Exercise 4.49

Proof for 4.103

To study the relation between \(e^{i(\hat{A}+\hat{B})\Delta t}\) and \( e^{i\hat{A}\Delta t}e^{i\hat{B}\Delta t}\), we firstly consider operation \(e^{i(\hat{A}+\hat{B})\Delta t}\) and expand it follows Taylor expansion as

(1)\[\begin{split} \begin{align} e^{i(\hat{A}+\hat{B})\Delta t} &= \sum_{n=0}^{\infty}\frac{(i\Delta t)^n(\hat{A}+\hat{B})^n}{n!} = 1 + i\hat{A}\Delta t+i\hat{B}\Delta t + \mathcal{O}(\Delta t^2)\\ \end{align} \end{split}\]

Meanwhile, we can also expand \( e^{i\hat{A}\Delta t}e^{i\hat{B}\Delta t}\) as

(2)\[\begin{split} \begin{align} e^{i\hat{A}\Delta t}e^{i\hat{B}\Delta t} =& \left[\sum_{n=0}^{\infty}\frac{(i\hat{A}\Delta t)^n}{n!} \right] \left[\sum_{n=0}^{\infty}\frac{(i\hat{B}\Delta t)^n}{n!} \right] \\ =& \left[1 + i\hat{A}\Delta t + \mathcal{O}(\Delta t^2)\right] \left[1 + i\hat{B}\Delta t + \mathcal{O}(\Delta t^2)\right] \\ =& 1 + i\hat{A}\Delta t + i\hat{B}\Delta t+ \mathcal{O}(\Delta t^2) \end{align} \end{split}\]

Compare eq. (1) and eq. (2), we should have

(3)\[ e^{i(\hat{A}+\hat{B})\Delta t} = e^{i\hat{A}\Delta t}e^{i\hat{B}\Delta t} + \mathcal{O}(\Delta t^2) \]

Proof for 4.104

To study the relation between \(e^{i(\hat{A}+\hat{B})\Delta t}\) and \( e^{i\hat{A}\Delta t/2}e^{\hat{B}\Delta t}e^{i\hat{A}\Delta t/2}\), we firstly consider operation \(e^{i(\hat{A}+\hat{B})\Delta t}\) and expand it follows Taylor expansion as

(4)\[\begin{split} \begin{align} e^{i(\hat{A}+\hat{B})\Delta t} &= \sum_{n=0}^{\infty}\frac{(\hat{A}+\hat{B})^n(i\Delta t)^n}{n!} \\ &= 1 + i(\hat{A}+\hat{B})\Delta t + \frac{(\hat{A}+\hat{B})^2(i\Delta t)^2}{2} + \mathcal{O}[(\Delta t)^3]\\ &= 1 + i{\hat{A}\Delta t+i\hat{B}\Delta t} + \frac{i^2}{2}(\hat{A}\Delta t+\hat{B}\Delta t)(\hat{A}\Delta t+\hat{B}\Delta t) + \mathcal{O}[(\Delta t)^3]\\ &= 1 + i{\hat{A}\Delta t+i\hat{B}\Delta t}+ \frac{i^2}{2}(\hat{A}\hat{A} + \hat{A}\hat{B}+ \hat{B}\hat{A}+ \hat{B}\hat{B})(\Delta t)^2 + \mathcal{O}[(\Delta t)^3]\\ \end{align} \end{split}\]

Meanwhile, we can also expand \(e^{i\hat{A}\Delta t/2}e^{\hat{B}\Delta t}e^{i\hat{A}\Delta t/2}\)​ as

