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I'm trying to solve this differential equation to obtain the density matrix elements $\rho_{22}$.

$$\frac{d\rho}{dt}=-i\Big(H_{0}(t)\rho(t)-\rho(t)H^{\dagger}_{0}(t)\Big)$$,

for non-Hermitian Hamiltonian $$H_{0}(t)=\begin{pmatrix}-v t & -{\it i}A \\{\it i}B & v t \\\end{pmatrix},$$

with $$H^{\dagger}_{0}(t)=\begin{pmatrix}-v t & -{\it i}B \\{\it i}A & v t \\\end{pmatrix},$$ my code works properly

        A=0.1;
        B=1;
        v=1;
        t1=-100;
        t2=0.5;    

    H0[t_] = {{-v*t , -I*A}, {I*B, v*t}};

    H0dag[t_] = {{-v*t , -I*B}, {I*A, v*t}};

Solnh = NDSolveValue[{D[rho[t], t] == -I*(H0[t].rho[t] - rho[t].H0dag[t]),rho[t1] == {{1, 0}, {0, 0}}}, rho, {t, t1, t2}];

    Sol1=Solnh[t2];    

    Print[Sol1[[2,2]]]

the results is $\rho_{22}=1.53105 $

I know that it is possible to transform the non-Hermitian Hamiltonain $H_0(t)$ to the Hermitian Hamiltonian by using a non unitary transformation

$$H_h=\begin{pmatrix}\sqrt{\frac{B}{A}} & 0 \\0 & 1 \\\end{pmatrix}.\begin{pmatrix}-v t & -{\it i}A \\{\it i}B & v t \\\end{pmatrix}.\begin{pmatrix}\sqrt{\frac{A}{B}} & 0 \\0 & 1 \\\end{pmatrix}=\begin{pmatrix}-v t & -{\it i}\sqrt{AB} \\{\it i}\sqrt{AB} & v t \\\end{pmatrix}$$.

By applying the same non-unitary transformation on the density matrix we have

$$\rho_h=\begin{pmatrix}\sqrt{\frac{b}{a}} & 0 \\0 & 1 \\\end{pmatrix}.\begin{pmatrix}\rho_{11} & \rho_{12} \\\rho_{21} & \rho_{22} \\\end{pmatrix}.\begin{pmatrix}\sqrt{\frac{a}{b}} & 0 \\0 & 1 \\\end{pmatrix}=\begin{pmatrix}\rho_{11} & \sqrt{\frac{B}{A}}\rho_{12} \\\sqrt{\frac{A}{B}}\rho_{21} & \rho_{22} \\\end{pmatrix}$$.

So $\rho_{22}$ does not change under transformation. Therefore if we solve the master equation for Hermitian Hamiltonian $$H_h=\begin{pmatrix}-v t & -{\it i}\sqrt{AB} \\{\it i}\sqrt{AB} & v t \\\end{pmatrix}=H_h^{\dagger}$$, i.e.,

$$\frac{d\rho_h}{dt}=-i\Big(H_{h}(t)\rho_h(t)-\rho_h(t)H_{h}(t)\Big)$$,

$\rho_{22}$ should be the same as the previous result.

My code is

            A=0.1;
            B=1;
            v=1;
            t1=-100;
            t2=0.5;
    
        Hh[t_] = {{-v*t , -I*Sqrt[A*B]}, {I*Sqrt[A*B], v*t}};
    
    Solh = NDSolveValue[{D[rhoh[t], t] == -I*(Hh[t].rhoh[t] - rhoh[t].Hh[t]),
rhoh[t1] == {{1, 0}, {0, 0}}}, rhoh, {t, t1, t2}];

        Sol2=Solh[t2];  
  
        Print[Sol2[[2,2]]]

which gives $\rho_{h,22}=0.153116 $. Why the results are different in both methods? If we check $\rho_{11}$ which does not changes under transformation, it remains intact and in both approach $\rho_{11}=0.8468$.

I was wondering if someone could help me. Thank you for your kind attention.

