Encoding a qubit into a cavity mode in circuit QED using phase estimation
Presented by Zhi Han
B. M. Terhal and D. Weigand, Encoding a Qubit into a Cavity Mode in Circuit QED Using Phase Estimation, Phys. Rev. A 93, 012315 (2016).
Overview
- GKP state
- Phase Estimation
- Implimentation
ℏ=1
\hbar = 1
Displacement operator
The displacement operator is a common operation in optics. Since momentum is the generator of translations,
D^q(a)∣x⟩:=e−iap^∣x⟩=∣x+a⟩
\hat D_q(a) |x\rangle:= e^{-i a \hat p}|x\rangle = |x + a\rangle
x
ψ(x)
e−iap^ψ(x)
GKP state: 0 and 1
∣0⟩∝∑n=−∞∞∣2nπ⟩q
|0\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_q
The ∣0⟩ and ∣1⟩ GKP states are defined to be
∣1⟩∝∑n=−∞∞∣(2n+1)π⟩q
|1\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_q
∣0⟩ and ∣1⟩ GKP states.
q
∣2n⟩
∣2n+1⟩
0
π
2π
3π
4π
5π
6π
X=e−iπp^
X = e^{-i \sqrt \pi \hat p}
GKP state: technical definition
GKP state is defined to be the +1 eigenspace of Sq,Sp where
S^p=e−i2πp^S^q=ei2πq^
\hat S_p = e^{-i 2\sqrt{\pi} \hat p}\\
\hat S_q = e^{i 2\sqrt{\pi}\hat q}
Set the eigenvalue to be Sq,Sp=+1. By definition, this implies ∣ψ⟩ is 2π periodic.
e−iap^∣q⟩q=∣q+a⟩q
e^{-i a \hat p}|q\rangle_q = |q + a \rangle_q
Sp∣ψ⟩q=e−i2πp^∣ψ⟩q=∣ψ⟩q
S_p|\psi\rangle_q = e^{-i 2\sqrt{\pi} \hat p}|\psi\rangle_q=|\psi\rangle_q
GKP state: + and -
∣0⟩∝∑n=−∞∞∣2nπ⟩q
|0\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_q
∣1⟩∝∑n=−∞∞∣(2n+1)π⟩q
|1\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_q
∣+⟩=21(∣0⟩+∣1⟩)
|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)
∣−⟩=21(∣0⟩−∣1⟩)
|-\rangle=\frac{1}{\sqrt{2}}(|0\rangle-|1\rangle)
∣+⟩∝∑n=−∞∞∣nπ⟩q
|+\rangle \propto \sum_{n=-\infty}^{\infty} |n\sqrt{\pi}\rangle_q
∣−⟩∝∑n=−∞∞(−1)n∣nπ⟩q
|-\rangle \propto \sum_{n=-\infty}^{\infty} (-1)^n|n\sqrt{\pi}\rangle_q
(−1)n∣n⟩
∣n⟩
∣+⟩ and ∣−⟩ GKP states.
q
0
π
2π
3π
4π
5π
6π
Z=eiπq^
Z = e^{i \sqrt \pi \hat q}
∣0⟩ and ∣1⟩ GKP states.
q
∣2n⟩
∣2n+1⟩
0
π
2π
3π
4π
5π
6π
(−1)n∣n⟩
∣n⟩
∣+⟩ and ∣−⟩ GKP states.
