground state energy $E_{\mathrm{gs}}$. Let $|\psi\rangle$ be any normalized quantum state (not necessarily a stationary state). Then,

\[E_{\mathrm{gs}} \geq \langle \psi | \hat{H} | \psi \rangle = \langle \hat{H} \rangle.\]

Assume that the energy spectrum of $\hat{H}$ is discrete with energies $E_1, E_2, \cdots$ and corresponding eigenstates $|E_1\rangle, |E_2 \rangle , \cdots$. We order the energies such that $E_1 \leq E_2 \leq E_3 \leq \cdots$. Obviously $E_1 = E_{\mathrm{gs}}$ is the ground state energy. Note, if any of the energies are degenerate, we just repeat them the appropriate number of times in the list of energies and choose the corresponding eigenstates to be a complete orthonormal basis for the degenerate eigenspace. For example, if the first excited is doubly degenerate then $E_2 = E_3$ are both the energy of the first excited state and we choose $|E_2 \rangle$ and $|E_3 \rangle$ to be orthonormal states in the two-dimensional eigenspace.

The eigenstates $|E_1\rangle, |E_2 \rangle , \cdots$ form an orthonormal basis and, since we assume that $|\psi\rangle$ is normalized, we can write

\[|\psi \rangle = \sum_j c_j |E_j\rangle,\]

for some coefficients $c_j$ such that $\sum_j |c_j|^2 = 1$.

Note that, had we not assumed that the energy spectrum is discrete, i.e. if there are also scattering states, then the scattering part of our eigenbasis would be continuous and we would have an integral over this part of the basis as well as the sum. This does not alter the structure of the proof, but just makes the equations look a bit more complicated.

We now calculate the expectation value of the energy as follows

\[\langle \psi |\hat{H} |\psi \rangle = \sum_{jk} c_j^* \langle E_j | \hat{H} |E_k \rangle c_k = \sum_{jk} E_k c_j^* c_k \langle E_j| E_k \rangle = \sum_{jk} E_k c_j^* c_k \delta_{jk} = \sum_j E_j |c_j|^2.\]

Now note that $E_{\mathrm{gs}} = E_1 \leq E_j$ for all $j$. Therefore,

\[\langle \psi |\hat{H} |\psi \rangle = \sum_j E_j |c_j|^2 \geq \sum_j E_{\mathrm{gs}} |c_j|^2 = E_{\mathrm{gs}} \sum_j |c_j|^2 = E_{\mathrm{gs}},\] where we used the normalization condition $\sum_j |c_j|^2 = 1$.

Let's give a simple example of how this can be used. Suppose we want to know the ground state energy of the Harmonic oscillator

\[\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2.\]

Unfortunately, we have forgotten to study section 4.viii and all quantum mechanics textbooks have suddenly disappeared from existence. We know that the ground state has no nodes and is an even function (See 4.iii.5 and 4.iii.6). We also know that it should decay to zero as $|x| \rightarrow \infty$ because it must be square integrable and the potential tends to infinity as $|x|\rightarrow \infty$. A reasonable guessansatz for the ground state would be a Gaussian function centered at $x=0$, i.e. $\psi(x) = Ae^{-bx^2}$. (Since you actually have studied section 4.viii you know that the ground state actually is of this form, so I am going to get the actual ground state energy rather than just an upper bound.)

The first thing we need to do is normalize our wavefunction, since the variational principle assumes we are working with normalized states. We have

\[\int_{-\infty}^{+\infty} | \psi(x)|^2 \, \mathrm{d}x = A^2 \int_{-\infty}^{+\infty} e^{-2bx^2} \,\mathrm{d}x = A^2 \sqrt{\frac{\pi}{2b}},\]

so we have $A = \left ( \frac{2b}{\pi} \right )^{\frac{1}{4}}$, or

\[\psi(x) = \left ( \frac{2b}{\pi} \right )^{\frac{1}{4}} e^{-bx^2}.\]

We now calculate $\langle \hat{H} \rangle$ as follows. First note that we can break up the calculation into two parts $\langle \hat{H} \rangle = \langle \hat{T} \rangle + \langle \hat{V} \rangle$, where

\[\hat{T} = \frac{\hat{p}^2}{2m}, \qquad\qquad \hat{V} = \frac{1}{2} m \omega^2 x^2,\]

are the kinetic and potential energies.

