## PHY456H1F, Quantum Mechanics II. My solutions to problem set 1 (ungraded).

Posted by peeterjoot on September 19, 2011

# Harmonic oscillator.

Consider

Since it’s been a while let’s compute the raising and lowering factorization that was used so extensively for this problem.

It was of the form

Why this factorization has an imaginary in it is a good question. It’s not one that is given any sort of rationale in the text ([1]).

It’s clear that we want and . The difference is then

That commutator is an value, but what was the sign? Let’s compute so we don’t get it wrong

So we have

Factoring out an produces the form of the Hamiltonian that we used before

The factors were labeled the uppering () and lowering () operators respectively, and written

Observe that we can find the inverse relations

**Question**

What is a good reason that we chose this particular factorization? For example, a quick computation shows that we could have also picked

I don’t know that answer. That said, this second factorization is useful in that it provides the commutator relation between the raising and lowering operators, since subtracting 1.11 and 1.6 yields

If we suppose that we have eigenstates for the operator of the form

then the problem of finding the eigensolution of reduces to solving this problem. Because commutes with , an eigenstate of is also an eigenstate of . Utilizing 1.12 we then have

so we see that is an eigenstate of with eigenvalue .

Similarly for the raising operator

and find that is also an eigenstate of with eigenvalue .

Supposing that there is a lowest energy level (because the potential has a lower bound of zero) then the state for which the energy is the lowest when operated on by we have

Thus

and

This seems like a small bit of slight of hand, since it sneakily supplies an integer value to where up to this point was just a label.

If the eigenvalue equation we are trying to solve for the Hamiltonian is

Then we must then have

## Part (a)

We’ve now got enough context to attempt the first part of the question, calculation of

We’ve calculated things like this before, such as

To continue we need an exact relation between and . Recall that was an eigenstate of with eigenvalue . This implies that the eigenstates and are proportional

or

so that

Similarly let

or

so that

We can now return to 1.19, and find

Consider half of this braket

Squaring, utilizing the Hermitian nature of the operator

## Part (b)

Find the ground state energy of the Hamiltonian for .

The new Hamiltonian has the form

where

The energy states of the Hamiltonian are thus

and the ground state of the modified Hamiltonian is thus

## Part (c)

Find the ground state energy of the Hamiltonian .

With a bit of play, this new Hamiltonian can be factored into

where

From 1.29 we see that we have the same sort of commutator relationship as in the original Hamiltonian

and because of this, all the preceding arguments follow unchanged with the exception that the energy eigenstates of this Hamiltonian are shifted by a constant

where the states are simultaneous eigenstates of the operator

The ground state energy is then

This makes sense. A translation of the entire position of the system should not effect the energy level distribution of the system, but we have set our reference potential differently, and have this constant energy adjustment to the entire system.

# Hydrogen atom and spherical harmonics.

We are asked to show that for any eigenkets of the hydrogen atom we have

The summary sheet provides us with the wavefunction

where is a real valued function defined in terms of Lagueere polynomials. Working with the expectation of the operator to start with we have

Recalling that the only dependence in is we can perform the integration directly, which is

We have the same story for the expectation which is

Our integral is then just

also zero. The expectation is a slightly different story. There we have

Within this last integral we can make the substitution

and the integral takes the form

Here we have the product of two even functions, times one odd function (), over a symmetric interval, so the end result is zero, completing the problem.

I wasn’t able to see how to exploit the parity result suggested in the problem, but it wasn’t so bad to show these directly.

# Angular momentum operator.

Working with the appropriate expressions in Cartesian components, confirm that for each component of angular momentum , if is in fact only a function of .

In order to proceed, we will have to consider a matrix element, so that we can operate on in position space. For that matrix element, we can proceed to insert complete states, and reduce the problem to a question of wavefunctions. That is

With we have

We are left with an sum of a symmetric product with the antisymmetric tensor so this is zero for all .

# References

[1] BR Desai. *Quantum mechanics with basic field theory*. Cambridge University Press, 2009.

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