q
0
π
2π
3π
4π
5π
6π
∣0⟩∝∑n=−∞∞∣2nπ⟩q
|0\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_q
∣1⟩∝∑n=−∞∞∣(2n+1)π⟩q
|1\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_q
∣+⟩∝∑n=−∞∞∣nπ⟩q
|+\rangle \propto \sum_{n=-\infty}^{\infty} |n\sqrt{\pi}\rangle_q
∣−⟩∝∑n=−∞∞(−1)n∣nπ⟩q
|-\rangle \propto \sum_{n=-\infty}^{\infty} (-1)^n|n\sqrt{\pi}\rangle_q
∣+⟩∝∑n=−∞∞∣2nπ⟩p
|+\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_p
∣−⟩∝∑n=−∞∞∣(2n+1)π⟩p
|-\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_p
∣0⟩∝∑n=−∞∞∣nπ⟩p
|0\rangle \propto \sum_{n=-\infty}^{\infty} |n\sqrt{\pi}\rangle_p
∣1⟩∝∑n=−∞∞(−1)n∣nπ⟩p
|1\rangle \propto \sum_{n=-\infty}^{\infty} (-1)^n|n\sqrt{\pi}\rangle_p
q basis
p basis
In a similar fashion, we can get the GKP states in the momentum basis instead of the position basis by taking the fourier transform.
∣0⟩ and ∣1⟩ GKP states.
∣+⟩ and ∣−⟩ GKP states.
q
0
π
2π
3π
4π
5π
6π
(−1)n∣n⟩
∣n⟩
p
∣2n⟩
∣2n+1⟩
0
π
2π
3π
4π
5π
6π
(−1)n∣n⟩
∣n⟩
p
0
π
2π
3π
4π
5π
6π
q
0
π
2π
3π
4π
5π
6π
∣2n⟩
∣2n+1⟩
∣0⟩∝∑n=−∞∞∣2nπ⟩q
|0\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_q
∣1⟩∝∑n=−∞∞∣(2n+1)π⟩q
|1\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_q
X=e−iπp^
X = e^{-i \sqrt \pi \hat p}
Z=eiπq^
Z = e^{i \sqrt \pi \hat q}
∣+⟩∝∑n=−∞∞∣2nπ⟩p
|+\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_p
∣−⟩∝∑n=−∞∞∣(2n+1)π⟩p
|-\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_p
X2=Sp=e−i2πp^
X^2 = S_p = e^{-i 2 \sqrt {\pi} \hat p}
Z2=Sq=ei2πq^
Z^2 = S_q = e^{i 2 \sqrt {\pi} \hat q}
X and Z gate
Sp=e−i2πp^
S_p = e^{-i 2 \sqrt {\pi} \hat p}
Sq=ei2πq^
S_q = e^{i 2 \sqrt {\pi} \hat q}
Bonus: why 2π?
eAeB=eBeAe[A,B]
e^A e^B = e^B e^A e^{[A, B]}
SqSp=ei2πq^e−i2πp^=e−i2πp^ei2πq^=1ei4π=SpSq
S_q S_p = e^{i 2 \sqrt {\pi} \hat q} e^{-i 2 \sqrt {\pi} \hat p} = e^{-i 2 \sqrt {\pi} \hat p} e^{i 2 \sqrt {\pi} \hat q} \underbrace{e^{i 4 \pi}}_{=1} = S_p S_q
QEC with GKP states
Imagine a shift error has a occured where the state has been displaced by e−iμqp^.
How can we detect and correct this error?
Sqe−iμqp^∣ψ⟩=ei2πq^e−iμqp^∣ψ⟩
S_q e^{-i \mu_q \hat p } |\psi\rangle =e^{i 2\sqrt{\pi} \hat q} e^{-i \mu_q \hat p } |\psi\rangle
=e−iμqp^ei2πq^ei2πμq∣ψ⟩
= e^{-i \mu_q \hat p } e^{i 2\sqrt{\pi} \hat q} e^{i 2 \sqrt \pi \mu_q}|\psi\rangle
=ei2πμqe−ivp^∣ψ⟩
=e^{i 2 \sqrt \pi \mu_q} e^{-i v \hat p } |\psi\rangle
Sq∣ψ′⟩=ei2πμq∣ψ′⟩
S_q |\psi'\rangle = e^{i 2 \sqrt \pi \mu_q} |\psi'\rangle
q
∣2n+μq⟩
0
π
2π
3π
4π
5π
6π
μq
∣2n⟩
Sp=e−i2πp^
S_p = e^{-i 2 \sqrt {\pi} \hat p}
Sq=ei2πq^
S_q = e^{i 2 \sqrt {\pi} \hat q}
∣0⟩∝∑n=−∞∞∣2nπ⟩q
|0\rangle \propto \sum_{n=-\infty}^{\infty} |2n\sqrt{\pi}\rangle_q
∣1⟩∝∑n=−∞∞∣(2n+1)π⟩q
|1\rangle \propto \sum_{n=-\infty}^{\infty} |(2n+1)\sqrt{\pi}\rangle_q
X^=e−iπp^
\hat X = e^{-i \sqrt \pi \hat p}
Z^=eiπq^
\hat Z = e^{i \sqrt \pi \hat q}
Gates
Stabilizers
States
Finitely squeezed GKP states
Problem: The infinitely squeezed GKP state is not normalizable.
