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Influence of environmental noise on the weak value amplification

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Abstract

Quantum systems are always disturbed by environmental noise. We have investigated the influence of the environmental noise on the amplification in weak measurements. Three typical quantum noise processes are discussed in this article. The maximum expectation values of the observables of the measuring device decrease sharply with the strength of the depolarizing and phase damping channels, while the amplification effect of weak measurement is immune to the amplitude damping noise. To obtain significantly amplified signals, we must ensure that the preselection quantum systems are kept away from the depolarizing and phase damping processes.

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Acknowledgments

This work was financially supported by the National Natural Science Foundation of China (Grants Nos. 11305118, and 11475084), and the Fundamental Research Funds for the Central Universities.

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Correspondence to Xuannmin Zhu.

Appendices

Appendix 1: The derivation of Equation (16)

The two pure states \(|\psi \rangle _s\) and \(|\psi _f\rangle _s\) could be rewritten as \(|\psi \rangle _s=\cos \frac{\theta _1}{2}|0\rangle _s+e^{i\phi _1}\sin \frac{\theta _1}{2}|1\rangle _s\) and \(|\psi _f\rangle _s = \cos \frac{\theta _2}{2}|0\rangle _s+e^{i\phi _2}\sin \frac{\theta _2}{2}|1\rangle _s\). The shifts of momentum and position in Eq. (15) could be rewritten as

$$\begin{aligned} \begin{aligned} \delta p'=\frac{g(\cos \theta _2+r\cos \theta _1)}{1+r\cos \theta _1\cos \theta _2+r\sin \theta _1\sin \theta _2\cos \phi _0e^{-2\varDelta ^2g^2}},\\ \delta q'=\frac{2gr\varDelta ^2\sin \theta _1\sin \theta _2\sin \phi _0}{1+r\cos \theta _1\cos \theta _2+r\sin \theta _1\sin \theta _2\cos \phi _0e^{-2\varDelta ^2g^2}}, \end{aligned} \end{aligned}$$
(44)

where \(\phi _0=\phi _1-\phi _2\).

First, we search for the extreme values of \(\delta p'\). By \(\frac{\partial \delta p' }{\partial \phi _0}=0\), we get \(\sin \phi _0=0\), \(\cos \phi _0=\pm 1\) and

$$\begin{aligned} \delta p'=\frac{g(\cos \theta _2+r\cos \theta _1)}{1+r\cos \theta _1\cos \theta _2\pm r\sin \theta _1\sin \theta _2e^{-2\varDelta ^2g^2}} \end{aligned}$$
(45)

Now, we use the method of Lagrange multipliers to obtain maximum values. Let \(x_1=\cos \theta _2\), \(x_2=\cos \theta _1\), \(y_1=\sin \theta _2\), and \(y_2=\sin \theta _1\) with the constraint \(x_1^2+y_1^2-1=0\) and \(x_2^2+y_2^2-1=0\), we have the Lagrange function

$$\begin{aligned} \begin{aligned} F_1=&\frac{(x_1+rx_2)g}{1+rx_1x_2\pm r y_1y_2e^{-2\varDelta ^2g^2}}\\&+\lambda _1(x_1^2+y_1^2-1)+\lambda _2(x_2^2+y_2^2-1). \end{aligned}\end{aligned}$$
(46)

Let \(\frac{\partial F_1}{\partial x_1}=\frac{\partial F_1}{\partial y_1}=0\), from \(y_1\frac{\partial F_1}{\partial x_1}=x_1\frac{\partial F_1}{\partial y_1}\), we have \(y_1=0\) or

$$\begin{aligned} y_1y_2=\mp (r+x_1x_2)e^{-2\varDelta ^2g^2}. \end{aligned}$$
(47)

For \(y_1=0\), we have the shift

$$\begin{aligned} \delta p '=\pm g. \end{aligned}$$
(48)