(5)\[\begin{split} \begin{align} e^{(\hat{A}\Delta t)/2}e^{\hat{B}\Delta t}e^{(\hat{A}\Delta t)/2} =& \left[\sum_{n=0}\frac{(\hat{A}\Delta t)^n}{2^n n!} \right] \left[\sum_{n=0}\frac{(\hat{B}\Delta t)^n}{n!} \right] \left[\sum_{n=0}\frac{(\hat{A}\Delta t)^n}{2^n n!} \right] \\ =& \left[1 + \frac{\hat{A}\Delta t}{2} + \frac{(\hat{A}\Delta t)^2}{8} +\mathcal{O}(\Delta t^3)\right] \left[ 1 + \hat{B}\Delta t + \frac{(\hat{B}\Delta t)^2}{2} +\mathcal{O}(\Delta t^3) \right] \left[ 1 + \frac{\hat{A}\Delta t}{2} + \frac{(\hat{A}\Delta t)^2}{8} +\mathcal{O}(\Delta t^3) \right] \\ =& \left[1 + \hat{B}\Delta t+ \frac{(\hat{B}\Delta t)^2}{2} + \frac{\hat{A}\Delta t}{2} + \frac{\hat{A}\hat{B}\Delta t^2}{2} + \frac{(\hat{A}\Delta t)^2}{8} + \mathcal{O}(\Delta t^3)\right] \\ &\cdot \left[ 1 + \frac{\hat{A}\Delta t}{2} + \frac{(\hat{A}\Delta t)^2}{8} +\mathcal{O}(\Delta t^3) \right] \\ =& 1 + \frac{\hat{A}\Delta t}{2} + \frac{(\hat{A}\Delta t)^2}{8} + \hat{B}\Delta t + \frac{\hat{B}\hat{A}\Delta t^2}{2} +\frac{(\hat{B}\Delta t)^2}{2} \\ &+\frac{\hat{A}\Delta t}{2} + \frac{\hat{A}^2\Delta t^2}{4}+\frac{\hat{A}\hat{B}\Delta t^2}{2} + \frac{(\hat{A}\Delta t)^2}{8}+\mathcal{O}(\Delta t^3) \\ =& 1 + \left(\frac{\hat{A}}{2} +\hat{B}+ \frac{\hat{A}}{2}\right)\Delta t + \left(\frac{\hat{A}^2}{8} +\frac{\hat{B}\hat{A}}{2}+ \frac{\hat{B}^2}{2} + \frac{\hat{A}^2}{4} +\frac{\hat{A}\hat{B}}{2} +\frac{\hat{A}^2}{8}\right)\Delta t + \mathcal{O}(\Delta t^3) \\ =& 1 + {(\hat{A}+\hat{B})\Delta t} + \frac{1}{2}(\hat{A}\hat{A} + \hat{A}\hat{B}+ \hat{B}\hat{A}+ \hat{B}\hat{B})\Delta t+ \mathcal{O}(\Delta t^3) \end{align} \end{split}\]

Note that \(\Delta t\) is a scalar and we can add them up together, and from eq. (4) and eq. (5) we could conclude that \(e^{(\hat{A}+\hat{B})\Delta t} = e^{(\hat{A}\Delta t)/2}e^{\hat{B}\Delta t}e^{(\hat{A}\Delta t)/2} +\mathcal{O}(\Delta t^3)\). Note that we could also have

(6)\[ e^{i(\hat{A}+\hat{B})\Delta t} = e^{i\hat{B}\Delta t/2}e^{i\hat{A}\Delta t}e^{i\hat{B}\Delta t/2} \]

Proof for 4.105

To study the relation between \(e^{(\hat{A}+\hat{B})\Delta t}\)​ and \( e^{\hat{A}\Delta t}e^{\hat{B}\Delta t}e^{-[\hat{A},\hat{B}]\Delta t^2/2}\)​, we firstly consider operation \(e^{(\hat{A}+\hat{B})\Delta t}\)​ and expand it follows Taylor expansion as