Idea: Replace each Dirac delta with a squeezed Gaussian state. Normalize the entire state with a Gaussian envelope.
(Terhal and Weigand 2016)
∣vacsq⟩=∫(πΔ2)1/4dqe−q2/(2Δ2)∣q⟩
\left|\operatorname{vac}_{\mathrm{sq}}\right\rangle=\int \frac{d q}{\left(\pi \Delta^{2}\right)^{1 / 4}} e^{-q^{2} /\left(2 \Delta^{2}\right)}|q\rangle
∣0⟩approx∝∑t=−∞∞e−2πΔ~2t2D(t2π)∣vacsq⟩
|0\rangle_{\mathrm{approx}} \propto \sum_{t=-\infty}^{\infty} e^{-2 \pi \tilde{\Delta}^{2} t^{2}} D(t \sqrt{2 \pi})\left|\mathrm{vac}_{\mathrm{sq}}\right\rangle
=∑t=−∞∞∫e−2πΔ~2t2e−(q−2tπ)2/(2Δ2)∣q⟩dq,
=\sum_{t=-\infty}^{\infty} \int e^{-2 \pi \tilde{\Delta}^{2} t^{2}} e^{-(q-2 t \sqrt{\pi})^{2} /\left(2 \Delta^{2}\right)}|q\rangle d q,
Δ= stdev/squeezing of mini peak
Δ~= stdev/squeezing of entire state
In the ∣+⟩ GKP state, roles of Δ~,Δ are interchanged.
Rest of the talk: Δ=Δ~
Infinite squeezing: Δ→0
Phase Estimation
U^∣ψ⟩=eiθ∣ψ⟩
\hat U |\psi\rangle = e^{i \theta} |\psi \rangle
Measuring the complex eigenvalue eiθ of a unitary operator U is called phase estimation.
Phase Estimation, standard
control
target
Image: Quskit
Phase Estimation
Input:
First register (control): n qubits to store the value 2nθ
Second register (target): ∣ψ⟩, an eigenvector of U
- For this section we will take U∣ψ⟩=e2πiθ∣ψ⟩.
Step 1: Hadamard the first register. ∣ψ1⟩=22n1(∣0⟩+∣1⟩)⊗n∣ψ⟩
Step 1: Hadamard the first register. ∣ψ1⟩=22n1(∣0⟩+∣1⟩)⊗n∣ψ⟩
Phase Estimation
Step 2: Apply controlled-U gates.
- The first qubit applies U once.
- The second qubit applies U twice.
- The third qubit applies U four times.