Let \(\frac{\partial F_1}{\partial x_2}=\frac{\partial F_1}{\partial y_2}=0\), from \(y_2\frac{\partial F_1}{\partial x_2}=x_2\frac{\partial F_1}{\partial y_2}\), we get

$$\begin{aligned} \frac{y_1}{y_2}=\mp \frac{r e^{-2\varDelta ^2g^2}(1+rx_1x_2)}{1-r^2x_2^2}. \end{aligned}$$
(49)

If we divide Eq. (47) by Eq. (49), we will get

$$\begin{aligned} y_2^2=\frac{(r+x_1x_2)(1-r^2x_2^2)}{r(1+rx_1x_2)}. \end{aligned}$$
(50)

Using the constraint \(x_2^2+y_2^2=1\), we have

$$\begin{aligned} x_2=-\frac{x_1}{r}~or~x_2=0. \end{aligned}$$
(51)

Substituting \(x_2=-\frac{x_1}{r}\) into Eq. (46), we have

$$\begin{aligned} \delta p'= 0. \end{aligned}$$
(52)

For \(x_2=0\), using the constraint \(x_2^2+y_2^2-1=0\), we get \(y_2=\pm 1\). Substituting \(y_2=\pm 1\) into Eq. (47), we have \(y_1=\mp e^{-2\varDelta ^2g^2}\), \(x_1=\pm \sqrt{1-r^2e^{-4\varDelta ^2g^2}}\). Substituting \(x_1\),\(x_2\),\(y_1\), and \(y_2\) into Eq. (46), we have

$$\begin{aligned} \delta p'=\pm \frac{g}{\sqrt{1-r^2e^{-4\varDelta ^2g^2}}}. \end{aligned}$$
(53)

Comparing the results in Eqs. (48), (52), and (53), we have the maximum shift of momentum

$$\begin{aligned} |\delta p'|_{\max }=\frac{g}{\sqrt{1-r^2e^{-4\varDelta ^2g^2}}}. \end{aligned}$$
(54)

The conditions of obtaining the maximum shift are \(\cos \theta _1=0\), \(\sin \theta _1=\pm 1\), \(\sin \theta _2=\pm re^{-2\varDelta ^2g^2}\), and \(\cos \theta _2=\pm \sqrt{1-r^2e^{-4\varDelta ^2g^2}}\).

Now we derive the maximum value of \(|\delta q'|\) by using the method of Lagrange multipliers. Let \(x_3=\cos \phi _0\), and \(y_3=\sin \phi _0\), from Eq. (44), we can get the Lagrange function

$$\begin{aligned} \begin{aligned} F_2=&\frac{2gr\varDelta ^2y_1y_2y_3}{1+rx_1x_2+ry_1y_2x_3e^{-2\varDelta ^2g^2}}+\lambda _1(x_1^2+y_1^2-1)\\&+\lambda _2(x_2^2+y_2^2-1)+\lambda _3(x_3^2+y_3^2-1). \end{aligned}\end{aligned}$$
(55)

Let \(\frac{\partial F_2}{\partial x_1}=\frac{\partial F_2}{\partial y_1}=0\), from \(y_1\frac{\partial F_2}{\partial x_1}=x_1\frac{\partial F_2}{\partial y_1}\), we have

$$\begin{aligned} y_2y_3(x_1+rx_2)=0. \end{aligned}$$
(56)

The solutions of this equation are \(y_2=0\), or \(y_3=0\), or

$$\begin{aligned} x_1=-rx_2. \end{aligned}$$
(57)

For \(y_2=0\) or \(y_3=0\), we have

$$\begin{aligned} \delta q'=0. \end{aligned}$$
(58)

For \(\frac{\partial F_2}{\partial x_2}=\frac{\partial F_2}{\partial y_2}=0\), from \(y_2\frac{\partial F_2}{\partial x_2}=x_2\frac{\partial F_2}{\partial y_2}\), we get \(y_2=0\), or \(y_3=0\), or

$$\begin{aligned} x_2=-rx_1. \end{aligned}$$
(59)

