3
$\begingroup$

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 ?

$\endgroup$
1
  • $\begingroup$ Hi, I see you've flagged this for migration to Quantum Computing SE. I'm not certain if this is on-topic per quantumcomputing.stackexchange.com/help/on-topic. Regardless, since neither of your tags exist on QCSE, the migration would be automatically rejected. The easiest method would just be to ask this question on QCSE yourself, with appropriate tags and possibly some edits. $\endgroup$ – Chris Apr 25 at 22:43
2
$\begingroup$

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 doe snot run over the same range of states.

What is hidden under the carpet here is how the osicllator 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 oen 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 aply anymore.

$\endgroup$
9
  • $\begingroup$ Thank you very much for your answer. Indeed in superconducting qubit we typically have an anharmonicity $x^4$. Do you think you could edit your question to explain how to regularize in this case (i.e consider that the system is Harmonic oscillator + a term $ \alpha x^4$). I am probably asking too much but I try anyway as it seems you know how to proceed =) $\endgroup$ – StarBucK Apr 26 at 10:03
  • $\begingroup$ The point is that any two-level system can be written as a cubit, using Pauli matrices or raising/lowering operators. Anharmonicity means that the oscillator is not harmonic anymore, so we can forget about the fact that it is an oscillator and consider only the lowest two levels. I guess the oscillator is useful here in the sense that you know the matrix elements $x_{01}$ and $p_{01}$. Not sure what other details you are looking for... $\endgroup$ – Roger Vadim Apr 26 at 10:07
  • $\begingroup$ For me if the question is only taking the two first level, we could it as well do it for the Harmonic oscillator with no problem by replacing $a^{\dagger} a \rightarrow |1\rangle \langle 1|$. If you use the anharmonicity in order to properly design the mapping it must help for something else. Like solving the "paradox" I talked about. I don't know if you see what I mean ? $\endgroup$ – StarBucK Apr 26 at 10:24
  • $\begingroup$ I don't think you need the anharmonicity to design the mapping. But it sems that you were misled by applying formulas for $a,a^\dagger$ that are valid only for the harmonic oscillator. In other words: you don't need to incorporate the anharmonicity in the math, but you need to keep in mind that it is there, that this is not an oscillator, so none of the oscillator formulas that involve more than the two lowest states can be used. $\endgroup$ – Roger Vadim Apr 26 at 11:30
  • $\begingroup$ So let me state the question differently. I have an Harmonic oscillator (and if necessary I add an interaction to make it anharmonic) that is interacting with something else. The interaction might involve any power of the operators $p$. When I want to approximate my (an)harmonic oscillator to its two first level, what do I do with the terms $p^n$ that might be involved in the interaction. Indeed if I replaced $a \to \sigma_-$, all even power in $p$ are proportional to identity: it will "kill" the associated interaction. How is the connection made in general ? $\endgroup$ – StarBucK Apr 26 at 11:40
-2
$\begingroup$

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

$\endgroup$
11
  • $\begingroup$ Hello. I don't understand what you mean. What is the mapping I should use to convert bosonic to fermionic operators. $\endgroup$ – StarBucK Apr 25 at 16:59
  • $\begingroup$ In my sense it doesn't have to be mapped. It's because p and x does not commute, therefore H=/=hbarwa-a+. $\endgroup$ – Kiw Apr 25 at 17:22
  • $\begingroup$ Quantum mechanics kinda starts from the axiom that position and momentum does not commute.(some might say it is a result, but still) $\endgroup$ – Kiw Apr 25 at 17:24
  • $\begingroup$ I agree with you but I don't see the connection with my question. The operators $x$ and $p$ as I defined them with annihilation & creation operator are not commuting so I don't see the problem. $\endgroup$ – StarBucK Apr 25 at 17:26
  • $\begingroup$ I was just looking at the last question. sorry. your contradiction probably starts from there. I don't get the point "cutting", do you mean you cut the states above first two level of HO so it only has two allowed states? $\endgroup$ – Kiw Apr 25 at 17:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.