(7)\[\begin{split} \begin{align} e^{(\hat{A}+\hat{B})\Delta t} &= \sum_{n=0}^{\infty}\frac{(\hat{A}+\hat{B})^n(\Delta t)^n}{n!} \\ &= 1 + (\hat{A}+\hat{B})\Delta t + \frac{(\hat{A}+\hat{B})^2(\Delta t)^2}{2} + \mathcal{O}[(\Delta t)^3]\\ &= 1 + \hat{A}\Delta t+\hat{B}\Delta t + \frac{(\Delta t)^2}{2}(\hat{A}+\hat{B})(\hat{A}+\hat{B}) + \mathcal{O}[(\Delta t)^3]\\ &= 1 + \hat{A}\Delta t+\hat{B}\Delta t + \frac{(\Delta t)^2}{2}(\hat{A}\hat{A}+\hat{A}\hat{B}+\hat{B}\hat{A}+\hat{B}\hat{B}) + \mathcal{O}[(\Delta t)^3]\\ \end{align} \end{split}\]

Meanwhile, we can also expand \(e^{\hat{A}\Delta t}e^{\hat{B}\Delta t}e^{-[\hat{A},\hat{B}]\Delta t^2/2}\) as

(8)\[\begin{split} \begin{align} e^{\hat{A}\Delta t}e^{\hat{B}\Delta t}e^{-\frac{1}{2}[\hat{A},\hat{B}]\Delta t^2} =& \left[\sum_{n=0}^{\infty}\frac{(\hat{A}\Delta t)^n}{n!} \right] \left[\sum_{n=0}^{\infty}\frac{(\hat{B}\Delta t)^n}{n!} \right] \left[\sum_{n=0}^{\infty}\frac{(-1)^{n}[\hat{A}, \hat{B}]^{n}[(\Delta t)^2]^n}{2^n n!} \right] \\ =& \left[ 1 + \hat{A}\Delta t + \frac{\hat{A}^2(\Delta t)^2}{2} +\mathcal{O}[(\Delta t)^3] \right]\cdot\left[ 1 + \hat{B}\Delta t + \frac{\hat{B}^2(\Delta t)^2}{2} +\mathcal{O}[(\Delta t)^3] \right] \cdot\left[ 1 - \frac{[\hat{A}, \hat{B}](\Delta t)^2}{2} +\mathcal{O}[(\Delta t)^4] \right]\\ =&\left[ 1 + \hat{B}\Delta t + \frac{\hat{B}^2(\Delta t)^2}{2} + \hat{A}\Delta t + \hat{A}\hat{B}(\Delta t)^2+ \frac{\hat{A}^2(\Delta t)^2}{2}+\mathcal{O}[(\Delta t)^3] \right] \cdot\left[ 1 - \frac{[\hat{A}, \hat{B}](\Delta t)^2}{2} +\mathcal{O}[(\Delta t)^4] \right] \\ =& 1 - \frac{[\hat{A}, \hat{B}](\Delta t)^2}{2} + \hat{B}\Delta t + \frac{\hat{B}^2(\Delta t)^2}{2} \\ &+ \hat{A}\Delta t + \hat{A}\hat{B}(\Delta t)^2+\frac{\hat{A}^2(\Delta t)^2}{2}+\mathcal{O}[(\Delta t)^3] \\ =& 1 + \hat{A}\Delta t + \hat{B}\Delta t \\ &+ \frac{(\Delta t)^2}{2} \left(\hat{A}^2 + \hat{B}^2 + 2\hat{A}\hat{B} - \hat{A}\hat{B} + \hat{B}\hat{A}\right) +\mathcal{O}[(\Delta t)^3] \\ =& 1 + \hat{A}\Delta t + \hat{B}\Delta t + \frac{(\Delta t)^2}{2} \left(\hat{A}\hat{A} + \hat{B}\hat{B} + \hat{A}\hat{B} + \hat{B}\hat{A}\right) +\mathcal{O}[(\Delta t)^3] \\ \end{align} \end{split}\]

Note that \(\Delta t\) is a scalar and we can add them up together, and from eq. (7) and eq. (8) we could conclude that \( e^{(\hat{A}+\hat{B})\Delta t}=e^{\hat{A}\Delta t}e^{\hat{B}\Delta t}e^{-\frac{1}{2}[\hat{A},\hat{B}]\Delta t^2} +\mathcal{O}[(\Delta t)^3]\).