Some of the calculable trial functions:

- \(\psi(x) = \cos\alpha x\), for \(|\alpha x|<\pi/2\), otherwise 0.
- \(\psi(x) = \alpha^2 - x^2\), for \(|x|<\alpha\), otherwise 0.
- \(\psi(x) = C \exp(-\alpha x^2/2)\).
- \(\psi(x) = C(\alpha - |x|)\), for \(|x|<\alpha\), otherwise 0.
- \(\psi(x) = C\sin\alpha x\), for \(|\alpha x|<\pi\), otherwise 0.

Pick a trial function.

Note

How to pick a trial function? For ground state energy, we should pick a function that has the same property as the real ground state. This requires some understanding of the problem we are dealing with.

Things to consider:

- The new problem is just a modification of a known solved problem. Then we can easily find out what really is different and interprete the new problem in terms of the old one.
- If the Hamiltonian have definite parity, the ground state wave function should pick up some parity which is usually even to make it the lowest energy.
- Continious function? A \(C^\infty\) Hamiltonian can only have continious functions as solutions for a finite system.
- Nodes deteremines the kinetic energy so check the nodes for ground state wave function.
- Check the behivior of the wave function at different limits. In most cases, the Shrödinger equation can be reduced to something solvable at some limits.
**One more thing, the trial function should make the problem calculable.**

Why don’t we just use a most general variational method to find out the ground state? Because we will eventually come back to the time-independent Shrödinger equation.

Suppose we have a functional form

\[E(\psi^*, \psi, \lambda) = \int dx \psi^* H \psi - \lambda \left( \int dx \psi^* \psi - 1 \right)\]

The reason we have this Lagrange multiplier method is that the wave function should be normalized and this multiplier provides the degree of freedom. We would only get a wrong result if we don’t include this DoF.

Variation of \(\psi^*\),

\[\delta E = \int dx \delta \psi^* H \psi - \int dx \delta \psi^* \psi = 0\]

Now what?

\[H \psi - \lambda \psi = 0\]

Not helpful.

For a potential \(V(x)=b x^n\), we can prove that virial theorem is valid for ground state if we use Gaussian trial function \(e^{- \alpha x^2/2}\).

A MMA proof is here.

Virial theorem is pretty interesting. It shares the same math with equipartition theorem.

This is a semi-classical method. It is semi classical because we will use the classical momentum

\[\hbar k(x) = \sqrt{2m (E - V(x))}\]

The following points are important for this method.

WKB start from a classical estimation of wave number at a certain energy \(E\) which is later quantified by the Bohr-Sommerfeld quantization rule.

Conservation law:

\[\frac{\partial}{\partial t}\rho + \nabla \cdot \vec j = 0\]where \(\rho = \psi^* \psi\), \(\vec j = -\frac{\hbar}{2 m i} \left( \psi \nabla \psi^* - \psi^* \nabla \psi \right)\). This can be derived from Shrödinger equation easily.

Phase: Wave function is generally \(A(x)\exp(\phi(x))\). However, \(\phi(x)\) should be the area of the phase function starting from some initial point. For example in WKB, \(k(x) = \phi'(x)\) and \(\phi(x) = \int \phi'(x')d x' = \int k(x') d x'\).

Using this general wave function and conservation law we find out that \(A(x) ~ \frac{1}{\sqrt{k(x)}}\). Then we can apply the two boundary conditions. However we will find two different wave functions given by two boundary conditions. Now we should connect them because \(\psi(a) = \psi(b)\) exactly. By comparing the two wave functions we can find something like Bohr-Sommerfeld quantization rule.

Correction at bouldary: However, this method requires that the potential varies slowly or equivalently the wave number varies slowly. Basicly we are just using the following approximation:

\[A'(x) = 0, k'(x) = 0\]For example when taking the derivative of wave function,

\[\psi'(x) = A'(x) e^{i\int \cdots} + A(x) k(x) e^{i\int \cdots} \approx A(x) k(x) e^{i\int \cdots}\]where we drop the term with \(A'(x)\). That is to say

\[|A'|\ll |A k| \Rightarrow |k'| \ll k^2\]But at boundary where \(E = V\), this is obviously not valid because \(k=0\). So we need to fix this problem.

The solution is to use first order of the potential in a Taylor expansion. Then solve the problem exactly. Finally we connect regions that is far out from the boundary, need the boundary and between the boundary.

If we can have a good boundary condition, then the energy spectrum given by WKB can be very good. Even we don’t have a good boundary condtion, the excited states given by this method are always close to the exact ones.

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