U2j∣ψ⟩=U2j−1U∣ψ⟩=U2j−1e2πiθ∣ψ⟩=⋯=e2πi2jθ∣ψ⟩
U^{2^{j}}|\psi\rangle=U^{2^{j}-1} U|\psi\rangle=U^{2^{j}-1} e^{2 \pi i \theta}|\psi\rangle=\cdots=e^{2 \pi i 2^{j} \theta}|\psi\rangle
Since U∣ψ⟩=e2πiθ∣ψ⟩, the action of 2j gates corresponding to the jth qubit is:
U2j∣ψ⟩=U2j−1U∣ψ⟩=U2j−1e2πiθ∣ψ⟩=⋯=e2πi2jθ∣ψ⟩
U^{2^{j}}|\psi\rangle=U^{2^{j}-1} U|\psi\rangle=U^{2^{j}-1} e^{2 \pi i \theta}|\psi\rangle=\cdots=e^{2 \pi i 2^{j} \theta}|\psi\rangle
Phase Estimation, standard
Step 2: Apply controlled unitary gates. Using the identity
U2j∣ψ⟩=U2j−1U∣ψ⟩=U2j−1e2πiθ∣ψ⟩=⋯=e2πi2jθ∣ψ⟩
U^{2^{j}}|\psi\rangle=U^{2^{j}-1} U|\psi\rangle=U^{2^{j}-1} e^{2 \pi i \theta}|\psi\rangle=\cdots=e^{2 \pi i 2^{j} \theta}|\psi\rangle
∣0⟩⊗∣ψ⟩+∣1⟩⊗e2πiθ∣ψ⟩=(∣0⟩+e2πiθ∣1⟩)⊗∣ψ⟩
|0\rangle \otimes|\psi\rangle+|1\rangle \otimes e^{2 \pi i \theta}|\psi\rangle=\left(|0\rangle+e^{2 \pi i \theta}|1\rangle\right) \otimes|\psi\rangle
∣ψ2⟩=22n1(∣0⟩+e2πiθ2n−1∣1⟩)⊗⋯⊗(∣0⟩+e2πiθ21∣1⟩)⊗(∣0⟩+e2πiθ20∣1⟩)⊗∣ψ⟩=22n1k=0∑2n−1e2πiθk∣k⟩⊗∣ψ⟩
\begin{aligned}
\left|\psi_{2}\right\rangle &=\frac{1}{2^{\frac{n}{2}}}\left(|0\rangle+e^{2 \pi i \theta 2^{n-1}}|1\rangle\right) \otimes \cdots \otimes\left(|0\rangle+e^{2 \pi i \theta 2^{1}}|1\rangle\right) \otimes\left(|0\rangle+e^{2 \pi i \theta 2^{0}}|1\rangle\right) \otimes|\psi\rangle \\
&=\frac{1}{2^{\frac{n}{2}}} \sum_{k=0}^{2^{n}-1} e^{2 \pi i \theta k}|k\rangle \otimes|\psi\rangle
\end{aligned}
k= the integer that represents a 2^n bit string. e.g. 3 = 11
Step 3: Apply inverse quantum fourier transform.
∣ψ2⟩=22n1(∣0⟩+e2πiθ2n−1∣1⟩)⊗⋯⊗(∣0⟩+e2πiθ21∣1⟩)⊗(∣0⟩+e2πiθ20∣1⟩)⊗∣ψ⟩=22n1k=0∑2n−1e2πiθk∣k⟩⊗∣ψ⟩
\begin{aligned}
\left|\psi_{2}\right\rangle &=\frac{1}{2^{\frac{n}{2}}}\left(|0\rangle+e^{2 \pi i \theta 2^{n-1}}|1\rangle\right) \otimes \cdots \otimes\left(|0\rangle+e^{2 \pi i \theta 2^{1}}|1\rangle\right) \otimes\left(|0\rangle+e^{2 \pi i \theta 2^{0}}|1\rangle\right) \otimes|\psi\rangle \\
&=\frac{1}{2^{\frac{n}{2}}} \sum_{k=0}^{2^{n}-1} e^{2 \pi i \theta k}|k\rangle \otimes|\psi\rangle
\end{aligned}
QFT∣x⟩=22n1(∣0⟩+e22πix∣1⟩)⊗(∣0⟩+e222πix∣1⟩)⊗…⊗(∣0⟩+e2n−12πix∣1⟩)⊗(∣0⟩+e2n2πix∣1⟩)
Q F T|x\rangle=\frac{1}{2^{\frac{n}{2}}}\left(|0\rangle+e^{\frac{2 \pi i}{2} x}|1\rangle\right) \otimes\left(|0\rangle+e^{\frac{2 \pi i}{2^{2}} x}|1\rangle\right) \otimes \ldots \otimes\left(|0\rangle+e^{\frac{2 \pi i}{2^{n-1}} x}|1\rangle\right) \otimes\left(|0\rangle+e^{\frac{2 \pi i}{2^{n}} x}|1\rangle\right)
Step 4: Measure. Obtain an integer 2nθ.