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  • $\begingroup$ Solnh[[t2]] could not have possibly worked. Since t2 is 0.5 you can not use 0.5 as index value. $\endgroup$
    – Nasser
    Commented Dec 24, 2022 at 19:41
  • $\begingroup$ @Nasser Sorry that is misprint $\endgroup$
    – Radmehr
    Commented Dec 24, 2022 at 19:46
  • $\begingroup$ @Radmehr: Your results imply that you multiply the second result by 10 or divide the first result by 10. I think somewhere in your code does the above transformation. $\endgroup$ Commented Dec 24, 2022 at 23:33
  • $\begingroup$ @TugrulTemel Thank you for your comment. Yes you are right. The results are different even by normalization of the density matrix. If we check $\rho_{11}$ which does not changes under transformation, it remains intact and in both approach $\rho_{11}=0.8468$. $\endgroup$
    – Radmehr
    Commented Dec 25, 2022 at 6:26

2 Answers 2

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Too long for comment.

Well, your new ode is different after transformation. Your claim is that the solution should remain the same due to how the math is supposed to work. This is really a math question. This is your ode in first case

ClearAll["Global`*"]
A = 0.1;
B = 1;
v = 1;
t1 = -100;
t2 = 0.5;

H0[t_] = {{-v*t, -I*A}, {I*B, v*t}};

H0dag[t_] = {{-v*t, -I*B}, {I*A, v*t}};
ode = D[rho[t], t] == -I*(H0[t] . rho[t] - rho[t] . H0dag[t]);
TraditionalForm[ode]
ic = rho[t1] == {{1, 0}, {0, 0}};

Solnh = NDSolveValue[{ode, ic}, rho, {t, t1, t2}];
Sol1 = Solnh[t2];
Print[Sol1[[2, 2]]]

enter image description here

And this is your second case

ClearAll["Global`*"]
A = 0.1;
B = 1;
v = 1;
t1 = -100;
t2 = 0.5;

Hh[t_] = {{-v*t, -I*Sqrt[A*B]}, {I*Sqrt[A*B], v*t}};
ode = D[rhoh[t], t] == -I*(Hh[t] . rhoh[t] - rhoh[t] . Hh[t]);
TraditionalForm[ode]
ic = rhoh[t1] == {{1, 0}, {0, 0}};

Solh = NDSolveValue[{ode, ic}, rhoh, {t, t1, t2}];
Sol2 = Solh[t2];
Print[Sol2[[2, 2]]]

enter image description here

You see the ode system is different in both cases.

Your claim is that the output should be the same, because the "math" should not have caused this to change.

So either NDSolve is wrong, or your math has some faulty assumptions in it.

Are your sure your math is correct and the result should be the same? May be you could check on this again since the difference is not negligible, I do not think this is due to round off errors, or may be it is? it is of order 0.0001 difference.

Did you try DSolve ?

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  • $\begingroup$ @ Nasser Thank you for your comment. As far as I know, in mathematics point of view both result should be the same. If you apply the transformation directly on the first differential equation (for non-Hermitian Hamiltonian) you get directly to the second differential equation for Hermitian Hamiltonian. Which means $\rho_{22}$ and $\rho_{11}$ do not change under transformation. $\endgroup$
    – Radmehr
    Commented Dec 24, 2022 at 20:15
  • 1
    $\begingroup$ @Radmehr well, if the math says the result should be the same, then I can only guess the reason is due to numerical round off, as the difference is of order 1/10000. That is my best guess. I do not see what else could cause this. $\endgroup$
    – Nasser
    Commented Dec 24, 2022 at 20:16
  • $\begingroup$ Thank you so much for your comments. $\endgroup$
    – Radmehr
    Commented Dec 24, 2022 at 20:20
  • $\begingroup$ I used NDSolve but the results are the same as NDSolvevalues $\endgroup$
    – Radmehr
    Commented Dec 25, 2022 at 6:53
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If you are sure your math is correct, you may try to increase accuracy and precision:

AccuracyGoal -> 24
PrecisionGoal -> 10

See more details here.

Having said that, really make sure that your analytical solution (algebraic transformation) is identical.

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  • $\begingroup$ Using Nasser's two versions, changing 0.1 to 1/10 and 0.5 to 1/2 (to allow use of higher than MachinePrecision) and including WorkingPrecision->32 (or 64) inside NDSolve results in bailing out after 10^4 steps. Including MaxSteps->10^5 results in bailing out after 10^5 steps, each with wildly different values in every case. $\endgroup$
    – Bill
    Commented Dec 25, 2022 at 2:51
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    $\begingroup$ @Adam Ciesielski Thank you for your comments. By changing the AccuracyGoal the results show large deviation. $\endgroup$
    – Radmehr
    Commented Dec 25, 2022 at 7:07

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