# Solving time-dependent Schrödinger equation using Lie algebra

I am trying to understand this chapter in order to solve the time-dependent Schrödinger equation by using On the time-dependent solutions of the Schrödinger equation (RG).

Example 1:

$$i \frac{\partial \psi}{\partial t}=\left\{{-\frac{1}{2}\frac{\partial^{2}}{\partial x^{2}}}+E_{0}x \sin \omega t\right\}\psi \tag{1}$$

$$i\frac{\partial \psi}{\partial t}=\left\{\sum_{i=1}^{4} a_{i}H_{i} \right\} \psi \tag{2}$$

where $$a_{1}=0$$, $$H_{1}=1$$, $$a_{2}=E_{0}\sin\omega t$$, $$H_2=x$$, $$a_{3}=0$$, $$H_{3}=\frac{\partial}{\partial x}$$, and $$a_{4}=-\frac{1}{2}$$, $$H_{4}=\frac{\partial^{2}}{\partial x^{2}}$$.

Solving above equation will be equal to solving the the evolution operator. $$i\frac{\partial U}{\partial t}=\dot{g_{1}}H_{1}U+\dot{g_{2}}H_{2}U+\dot{g_{3}}H_{3}U+\dot{g_{4}}H_{4}U \tag{3}$$

where $$\psi(x,t)=U(t)\psi(x,0) \tag{4}.$$

Now the next step was to find $$g$$s. For that purpose I started from $$\psi (x,t)=e^{g_{1}H_{1}+g_{2}H_{2}+g_{3}H_{3}+g_{4}H_{4}} \psi(x,0) \tag{5}$$ Taking derivative of $$(5)$$ w.r.t $$t$$ and then equating to equation $$(1)$$ as $$\dot{g_{1}}+\dot{g_{2}}x+\dot{g_3} \frac{\partial}{\partial x} +\dot{g_{4}}\frac{\partial^{2}}{\partial x^{2}}=-\frac{1}{2}\frac{\partial^2}{\partial x^2}+E_{0}x\sin \omega t$$. By comparing the coefficients came to the conclusion that $$i\dot{g_{4}}=-\frac{1}{2}$$ $$i\dot{g_{2}}=-E_{0}\sin \omega t$$ $$i\dot{g_{3}}=0$$ $$i\dot{g_{1}}=0$$ However, in the mentioned paper, $$\dot{g_{3}}$$ and $$\dot{g_1}$$ is not zero. Where am I wrong?

Your evolution operator (3) is very-very wrong, since $$H_3, H_4$$ visibly fail to commute with U. By contrast, your paper's (3) is right, $$i\frac{\partial U}{\partial t}=i(\dot{g_{1}}H_{1}U+\dot{g_{2}}H_{2}U+ \dot{g_{3}}UH_{3} +\dot{g_{4}}UH_{4}) \tag{3}$$ $$= i(\dot{g_{1}}H_{1} +\dot{g_{2}}H_{2} + \dot{g_{3}}UH_{3} U^{-1}+\dot{g_{4}}UH_{4}U^{-1}) U\\ =i(\dot{g_{1}}H_{1} +\dot{g_{2}}H_{2} + \dot{g_{3}} e^{g_2x} \partial_x e^{-g_2x}+\dot{g_{4}}e^{g_2x} \partial_x^2 e^{-g_2x}) U .$$

Needless to say, your starting point (5) is completely wrong. The correct starting point as your reference stresses, is, instead, $$U(t)= e^{g_1} e^{g_2 x} e^{g_3 \partial_x} e^{g_4 \partial_x^2} .$$

Can you take it from here to their correct (4)?

NB on request

Recall $$\partial_x e^{-g_2 x}= e^{-g_2 x}(-g_2 + \partial_x ),\implies\\ \partial_x^2 e^{-g_2 x}= e^{-g_2 x}( g_2^2 -2g_2 \partial_x + \partial_x^2).$$ The term linear in $$\partial_x$$ must vanish, yielding your source's $$\dot g_3-2\dot g_4 g_2=0. \tag{4c}$$

NB on further request

You must have made a mistake. Remember, cf (8), that the previous order of the Hs has effectively been reversed. In explicit terms this second U effectively uses the order of the inverse of the previous one. So your analog of (3) is different, and you have, instead, $$\partial_t U= U\Bigl ( U^{-1}(\dot g_1 H_1 +\dot g _2 H_2 )U + \dot{g}_3 H_3+ \dot g_4 H_4 \Bigr ).$$ They predictably assume that the reader has internalized the strategy and can easily deal with a transposition.

• Thank you so much for your answer and guiding me to the right path. After following your steps, i took first and second derivatives of $e^{-g_2 x}$ and then comparing coefficients on both sides, i got 4(a), 4(b), and 4(d). I an still confused how to get number 4(c) on the paper. I really appreciate your help guiding me on this. I am new in this filed and never worked with evolution operator before.
– Sam
Commented Feb 28, 2022 at 9:11
• Is not $\partial _x^2 e^{-g_2 x} = g_2^2 e^{-g_2 x}-2g_2e^{-g_2 x} + 2e^{-g_2 x} \partial _x^2$. After following the steps this is what I got $$\partial _x^2 e^{-g_2 x} = \partial_x (\partial_x e^{-g_2 x}) = \partial_x(-g_2e^{-g_2 x} + e^{-g_2 x} \partial_x)$$ $$= \partial_x(-g_2 e^{-g_2 x} ) + \partial_x ( \partial_x(-g_2e^{-g_2 x} \partial_x)$$ $$= g_2^2 e^{-g_2 x}-g_2e^{-g_2 x} \partial_x + \partial_x(e^{-g_2 x}) \partial_x + e^{-g_2 x} \partial_x^2$$
– Sam
Commented Feb 28, 2022 at 14:16
• Can you refer a book/notes from where I can learn about this material?
– Sam
Commented Feb 28, 2022 at 14:30
• No. Your evaluation is very wrong. Note $\partial_x e^{-g_2 x}=e^{-g_2 x} (\partial_x -g_2)$ so that $\partial_x^2 e^{-g_2 x}=e^{-g_2 x} (\partial_x -g_2)^2$. Commented Feb 28, 2022 at 14:35
• Yes, I got that part where $\partial_x e^{-g_2 x} = e^{-g_2 x} (\partial_x -g_2)$ but still confused how you got $\partial_x^2 e^{-g_2 x}= e^{-g_2 x} (\partial_x -g)^2$. Why there is no square for $e^{-g_2 x}$ term?
– Sam
Commented Feb 28, 2022 at 14:44