∣ψ4⟩=∣2nθ⟩⊗∣ψ⟩
\left|\psi_{4}\right\rangle=\left|2^{n} \theta\right\rangle \otimes|\psi\rangle
For this problem, x=2nθ.
Standard phase estimation with a hybrid setup.
Limitations:
- Requires high number of photons
- Difficulty with controlled U2n
control
target
(Terhal and Weigand 2016)
Main idea of the paper
U^∣φ⟩→eiθ∣ψ⟩
\hat U |\varphi\rangle \to e^{i \theta} |\psi \rangle
If we perform phase estimation on a unitary operator U^ with an arbitrary state ∣φ⟩. Does the state go to the eigenvector ∣ψ⟩? If so, can we use phase estimation to generate exotic states such as the GKP state?
H^∣φ⟩→E∣ψ⟩
\hat H |\varphi\rangle \to E |\psi \rangle
When I measure a arbitrary state ∣φ⟩ with an hermitian operator H^, the state goes to the eigenvector ∣ψ⟩.
For the rest of this talk, I will refer to U=Sp=e−i2πp^
Phase estimation with repetition
control
target
(Terhal and Weigand 2016)
I^⊗∣0⟩⟨0∣+U^⊗∣1⟩⟨1∣
\hat I \otimes |0\rangle\langle 0| + \hat U \otimes |1\rangle \langle 1|
diag(1,eiφ)=[100eiφ]
\text{diag}(1, e^{i\varphi}) = \begin{bmatrix}
1 & 0 \\
0 & e^{i\varphi}
\end{bmatrix}
Controlled-U:
Diag:
∣ψ⟩⊗∣0⟩+eiθ∣ψ⟩⊗∣1⟩=∣ψ⟩⊗(∣0⟩+eiθ∣1⟩)
|\psi\rangle \otimes |0\rangle+ e^{i \theta}|\psi\rangle\otimes |1\rangle = |\psi\rangle \otimes \left(|0\rangle+e^{i \theta}|1\rangle\right)
control
target
⟨+∣(∣0⟩+eiθ∣1⟩)=21(1+cos(θ))=Pr(+∣θ)
\langle +| \left(|0\rangle+e^{i \theta}|1\rangle\right) = \frac{1}{2}(1 + \cos(\theta)) = \text{Pr}(+|\theta)
Consider φ=0. On the control qubit, measure in ∣+⟩ basis.
⟨+∣(∣0⟩+ei(θ+φ)∣1⟩)=21(1−sin(θ))=Pr(+∣θ)
\langle +| \left(|0\rangle+e^{i (\theta + \varphi)}|1\rangle\right) = \frac{1}{2}(1 - \sin(\theta)) = \text{Pr}(+|\theta)
Consider φ=π/2. On the control qubit, measure in ∣+⟩ basis.
21(1−sin(θ))=Pr(+∣θ)φ=π/2
\frac{1}{2}(1 - \sin(\theta)) = \text{Pr}(+|\theta)_{\varphi=\pi/2}
21(1+cos(θ))=Pr(+∣θ)φ=0
\frac{1}{2}(1 + \cos(\theta)) = \text{Pr}(+|\theta)_{\varphi = 0}
Can be obtained experimentally
Enough to resolve ambiguity in theta
Adaptive Phase estimation
control
target
21(1+cos(θ+φ))=Prφ(+∣θ)
\frac{1}{2}(1 + \cos(\theta + \varphi)) = \text{Pr}_{\varphi}(+|\theta)
Idea: Optimize φ to gain the most information possible about θ by considering the derivative dθPrφ(+∣θ).