For the kinetic term we have, in the position representation,

\[\langle \hat{T} \rangle = -\frac{\hbar^2}{2m}\int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* \frac{\mathrm{d}^2 \psi(x)}{\mathrm{d}x^2},\]

and for the potential term we have,

\[\langle \hat{V} \rangle = \frac{1}{2}m \omega^2 \int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* x^2 \psi(x).\]

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

\[\langle \hat{H} \rangle = \frac{\hbar^2 b}{2m} + \frac{m\omega^2}{8b}.\]

Now, this is an upper bound on the ground state energy for all possible values of $b$, so we can minimize over $b$. To do this, we calculate

\[\frac{\mathrm{d} \langle \hat{H} \rangle}{\mathrm{d}b} = \frac{\hbar^2}{2m} - \frac{m\omega^2}{8b^2}.\]

Setting this equal to zero gives

\[b = \frac{m\omega}{2\hbar},\]

which gives

\[E_{\mathrm{gs}} \geq \langle \hat{H} \rangle = \frac{\hbar \omega}{4} + \frac{\hbar \omega}{4} = \frac{1}{2} \hbar \omega.\]

This is indeed the exact ground state energy of the Harmonic oscillator, which we expected as the exact ground state happens to have the same form as our ansatz.

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 guessansatz.

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 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 guessingansatzing 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.

For some potentials, it may be useful to use an ansatz $\psi(x)$ that has a discontinuity in its derivative. As we know, solutions with discontinuous derivatives do occur as solutions to the TISE when the potential has a jump discontinuity or a delta-function term. This kind of ansatz needs to be treated with some care because the kinetic energy term in the expectation value of the Hamiltonian

\[\langle \hat{T} \rangle = -\frac{\hbar^2}{2m}\int_{-\infty}^{+\infty} \mathrm{d}x \, \psi(x)^* \frac{\mathrm{d}^2 \psi(x)}{\mathrm{d}x^2},\]

depends on the second derivative, which is, strictly speaking, undefined at points where the first derivative is discontinuous. There are two ways of handling this. The first is to recognize that, since we are integrating the above expression over $x$, it is OK to have delta-functions in the second derivative. The second is to replace the expression with one that only involves first derivatives. To do this, we first note that

\[\langle \psi | \hat{p}^2 | \psi \rangle = \langle \psi | \hat{p} \hat{p} | \psi \rangle = \langle \hat{p} \psi | \hat{p} \psi \rangle, \]

where we have used the fact that $\hat{p}$ is a Hermitian operator.

Now, in the position representation, $\hat{p} = -i\hbar \frac{\mathrm{d}}{\mathrm{d}x}$, so

\[\hat{p}|\psi\rangle = -i\hbar \int_{-\infty}^{+\infty} \mathrm{d}x\, \frac{\mathrm{d}\psi}{\mathrm{d}x} \, |x \rangle,\]

and hence

\[\langle \psi | \hat{p}^2 | \psi \rangle = \hbar^2 \int_{-\infty}^{+\infty} \mathrm{d}x\, \left | \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |^2 , \]

so the kinetic energy term will be

\[ \langle \hat{T} \rangle = \frac{\hbar^2}{2m} \int_{-\infty}^{+\infty} \mathrm{d}x\, \left | \frac{\mathrm{d}\psi}{\mathrm{d}x} \right |^2.\]

You can either use this method or handle the delta-functions in the second derivative correctly, but a naive calculation with the second derivative might lead to an incorrect result. This is best illustrated by an example.

Consider the harmonic oscillator Hamiltonian \[\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 x\] and suppose that our ansatz for the ground state is $\psi(x) = Ae^{-b|x|}$. This is plotted below for the case $b=1$.

The ansatz has a cusp at $x=0$, so its first derivative is discontinuous at $x=0$.

The first step is to normalize the wavefunction

\[\int_{-\infty}^{+\infty} \mathrm{d}x\, |\psi(x)|^2 = A^2 \int_{-\infty}^{+\infty} \mathrm{d}x\, e^{-2b|x|} = A^2 \left [ \int_{-\infty}^{0} \mathrm{d}x\, e^{2bx} + \int_{0}^{+\infty} \mathrm{d}x\, e^{-2bx} = \frac{A^2}{b},\right ]\]

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. 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.

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.

1. For the harmonic oscillator Hamiltonian \[\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2 \hat{x}^2,\] show that the ansatz \[\psi(x) = \left ( \frac{2b}{\pi} \right )^{\frac{1}{4}} e^{-bx^2},\] has \[\langle \hat{T}\rangle = \frac{\hbar^2 b}{2m}, \quad \mathrm{and} \quad \langle \hat{V} \rangle = \frac{m\omega^2}{8b}.\]
2. For the same Hamiltonian, consider the ansatz $\psi(x) = \sqrt{b} e^{-b|x|}$. Show that the potential energy term is \[\langle \hat{V} \rangle = \frac{m\omega^2}{4b^2}.\]