Yes, you are on the right track. The series you have there is called Dyson's series.

First note that the $n$'th term looks like $$ U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$

The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:

\begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right] \end{align}

Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)

Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as

\begin{align} \mathcal{T}[H(t_1) H(t_2)] = \begin{cases} H(t_1) H(t_2) & t_2 > t_1\\ H(t_2) H(t_1) & t_2 < t_1 \end{cases} \end{align}

Putting everything together we have

$$ U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)] $$ Frequently, this is denoted symbolically as

$$ U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right) $$ This notation is understood as representing the power series.

Yes, you are on the right track. The series you have there is called Dyson's series.

First note that the $n$'th term looks like $$ U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$

The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:

\begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right] \end{align}

Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)

Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as

\begin{align} \mathcal{T}[H(t_1) H(t_2)] = \begin{cases} H(t_1) H(t_2) & t_1 > t_2\\ H(t_2) H(t_1) & t_1 < t_2 \end{cases} \end{align}

Putting everything together we have

$$ U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)] $$ Frequently, this is denoted symbolically as

$$ U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right) $$ This notation is understood as representing the power series.

Yes, you are on the right track. The series you have there is called Dyson's series.

First note that the $n$'th term looks like $$ U_n = (-\frac{i}{\hbar})^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$

The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:

\begin{align} U_n =(-\frac{i}{\hbar})^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{(-\frac{i}{\hbar})^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right] \end{align}

Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)

Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures thatthe Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as

\begin{align} \mathcal{T}[H(t_1)H(t_2)] = \begin{cases} H(t_1)H(t_2) & t_2 > t_1\\ H(t_2)H(t_1) & t_2 < t_1 \end{cases} \end{align}

Putting everything together we have

$$ U(t,t') = 1 + \sum_{n=1}^\infty \frac{(-\frac{i}{\hbar})^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)] $$ Frequently, this is denoted symbolically as

$$ U(t,t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right) $$ This notation is understood as representing the power series.

Yes, you are on the right track. The series you have there is called Dyson's series.

First note that the $n$'th term looks like $$ U_n = \left(-\frac{i}{\hbar}\right)^n\int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n} H(t_1)\cdots H(t_n) $$

The order of the Hamiltonians is important, since we work with operators. Each term in the series possess a nice symmetry, allowing us to write:

\begin{align} U_n = \left(-\frac{i}{\hbar}\right)^n \int_0^t dt_1 \cdots\int_0^{t_{n-1}} dt_{n}\ H(t_1)\cdots H(t_n) = \frac{\left(-\frac{i}{\hbar}\right)^n}{n!}\int_0^t dt_1 \cdots\int_0^t dt_{n} \mathcal{T}\left[H(t_1)\cdots H(t_n)\right] \end{align}

Two things happened: first, we "overcount" by making the upper limits equal to $t$ on all the integrals. This is compensated by the factor of $\frac{1}{n!}$. You'll need to convince yourself why this factor is needed ;)

Second, by this change of integration area we mess up the ordering of the Hamiltonians in the process. This is where the time-ordering symbol $\mathcal{T}$ comes in. Basically, this operator ensures that the Hamiltonians are always ordered in the correct way. For instance for $n=2$ it operates as

\begin{align} \mathcal{T}[H(t_1) H(t_2)] = \begin{cases} H(t_1) H(t_2) & t_2 > t_1\\ H(t_2) H(t_1) & t_2 < t_1 \end{cases} \end{align}

Putting everything together we have

$$ U(t, t') = 1 + \sum_{n=1}^\infty \frac{\left(-\frac{i}{\hbar}\right)^n}{n!} \int_{t'}^t dt_1 \cdots\int_{t'}^t dt_n \mathcal{T}[H(t_1)\cdots H(t_n)] $$ Frequently, this is denoted symbolically as

$$ U(t, t') = \mathcal{T}\exp\left(-\frac{i}{\hbar} \int_{t'}^t H(t_1) dt_1\right) $$ This notation is understood as representing the power series.