\frac{}{}
(Terhal and Weigand 2016)
∣α⟩→∣α−π/2⟩+(−1)xeiφ∣α+π/2⟩
|\alpha\rangle \rightarrow|\alpha-\sqrt{\pi / 2}\rangle+(-1)^{x} e^{i \varphi}|\alpha+\sqrt{\pi / 2}\rangle
Consider a coherent state, which is not a eigenstate of
U=Sp. After one round of phase estimation (with repetition), x=0,1
This will create a sequence of coherent states on a line. The filter is Binomial, and approximately Gaussian.
(Terhal and Weigand 2016)
(Terhal and Weigand 2016)
M
(Terhal and Weigand 2016)
(Terhal and Weigand 2016)
Why does this work?
Displaced GKP states
GKP0,μq,μp⟩GKP1,μq,μp⟩:=Dq(μq)Dp(μp)∣GKP0⟩:=Dq(μq)Dp(μp)∣GKP1⟩
\begin{aligned}
\left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle&:= D_{q}\left(\mu_{q}\right) D_{p}\left(\mu_{p}\right)\left|G K P_{0}\right\rangle \\
\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle&:=D_{q}\left(\mu_{q}\right) D_{p}\left(\mu_{p}\right)\left|G K P_{1}\right\rangle
\end{aligned}
Def. A displaced GKP state is a GKP state with two displacement operators Dq(μq),Dp(μp) and μq,μp∈[0,π).
Displaced GKP states
S^qGKPμq,μp⟩S^pGKPμq,μp⟩=ei2πμqGKPμq,μp⟩=e−i2πμpGKPμq,μp⟩
\begin{aligned}
\hat{S}_{q}\left|G K P_{\mu_{q}, \mu_{p}}\right\rangle &=e^{i 2 \sqrt{\pi} \mu_{q}}\left|G K P_{\mu_{q}, \mu_{p}}\right\rangle \\
\hat{S}_{p}\left|G K P_{\mu_{q}, \mu_{p}}\right\rangle &=e^{-i 2 \sqrt{\pi} \mu_{p}}\left|G K P_{\mu_{q}, \mu_{p}}\right\rangle
\end{aligned}
They are the eigenstates of the Sq,Sp operators.
q
∣2n+μq⟩
0
π
2π
3π
4π
5π
6π
μq
∣2n⟩
Sqe−iμqp^∣ψ⟩=ei2πq^e−iμqp^∣ψ⟩
S_q e^{-i \mu_q \hat p } |\psi\rangle =e^{i 2\sqrt{\pi} \hat q} e^{-i \mu_q \hat p } |\psi\rangle
=e−iμqp^ei2πq^ei2πμq∣ψ⟩
= e^{-i \mu_q \hat p } e^{i 2\sqrt{\pi} \hat q} e^{i 2 \sqrt \pi \mu_q}|\psi\rangle
=ei2πμqe−iμqp^∣ψ⟩
=e^{i 2 \sqrt \pi \mu_q} e^{-i \mu_q \hat p } |\psi\rangle
Sq∣ψ′⟩=ei2πμq∣ψ′⟩
S_q |\psi'\rangle = e^{i 2 \sqrt \pi \mu_q} |\psi'\rangle
If ∣ψ⟩ is a displaced GKP state, we can do error correction:
Could something like this hold for general ∣ψ⟩?