As r does not always equal to 1, to satisfy Eqs. (57) and (59), we have \(x_1=x_2=0\) and \(y_1y_2=\pm 1\). Then the Lagrange function changes into

$$\begin{aligned} F_2(x_3,y_3,\lambda _3)=\pm \frac{2gr\varDelta ^2y_3}{1\pm rx_3e^{-2\varDelta ^2g^2}}. \end{aligned}$$
(60)

Let \(\frac{\partial F_2}{\partial x_3}=\frac{\partial F_2}{\partial y_3}=0\), from \(y_3\frac{\partial F_2}{\partial x_3}=x_3\frac{\partial F_2}{\partial y_3}\), we have

$$\begin{aligned} x_3=\mp re^{-2\varDelta ^2g^2},\quad y_3=\pm \sqrt{1-r^2e^{-4\varDelta ^2g^2}}. \end{aligned}$$
(61)

Substituting \(x_3\) and \(y_3\) into Eq. (60), we obtain the shifts of position

$$\begin{aligned} \delta q'=\pm \frac{2gr\varDelta ^2}{\sqrt{1-r^2e^{-4\varDelta ^2g^2}}}. \end{aligned}$$
(62)

Comparing the results in Eqs. (58) and (62), we have the maximum position shift

$$\begin{aligned} |\delta q'|_{\max }=\frac{2gr\varDelta ^2}{\sqrt{1-r^2e^{-4\varDelta ^2g^2}}}. \end{aligned}$$
(63)

The conditions of attaining the maximum shift are \(\sin \theta _1=\pm 1\), \(\sin \theta _2=\mp 1\), \(\cos \phi _0=\mp re^{-2\varDelta ^2g^2}\).

Appendix 2: The derivation of Equation (25)

For the pure state \(|\psi \rangle _s=\cos \frac{\theta _1}{2}|0\rangle _s+e^{i\phi _1}\sin \frac{\theta _1}{2}|1\rangle _s\) and the postselection state \(|\psi _f\rangle _s = \cos \frac{\theta _2}{2}|0\rangle _s+e^{i\phi _2}\sin \frac{\theta _2}{2}|1\rangle _s\), Eq. (24) could be rewritten as

$$\begin{aligned} \left<O'\right>=\frac{(1+\cos \theta _1\cos \theta _2-\sin \theta _1\sin \theta _2\cos \phi _0)\sin ^2g}{1+\cos \theta _1\cos \theta _2+\sin \theta _1\sin \theta _2\cos \phi _0\cos 2g}, \end{aligned}$$
(64)

where \(\phi _0=\phi _1-\phi _2\). Let \(\frac{\partial \left<O'\right>}{\partial \phi _0}=0\), we have \(\sin \phi _0=0\), \(\cos \phi _0=\pm 1\). Substituting \(\cos \phi _0=\pm 1\) into Eq. (64), we obtain

$$\begin{aligned} \left<O'\right>=\frac{(1+\cos \theta _1\cos \theta _2\mp \sin \theta _1\sin \theta _2)\sin ^2g}{1+\cos \theta _1\cos \theta _2\pm \sin \theta _1\sin \theta _2\cos 2g}. \end{aligned}$$
(65)

Let \(t=\cos \frac{\theta _1\mp \theta _2}{2}/\cos \frac{\theta _1\pm \theta _2}{2}\), we have

$$\begin{aligned} \left<O'\right>=\frac{\sin ^2g}{1+t^2+(t^2-1)\cos 2g}. \end{aligned}$$
(66)

From \(\frac{\partial \left<O'\right>}{\partial t}=0\), we have \(t=0\), and the maximum measured value

$$\begin{aligned} \left<O'\right>_{\max }=\frac{\sin ^2g}{1-\cos 2g}=1. \end{aligned}$$
(67)

The conditions of obtaining the maximum measured value are \(\phi _1-\phi _2=0\) and \(\theta _1-\theta _2=\pi \), or \(\phi _1-\phi _2=\pi \), and \(\theta _1+\theta _2=\pi \). In other words, when the preselection and postselection states are orthogonal to each other, the measured value of qubit measuring device attains its maximum value.

