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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 ?

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3 Answers 3

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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.

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  • $\begingroup$ 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, 2021 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... $\endgroup$ Apr 26, 2021 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, 2021 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. $\endgroup$ Apr 26, 2021 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, 2021 at 11:40
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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...

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

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  • $\begingroup$ 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, 2021 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, 2021 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, 2021 at 17:24
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    $\begingroup$ If you keep only the first two levels, you have changed the system drastically. there is no reason why any properties should survive the chop. In particlular $[x,p]=i\hbar$ cannot be represented by a finite dimensional Hilbert space. $\endgroup$
    – mike stone
    Apr 25, 2021 at 18:19
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    $\begingroup$ No. It is brutal surgery. Theusual justification for two level systems is that you only use one frequency and all other energy levels are different and therefore not affcetd. This is not true for a harmonic oscillator, so the approximation is bad one. $\endgroup$
    – mike stone
    Apr 25, 2021 at 19:44

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