U^∣φ⟩→eiθ∣ψ⟩
\hat U |\varphi\rangle \to e^{i \theta} |\psi \rangle
Every CV state can be expressed as a superposition of displaced GKP states. For m∈Z,μq∈[0,π),
∣ψcv⟩q=π1∫0π∫0πdμqdμpa0(μq,μp)GKP0,μq,μp⟩+π1∫0π∫0πdμqdμpa1(μq,μp)GKP1,μq,μp⟩
\begin{aligned}
|\psi_{cv}\rangle_{q} &=\frac{1}{\sqrt \pi} \int_{0}^{\sqrt \pi} \int_{0}^{\sqrt \pi} d \mu_{q} d \mu_{p} \, a_0(\mu_q, \mu_p) \left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle \\
&+\frac{1}{\sqrt \pi} \int_{0}^{\sqrt \pi} \int_{0}^{\sqrt \pi} d \mu_{q} d \mu_{p} \, a_1(\mu_q, \mu_p)\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle
\end{aligned}
If we collapse the superposition μq,μp through phase estimation,
∣ψL⟩=a0(μq,μp)GKP0,μq,μp⟩+a1(μq,μp)GKP1,μq,μp⟩
\left|\psi_{L}\right\rangle=a_{0}(\mu_q, \mu_p)\left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle+a_{1}(\mu_q, \mu_p)\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle
a0a1:=m even ∑[ψ(μq+mαq)e−imαqμp]:=m odd ∑[ψ(μq+mαq)e−imαqμp]
\begin{aligned}
a_{0} &:=\sum_{m \text { even }}\left[\psi\left(\mu_{q}+m \alpha_{q}\right) e^{-i m \alpha_{q} \mu_{p}}\right] \\
a_{1} &:=\sum_{m \text { odd }}\left[\psi\left(\mu_{q}+m \alpha_{q}\right) e^{-i m \alpha_{q} \mu_{p}}\right]
\end{aligned}
∣ψcv⟩q=π1∫0π∫0πdμqdμpa0(μq,μp)GKP0,μq,μp⟩+π1∫0π∫0πdμqdμpa1(μq,μp)GKP1,μq,μp⟩
\begin{aligned}
|\psi_{cv}\rangle_{q} &=\frac{1}{\sqrt \pi} \int_{0}^{\sqrt \pi} \int_{0}^{\sqrt \pi} d \mu_{q} d \mu_{p} \, a_0(\mu_q, \mu_p) \left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle \\
&+\frac{1}{\sqrt \pi} \int_{0}^{\sqrt \pi} \int_{0}^{\sqrt \pi} d \mu_{q} d \mu_{p} \, a_1(\mu_q, \mu_p)\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle
\end{aligned}
This explains why the protocol in (Terhal and Weigand 2016) works so well:
- Displaced GKP states are the eigenstates of Sq,Sp operators
- Every CV state is a superposition of displaced GKP states.
- Phase estimation = collapsing the superposition.
- Example with two mode squeezing.
We can express a two mode squeezed state into the GKP basis.
- Step 1. Write the wavefunction. ψ(x)=ψ(mπ+μq)
- Step 2. Perform phase estimation to measure all μq,μp.
- Step 3. Obtain the remaining coefficients in the computational basis.
- Step 4. Convert the coefficients to the Bell basis.