Appendix 3: The derivation of Equation (27)

Let \(|\psi \rangle _s=\cos \frac{\theta _1}{2}|0\rangle _s+e^{i\phi _1}\sin \frac{\theta _1}{2}|1\rangle _s\) and \(|\psi _f\rangle _s= \cos \frac{\theta _2}{2}|0\rangle _s+e^{i\phi _2}\sin \frac{\theta _2}{2}|1\rangle _s\), the Eq. (26) could be rewritten as

$$\begin{aligned} \left<O''\right>=\frac{(1-r+r(1+\cos \theta _1\cos \theta _2-\sin \theta _1\sin \theta _2\cos \phi _0))\sin ^2g}{1-r+r(1+\cos \theta _1\cos \theta _2+\sin \theta _1\sin \theta _2\cos \phi _0\cos 2g)}, \end{aligned}$$
(68)

where \(\phi _0=\phi _1-\phi _2\). From \(\frac{\partial \left<O''\right>}{\partial \phi _0}=0\), we have \(\sin \phi _0=0\), \(\cos \phi _0=\pm 1\). Substituting \(\cos \phi _0=\pm 1\) into Eq. (68), we have

$$\begin{aligned} \left<O''\right>=\frac{\left( 1-r+2r\cos ^2\frac{\theta _1\pm \theta _2}{2}\right) \sin ^2g}{1-r+2r\left( \cos ^2\frac{\theta _1\mp \theta _2}{2}-\sin ^2g\left( \cos ^2\frac{\theta _1\pm \theta _2}{2}-\cos ^2\frac{\theta _1\mp \theta _2}{2}\right) \right) }. \end{aligned}$$
(69)

The value \(\left<O''\right>\ge 0\) is the monotonically decreasing function with \(\cos ^2\frac{\theta _1\mp \theta _2}{2}\). Therefore, let \(\cos ^2\frac{\theta _1\mp \theta _2}{2}=0\), we obtain the maximum measured value of the qubit measuring device

$$\begin{aligned} \left<O''\right>_{\max }=1. \end{aligned}$$
(70)

The maximum measured value is obtained under the conditions \(\phi _1-\phi _2=0\) and \(\theta _1-\theta _2=\pi \), or \(\phi _1-\phi _2=\pi \), and \(\theta _1+\theta _2=\pi \) for which cases the PPS are orthogonal.

Appendix 4: The derivation of Equation (33)

For \(|\psi _i\rangle _s=\cos \frac{\theta _1}{2}|0\rangle _s+e^{i\phi _1}\sin \frac{\theta _1}{2}|1\rangle _s\) and \(|\psi _f\rangle _s= \cos \frac{\theta _2}{2}|0\rangle _s+e^{i\phi _2}\sin \frac{\theta _2}{2}|1\rangle _s\), Eq. (32) could be rewritten as

$$\begin{aligned} \begin{aligned} \delta p'=\frac{(\cos \theta _1+\cos \theta _2)g}{1+\cos \theta _1\cos \theta _2+(1-\gamma )\sin \theta _1\sin \theta _2\cos \phi _0e^{-2\varDelta ^2g^2}},\\ \delta q'= \frac{2(1-\gamma )g\varDelta ^2e^{-2\varDelta ^2g^2}\sin \theta _1\sin \theta _2\sin \phi _0}{1+\cos \theta _1\cos \theta _2+(1-\gamma )\sin \theta _1\sin \theta _2\cos \phi _0e^{-2\varDelta ^2g^2}} \end{aligned}\end{aligned}$$
(71)

where \(\phi _0=\phi _1-\phi _2\). First we search for the extreme values of \(\delta p'\), by \(\frac{\partial \delta p' }{\partial \phi _0}=0\), we get \(\sin \phi _0=0\), \(\cos \phi _0=\pm 1\) and

