Mimicking the Hadamard discrete-time quantum walk with a time-independent Hamiltonian

Abstract

The discrete-time quantum walk dynamics can be generated by a time-dependent Hamiltonian, repeatedly switching between the coin and the shift generators. We change the model and consider the case where the Hamiltonian is time-independent, including both the coin and the shift terms in all times. The eigenvalues and the related Bloch vectors for the time-independent Hamiltonian are then compared with the corresponding quantities for the effective Hamiltonian generating the quantum walk dynamics. Restricted to the non-localized initial quantum walk states, we optimize the parameters in the time-independent Hamiltonian such that it generates a dynamics similar to the Hadamard quantum walk. We find that the dynamics of the walker probability distribution and the corresponding standard deviation, the coin-walker entanglement, and the quantum-to-classical transition of the discrete-time quantum walk model can be approximately generated by the optimized time-independent Hamiltonian. We, further, show both dynamics are equivalent in the classical regime, as expected.

Appendices

Shift operator in the Fourier basis

The action of the Shift operator (2) on the Fourier basis is calculated as

$$\begin{aligned} S\;|\tilde{k}\rangle \otimes |s\rangle&= S \frac{1}{\sqrt{d}}\sum _n e^{- i \tilde{k} n } |n\rangle \otimes |s\rangle \nonumber \\&= \frac{1}{\sqrt{d}}\sum _n e^{- i \tilde{k} n } S|n\rangle \otimes |s\rangle \nonumber \\&= \frac{1}{\sqrt{d}}\sum _n e^{- i \tilde{k} n } |n+(-1)^s\rangle \otimes |s\rangle \nonumber \\&= \frac{1}{\sqrt{d}}\sum _{n'} e^{- i \tilde{k} [n' - (-1)^s] } |n'\rangle \otimes |s\rangle \end{aligned}$$

(A1)

$$\begin{aligned}&= e^{i \tilde{k} (-1)^s } \frac{1}{\sqrt{d}}\sum _{n'} e^{- i \tilde{k} n' } |n'\rangle \otimes |s\rangle \nonumber \\&= e^{i \tilde{k} (-1)^s } |\tilde{k}\rangle \otimes |s\rangle , \end{aligned}$$

(A2)

where in (A1) n is replaced by $n' = n + (-1)^s$, and Eq. (3) in Sect. 2 is then proved. The shift operator S is diagonal in the Fourier basis, hence,

$$\begin{aligned} S&= \sum _k \biggl (|\tilde{k}\rangle \langle \tilde{k}| \otimes |0\rangle \langle 0| e^{i\tilde{k}} + |\tilde{k}\rangle \langle \tilde{k}| \otimes |1\rangle \langle 1| e^{-i\tilde{k}} \biggr ) \nonumber \\&= \sum _k |\tilde{k}\rangle \langle \tilde{k}| \otimes \biggl ( e^{i\tilde{k}} |0\rangle \langle 0| + e^{-i\tilde{k}} |1\rangle \langle 1| \biggr ) \nonumber \\&= \sum _k |\tilde{k}\rangle \langle \tilde{k}| \otimes \begin{pmatrix} e^{i\tilde{k}} &{}\quad 0 \\ 0 &{}\quad e^{-i\tilde{k}} \end{pmatrix} \nonumber \\&= \sum _k |\tilde{k}\rangle \langle \tilde{k}| \otimes e^{i\tilde{k}\sigma _z}, \end{aligned}$$

(A3)

proving Eq. (4) in Sect. 2.

Bloch vectors

The Bloch vectors in Eq. (7) are obtained by substituting Eq. (1) together with

$$\begin{aligned} e^{i\tilde{k}\sigma _z} = \begin{pmatrix} e^{i\tilde{k}} &{}\quad 0 \\ 0 &{}\quad e^{-i\tilde{k}} \end{pmatrix}, \end{aligned}$$

(B1)

and

$$\begin{aligned} e^{-i \epsilon _\theta (\tilde{k}) \varvec{d}_\theta (\tilde{k}).\varvec{\sigma }} = \cos \epsilon _\theta (\tilde{k}) \; \mathbb {1}_2 - i \sin \epsilon _\theta (\tilde{k}) \begin{pmatrix} d_z &{}\quad d_x-id_y \\ d_x+id_y &{}\quad -d_z \\ \end{pmatrix}, \end{aligned}$$

(B2)

in Eq. (6) where $\varvec{d}_\theta (\tilde{k})$ is supposed to be a unite vector with the components $(d_x,d_y,d_z)$ and $\mathbb {1}_2$ is the identity of the two-dimensional spin Hilbert space. Equation (6) also gives the dispersion relation.

Circuit QED

The Hamiltonian of the system is given by the Jaynes-cummings model including the terms corresponding to the two-level atom, the quantized field and the atom-field interaction [50]

$$\begin{aligned} {H}_\mathrm {JC} = \frac{1}{2} \omega _q \sigma _z + \omega _r a^{\dagger } a + g_\mathrm {qr} (a^{\dagger } \sigma ^- + a \sigma ^+), \end{aligned}$$

(C1)

where $\sigma ^+$ ($\sigma ^-$) is the rising (lowering) operator of the atom. To control the state of the qubit (realizing the spin rotation), the system is irradiated by a microwave field with a frequency close to the qubit’s frequency

$$\begin{aligned} {H}_\mathrm{d} = \varepsilon (t)( a^{\dagger }e^{-i\omega _\mathrm{d} t} + ae^{i\omega _\mathrm{d} t} ). \end{aligned}$$

(C2)

In the large detuning regime ($g_\mathrm {qr}\ll \varDelta =\omega _q-\omega _r$), by applying the unitary transformation $U = e^{(a\sigma ^+ - a^{\dagger }\sigma ^-)g_\mathrm {qr}/\varDelta }$ on the total Hamiltonian and expanding the result up to the second order in $g/\varDelta$, we obtain

$$\begin{aligned} U ({H}_\mathrm {JC} + {H}_\mathrm{d}) U^{\dagger } \approx&\; \frac{1}{2} \left( \omega _q + \frac{g_\mathrm {qr}^2}{\varDelta }\right) \sigma _z + (\omega _r + \frac{g_\mathrm {qr}^2}{\varDelta } \sigma _z) a^{\dagger } a + \nonumber \\&\varepsilon (t) \left[ \left( a^{\dagger } + \frac{g_\mathrm {qr}}{\varDelta } \sigma ^{+}\right) e^{- i \omega _\mathrm{d} t} + \left( a + \frac{g_\mathrm {qr}}{\varDelta } \sigma ^{-}\right) e^{ i \omega _\mathrm{d} t}\right] . \end{aligned}$$

(C3)

Switching to the frame rotating at the drive frequency $\omega _\mathrm{d}$ by applying the transformation $V=e^{i \omega _\mathrm{d} a^{\dagger }at}$ leads to Eq. (27) in Sect. 6.