Procedure to cut an Harmonic oscillator to two first level to obtain a qubit

Question

Let us consider a (quantum) Harmonic oscillator:

$$H=\frac{p^2}{2m}+\frac{1}{2} m \omega^2 x^2$$

Using the annihilation/creation operators defined as:

$$a=\sqrt{\frac{\hbar}{2 m \omega}}(x+\frac{i}{m \omega}p)$$ $$a^{\dagger}=\sqrt{\frac{\hbar}{2 m \omega}}(x-\frac{i}{m \omega}p)$$

I end up with $H=\hbar \omega a^{\dagger} a$

Now, let's assume that for any reason I know that the dynamic should be restricted to the two first levels only. One way that I should we could do is to consider: $a \to \sigma_-, a^{\dagger} \to \sigma_+$

Then, we find:

$$H=\hbar \omega \sigma_+ \sigma_-=\hbar \omega |1\rangle \langle 1|$$

But what confuses me is that if I do the replacement in $p$ and $x$, I don't find the same. Indeed we have:

$$p=i\sqrt{\frac{\hbar m \omega}{2}}(a^{\dagger}-a)$$ $$x=\sqrt{\frac{\hbar}{2 m \omega}}(a^{\dagger}+a)$$

Performing the replacement there, I find:

$$p=\sqrt{\frac{\hbar m \omega}{2}}\sigma_y$$ $$x=\sqrt{\frac{\hbar}{2 m \omega}}\sigma_x$$

And as $\sigma_y^2=\sigma_x^2=I$, I would end up with $H \propto I$ from the first Hamiltonian I wrote.

Another annoying example. Imagine that my oscillator as in addition an interacting term of the form:

$$H_{\text{int}}=p^2*F(t)$$

Where $F(t)$ is a driving field for instance. With the two level approximation and the mapping considered, as $p^2 \propto I$, the interaction would disappear.

What is the procedure to cut to its two first level an Harmonic oscillator to get a qubit. I feel like I don't understand anymore.

Typically to what the momentum and position should be mapped to ?

How to solve the "paradox" I get to ?

Answer 1

The error here is identifying the raising/lowering operators for the two-level system $$ a=|0\rangle\langle 1|,a^\dagger=|1\rangle\langle 0| $$ with the raising lowering operators of the oscillator, $b, b^\dagger$. These are not the same since they do not act in the same Hilbert space. One could use the matrix elements of momentum and position to construct the corresponding operators for the two-level system using standard procedure $$ \hat{A} = \sum_{n,m}|n\rangle A_{nm}\langle m|, $$ but these will not be the same as for the oscillator, since the sum does not run over the same range of states.

What is hidden under the carpet here is how the oscillator can be made into a two-level system physically. This is related to a well-known question of whether one can make a laser from a harmonic oscillator. The answer is no, because from any level you can go up, absorbing a photon, just as well as you can go down emitting it, so the population inversion cannot be achieved.

If one introduces a realistic way of cutting the oscillator to just two levels (i.e., more than just a mathematical trick), e.g., by introducing a non-linear potential term, $V(x)=\alpha x^4$, the things become clearer... but nice representations of momentum and position in terms of creation and annihilation operators do not apply anymore.

Answer 2

The problem is with your truncation arrows. Truncation is not a functor.

Set $m=1, ~\omega=1$ and $\hbar=1$ w.l.o.g., to save yourself confusing elaboration. You may reinsert your nondimensionalized units at will.

In the Fock state basis, the infinite space of oscillator eigenvalues, the truncation operator to the ground and first excited states is just $$ P=P^2=\operatorname{diag}(1,1,0,0,....) \implies\\ PaP=\begin{pmatrix}0&1\\ 0&0 \end{pmatrix}, ~~~Pa^\dagger P = (Pa P)^\dagger,\leadsto \\ \sqrt{2} PxP= \sigma_x, ~~~~\sqrt{2} PpP= \sigma_y, $$ where the part of the infinite matrix truncated out, all zero, is omitted, to leave just the 2x2 block on the upper left of the matrix, as you observed.

Now note $$ Px^2 P = \operatorname{diag}(1/2,3/2,0,0...) \neq {1\!\! 1 \over 2}= (PxP)^2 , $$ and similarly for PpP.

The truncated hamiltonian is not equal to the hamiltonian of the truncated canonical variables, nor should it be...

Answer 3

You've constructed Operators wrong. They must have canonical commutation relation. In other words, (a-)*(a+) =/= H.