$$\begin{aligned} \delta p'=\frac{2gt}{1+t^2\pm (1-\gamma )(1-t^2)e^{-2\varDelta ^2g^2}}, \end{aligned}$$
(72)

where \(t={\cos \frac{\theta _1-\theta _2}{2}}/{\cos \frac{\theta _1+\theta _2}{2}}\). From \(\frac{\partial \delta p'}{\partial t}=0\), we have

$$\begin{aligned} t^2=\frac{1\pm (1-\gamma )e^{-2\varDelta ^2g^2}}{1\mp (1-\gamma )e^{-2\varDelta ^2g^2}}. \end{aligned}$$
(73)

Substituting t into Eq. (72), we get the maximum shift of momentum is

$$\begin{aligned} |\delta p'|_{\max }=\frac{g}{\sqrt{1-(1-\gamma )e^{-4\varDelta ^2g^2}}}, \end{aligned}$$
(74)

and the conditions of obtaining the maximum shifts are \(\phi _1-\phi _2=0\) and \({\cos \frac{\theta _1-\theta _2}{2}}/{\cos \frac{\theta _1+\theta _2}{2}}=\pm \sqrt{1+(1-\gamma )e^{-4\varDelta ^2g^2}/{1-(1-\gamma )e^{-4\varDelta ^2g^2}}}\), or \(\phi _1-\phi _2=\pi \) and \({\cos \frac{\theta _1-\theta _2}{2}}/{\cos \frac{\theta _1+\theta _2}{2}}=\pm \sqrt{1-(1-\gamma )e^{-4\varDelta ^2g^2}/{1+(1-\gamma )e^{-4\varDelta ^2g^2}}}\).

Now we use the method of Lagrange multipliers to derive the extreme values of \(\delta q'\). Let \(x_1=\cos \theta _1\), \(x_2=\cos \theta _2\), \(x_3=\cos \phi _0\), \(y_1=\sin \theta _1\), \(y_2=\sin \theta _2\), \(y_3=\sin \phi _0\), from Eq. (71), we have the Lagrange function

$$\begin{aligned} \begin{aligned} F_3=&\frac{2g\varDelta ^2(1-\gamma )e^{-2\varDelta ^2g^2}\varDelta ^2y_1y_2y_3}{1+x_1x_2+(1-\gamma )y_1y_2x_3e^{-2\varDelta ^2g^2}}+\lambda _1(x_1^2+y_1^2-1)\\&+\lambda _2(x_2^2+y_2^2-1)+\lambda _3(x_3^2+y_3^2-1). \end{aligned}\end{aligned}$$
(75)

Let \(\frac{\partial F_3}{\partial x_1}=\frac{\partial F_3}{\partial y_1}=0\), we have

$$\begin{aligned} y_2y_3(x_1+x_2)=0. \end{aligned}$$
(76)

The solution of this equation is \(y_2=0\), or \(y_3=0\), or

$$\begin{aligned} x_1=-x_2. \end{aligned}$$
(77)

For \(y_2=0\) or \(y_3=0\), we have

$$\begin{aligned} \delta q'=0. \end{aligned}$$
(78)

For \(x_1=-x_2\), we have

$$\begin{aligned} F_3=\pm \frac{2g\varDelta ^2(1-\gamma )e^{-2\varDelta ^2g^2}y_3}{1\pm (1-\gamma )x_3e^{-2\varDelta ^2g^2}}. \end{aligned}$$
(79)

Let \(\frac{\partial F_3}{\partial x_3}=\frac{\partial F_3}{\partial y_3}=0\), we have

$$\begin{aligned} \begin{aligned} x_3&=\mp (1-\gamma )e^{-2\varDelta ^2g^2},\\ y_3&=\pm \sqrt{1-(1-\gamma )^2e^{-4\varDelta ^2g^2}}. \end{aligned}\end{aligned}$$
(80)