ψ(q1,q2)=π1exp(−41(e2ζ(q1+q2)2+e−2ζ(q1−q2)2))
\psi\left(q_{1}, q_{2}\right)=\frac{1}{\sqrt{\pi}} \exp \left(-\frac{1}{4}\left(e^{2 \zeta}\left(q_{1}+q_{2}\right)^{2}+e^{-2 \zeta}\left(q_{1}-q_{2}\right)^{2}\right)\right)
∣ψcv⟩q=π1∫0π∫0πdμqdμpa0(μq,μp)GKP0,μq,μp⟩+π1∫0π∫0πdμqdμpa1(μq,μp)GKP1,μq,μp⟩
\begin{aligned}
|\psi_{cv}\rangle_{q} &=\frac{1}{\sqrt \pi} \int_{0}^{\sqrt \pi} \int_{0}^{\sqrt \pi} d \mu_{q} d \mu_{p} \, a_0(\mu_q, \mu_p) \left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle \\
&+\frac{1}{\sqrt \pi} \int_{0}^{\sqrt \pi} \int_{0}^{\sqrt \pi} d \mu_{q} d \mu_{p} \, a_1(\mu_q, \mu_p)\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle
\end{aligned}
∣ψL⟩=a00GKP0,μq,μp⟩GKP0,μq,μp⟩+a01GKP0,μq,μp⟩GKP1,μq,μp⟩+a10GKP1,μq,μp⟩GKP0,μq,μp⟩+a11GKP1,μq,μp⟩GKP1,μq,μp⟩
\begin{aligned}
\left|\psi_{L}\right\rangle &=a_{00}\left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle\left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle+a_{01}\left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle \\
&+a_{10}\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle\left|G K P_{0, \mu_{q}, \mu_{p}}\right\rangle+a_{11}\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle\left|G K P_{1, \mu_{q}, \mu_{p}}\right\rangle
\end{aligned}
a00=m1 even ∑m2 even ∑[ψ(μq1+m1π)e−im1πμp1ψ(μq2+m2π)e−im2πμp2]a01=m1 even ∑m2 odd ∑[ψ(μq1+m1π)e−im1πμp1ψ(μq2+m2π)e−im2πμp2]a10=m1 odd ∑m2 even ∑[ψ(μq1+m1π)e−im1πμp1ψ(μq2+m2π)e−im2πμp2]a11=m1 odd ∑m2 odd ∑[ψ(μq1+m1π)e−im1πμp1ψ(μq2+m2π)e−im2πμp2]
\begin{aligned}
&a_{00}=\sum_{m_{1} \text { even }} \sum_{m_{2} \text { even }}\left[\psi\left(\mu_{q_{1}}+m_{1} \sqrt \pi \right) e^{-i m_{1} \sqrt \pi \mu_{p_{1}}} \psi\left(\mu_{q_{2}}+m_{2} \sqrt \pi \right) e^{-i m_{2} \sqrt \pi \mu_{p_{2}}}\right] \\
&a_{01}=\sum_{m_{1} \text { even }} \sum_{m_{2} \text { odd }}\left[\psi\left(\mu_{q_{1}}+m_{1} \sqrt \pi \right) e^{-i m_{1} \sqrt \pi \mu_{p_{1}}} \psi\left(\mu_{q_{2}}+m_{2} \sqrt \pi \right) e^{-i m_{2} \sqrt \pi \mu_{p_{2}}}\right] \\
&a_{10}=\sum_{m_{1} \text { odd }} \sum_{m_{2} \text { even }}\left[\psi\left(\mu_{q_{1}}+m_{1} \sqrt \pi \right) e^{-i m_{1} \sqrt \pi \mu_{p_{1}}} \psi\left(\mu_{q_{2}}+m_{2} \sqrt \pi \right) e^{-i m_{2} \sqrt \pi \mu_{p_{2}}}\right] \\
&a_{11}=\sum_{m_{1} \text { odd }} \sum_{m_{2} \text { odd }}\left[\psi\left(\mu_{q_{1}}+m_{1} \sqrt \pi \right) e^{-i m_{1} \sqrt \pi \mu_{p_{1}}} \psi\left(\mu_{q_{2}}+m_{2} \sqrt \pi \right) e^{-i m_{2} \sqrt \pi \mu_{p_{2}}}\right]
\end{aligned}
phase diagram... we find, for specific values of μq,μp the result is a Bell state.
Questions?
Encoding a qubit into a cavity mode in circuit QED using phase estimation Presented by Zhi Han B. M. Terhal and D. Weigand, Encoding a Qubit into a Cavity Mode in Circuit QED Using Phase Estimation , Phys. Rev. A 93 , 012315 (2016).
gkp
By Zhi Han
gkp
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