# Diagonalizing a polynomial of operators (Quantum Mechanics)

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1. Apr 2, 2016

### Ghazaleh_Zm

The problem asks for the diagonalization of (a(p^2)+b(x^2))^n, where x and p are position and momentum operators with the commutation relation [x,p]=ihbar. a and b are real on-zero numbers and n is a positive non-zero integer.

I know that it is not a good way to use the matrix diagonalization method, so I need the method using Dirac notation.

2. Apr 2, 2016

### blue_leaf77

An operator is diagonal when it is written using the basis of its own eigenstates. Now you have an operator of the form $(ap^2+bx^2)^n$. The eigenstate of this operator is also the eigenstate of $ap^2+bx^2$ since $[(ap^2+bx^2)^n,ap^2+bx^2]=0$. Does $ap^2+bx^2$ remind you of the Hamiltonian of a certain quantum system?

Last edited: Apr 2, 2016
3. Apr 2, 2016

### Ghazaleh_Zm

Isn't it Hamiltonian of the Harmonic oscillator?
Then should I use the energy eigenstates? and transform x and p to ladder operators?

Thanks a lot!

4. Apr 2, 2016

### blue_leaf77

Well, not exactly since the coefficients $a$ and $b$ are taken to be general here. Nevertheless, you can still adopt the same method to find the eigenvalues, e.g. in the harmonic oscillator, the eigenvalues of the Hamiltonian are half-odd multiples of $\hbar \omega$. In this problem, $\hbar \omega$ must be replaced by something else.
If you are familiar with the harmonic oscillator problem, you don't have to start over again from the factorization of the Hamiltonian into the ladder operators, just make some replacement of the constants.

5. Apr 2, 2016

6. Apr 2, 2016

### blue_leaf77

What do you mean by operator algebra? As you may already know, the eigenstates of the harmonic oscillator-like Hamiltonian form an infinite set, therefore I don't recommend writing that operator in matrix form. An equivalent form of a diagonal representation is the so-called spectral representation. The spectral representation of an operator $O$ is obtained by first computing
$$\sum_i |u_i\rangle \langle u_i| O \sum_j |u_j\rangle \langle u_j|$$
where $\sum_i |u_i\rangle$ is the eigenstate of $O$.

Sanwiching in position basis braket like $\langle x|(ap^2+bx^2)^n|x \rangle$? It obviously won't be diagonal.

7. Apr 2, 2016

### Ghazaleh_Zm

I did the following procedure. Just please tell me its correct or not. Thank you!

because the operator $(Ap^2+Bx^2)^n$ is in the form of general harmonic oscillator Hamiltonian, the basis set in which it would be diagonal is energy eigenstates $|n\rangle \langle n|$ . If we equate $(Ap^2+Bx^2) = (\frac{p^2}{2m}+\frac{mω^2 x^2}{2})=H$ and use:
$$x^2 = \frac{\hbar}{2mω} (a†+a)^2 = A \frac{\hbar}{ω} (a†+a)^2$$
and
$$p^2 = \frac{-m\hbar ω^2}{2} (a†-a)^2 = -B \frac{\hbar}{ω} (a†-a)^2$$
then:
$$Ap^2+Bx^2=-AB \frac{\hbar}{ω} (a†-a)^2 + AB\frac{\hbar}{ω} (a†+a)^2 = 2AB\frac{\hbar}{ω} (a†a+aa†) = 2AB\frac{\hbar}{ω}$$
which is the eigenvalue of the general Hamiltonian operator. Therefor the diagonalized general Hamiltonian operator will be:
$$2AB\frac{\hbar}{ω} |n\rangle \langle n|$$
therefore generalized Hamiltonian operator to the power of n can be diagonalized as:
$$(2AB\frac{\hbar}{ω})^n |n\rangle \langle n|$$
and to find the eigenstates, this eigenvalues should be pluged in to the Hermite polynomial $det \left| (Ap^2+Bx^2)^n - (2AB\frac{\hbar}{ω})^n \hat{1} \right| = 0$
because the question did not ask for the eigenstate, just asked to explain how to obtain the eigenstates!

8. Apr 2, 2016

### blue_leaf77

$a^\dagger a + a a^\dagger \neq 1$. Moreover, you should not introduce $\omega$ in the last expression of your calculation as this quantity was not given at the start. Consider the fact that you can express $m$ in term of $A$ alone. From this and $B = m\omega^2/2$, you can write $\omega$ in terms of $A$ and $B$. You can eventually bypass all those steps by simply adapting the eigenvalue of a harmonic oscillator, $E_n = \hbar \omega (n+1/2)$, to the present problem where $A$ and $B$ replaces the constants.