Substituting \(x_3\) and \(y_3\) into Eq. (79), we get the shift

$$\begin{aligned} \delta q'=\pm \frac{2g\varDelta ^2(1-\gamma )e^{-2\varDelta ^2g^2}}{\sqrt{1-(1-\gamma )^2e^{-4\varDelta ^2g^2}}}. \end{aligned}$$
(81)

Comparing Eq. (78) with Eq. (81), we get the maximum shift

$$\begin{aligned} |\delta q'|_{\max }=\frac{2g\varDelta ^2(1-\gamma )e^{-2\varDelta ^2g^2}}{\sqrt{1-(1-\gamma )^2e^{-4\varDelta ^2g^2}}}. \end{aligned}$$
(82)

The maximum shift is obtained on the conditions \(\cos \theta _1=-\cos \theta _2\), \(\sin \theta _1=\pm \sin \theta _2\), and \(\cos \phi _0=\mp (1-\gamma )e^{-2\varDelta ^2g^2}\).

Appendix 5: The derivation of Equation (35)

The preselection state \(|\psi _i\rangle _s=\cos \frac{\theta _1}{2}|0\rangle _s+e^{i\phi _1}\sin \frac{\theta _1}{2}|1\rangle _s\) and the postselection state \(|\psi _f\rangle _s = \cos \frac{\theta _2}{2}|0\rangle _s+e^{i\phi _2}\sin \frac{\theta _2}{2}|1\rangle _s\) could be regarded as two vectors in Bloch sphere, with \(\theta _1\in [0,\pi )\), \(\theta _2\in [0,\pi )\), \(\phi _1\in [0,2\pi )\), and \(\phi _2\in [0,2\pi )\). Equation (34) could be rewritten as

$$\begin{aligned} \left<O'\right>=\frac{(1+\cos \theta _1\cos \theta _2-(1-\gamma )\sin \theta _1\sin \theta _2\cos \phi _0)\sin ^2g}{1+\cos \theta _1\cos \theta _2+(1-\gamma )\sin \theta _1\sin \theta _2\cos \phi _0\cos 2g}, \end{aligned}$$
(83)

where \(\phi _0=\phi _1-\phi _2\in (-2\pi ,2\pi )\). Since \(\frac{\partial \left<O'\right>}{\partial \cos \phi _0}\le 0\), and \(\left<O'\right>\ge 0\), when \(\cos \phi _0=-1\), the expectation value of O reaches its maximum value

$$\begin{aligned} \left<O'\right>=\frac{(1+\cos \theta _1\cos \theta _2+(1-\gamma )\sin \theta _1\sin \theta _2)\sin ^2g}{1+\cos \theta _1\cos \theta _2-(1-\gamma )\sin \theta _1\sin \theta _2\cos 2g}. \end{aligned}$$
(84)

Let \(t={\cos \frac{\theta _1-\theta _2}{2}}/{\cos \frac{\theta _1+\theta _2}{2}}\), we have

$$\begin{aligned} \left<O'\right>=\frac{(1+t^2+(1-\gamma )(1-t^2))\sin ^2g}{1+t^2-(1-\gamma )(1-t^2)\cos 2g}. \end{aligned}$$
(85)

Since \(\frac{\partial \left<O'\right>}{\partial t^2}\le 0\), when \(t^2=0\), \(\left<O'\right>\) reaches its maximum value

$$\begin{aligned} \left<O'\right>_{\max }=\frac{(2-\gamma )\sin ^2g}{\gamma +2(1-\gamma )\sin ^2g}. \end{aligned}$$
(86)

The condition of obtaining maximum value is \(\phi _1-\phi _2=\pi \), \(\theta _1+\theta _2=\pi \), and the PPS are orthogonal.

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Zhu, X., Zhang, YX. Influence of environmental noise on the weak value amplification. Quantum Inf Process 15, 3421–3441 (2016). https://doi.org/10.1007/s11128-016-1340-x

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