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The short answer: the unitary time-evolution operator in quantum mechanics is

$$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right), $$

where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer...

The exponential of a matrix is defined by

$$\mathbf{M} = \sum\limits_{n=0}^{\infty} \frac{\mathbf{M}^n}{n!}.$$

(Aside: The exponential of a matrix always converges for finite dimensional matrices.)

Suppose that $\mathbf{M}$ depends on time; i.e. $\mathbf{M} \equiv \mathbf{M}(t)$. Then after a little bit of algebra, one can show that

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \int\limits_0^1 \mathrm ds \exp\left(s\mathbf{M}(t)\right)\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\exp\left((1-s)\mathbf{M}(t)\right). $$

Using Baker-Campbell-Hausdorff,

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\left[\mathbf{M}(t),\left[\dots,\left[\mathbf{M}(t),\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\right]\right]\dots\right]\exp\left(\mathbf{M}(t)\right), $$

where the $\text{n}^{\text{th}}$ term has $n$ commutators with $\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)$.

What you want is the following: a unitary time evolution operator $U(0,t)$ that satisfies

$$ \frac{\mathrm d}{\mathrm dt} U(0,t) = \frac{i}{\hbar} H(t) U(0,t), $$

because $U(0,t)\psi(0) = \psi(t)$ for a state $\psi$ (assuming t > 0). Substitution of $\psi(t)$ yields the TDSE. Naively solving the differential equation for $U$ gives $U(0,t) = \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right)$. However, expansion of the exponential using the aforementioned formula involves commutators of the Hamiltonian with itself at different times. Because $H$ is an operator, it is not guaranteed to commute with itself at different times. While you can write the time evolution operator in this way, the evolution operator itself will not obey the TDSE, i.e. you will have to use the full matrix expansion to obtain the correct time-dependent evolution equation for the state $\psi(t)$. The correct way to solve this differential equation--such that the unitary operator acting on $\psi$ obeys the TDSE--is recursively. The formal solution is

$$U(0,t) = 1 + \frac{i}{\hbar}\int\limits_0^{t}\mathrm dt' H(t')U(0,t').$$

The differential equation has now been converted to an integral equation. Continue to substitute the solution for $U$ into the above equation. Note that $0 < t' < t$. For the second iteration, you will find that $0 < t'' < t' < t$. This pattern continues ad infinitum. This provides a time-ordering to the expansion, which ensures that the Hamiltonians act on the state $\psi$ in the correct order in time. This yields the time-ordered exponential:

$$ \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right) = \sum\limits_{n=0}^{\infty} \left ( \frac{i}{\hbar} \right )^n \int\limits_{t''}^t \mathrm dt' \dots \int\limits_0^{t^{(n-1)}}\mathrm dt^{(n)} H(t')H(t'')\dots H(t^{(n-1)})H(t^{(n)}) $$

It is this unitary operator which a) evolves $\psi$ from an earlier time to a later time and b) obeys the TDSE on its own. Altogether:

$$ \frac{\mathrm d}{\mathrm dt}\psi(t) = \frac{\mathrm d}{\mathrm dt}U(0,t)\psi(0) = \frac{i}{\hbar}H(t)U(0,t)\psi(0) = \frac{i}{\hbar}H(t)\psi(t). $$

NB: the operator without time-ordering is a unitary operator for time-translation, but the time-translation in infinitesimal time-steps zig-zags through time instead of taking a time-ordered path. Because that operator has an expansion in an infinite number of nested commutators, its action on $\psi$ does not yield a TDSE at first glance, but I think it should be equivalent to the action of the time-ordered exponential on the state after an infinite number of algebraic manipulations (correct me if I am wrong on this point). The extra factor of $\frac{1}{n!}$ in the non-ordered exponential (c.f. the formula involving the derivative of $\mathbf{M}(t)$ above) accounts for the over-counting of equivalent paths through time.

The short answer: the unitary time-evolution operator in quantum mechanics is

$$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right), $$

where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer...

The exponential of a matrix is defined by

$$\exp(\mathbf{M}) = \sum\limits_{n=0}^{\infty} \frac{\mathbf{M}^n}{n!}.$$

(Aside: The exponential of a matrix always converges for finite dimensional matrices.)

Suppose that $\mathbf{M}$ depends on time; i.e. $\mathbf{M} \equiv \mathbf{M}(t)$. Then after a little bit of algebra, one can show that

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \int\limits_0^1 \mathrm ds \exp\left(s\mathbf{M}(t)\right)\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\exp\left((1-s)\mathbf{M}(t)\right). $$

Using Baker-Campbell-Hausdorff,

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\left[\mathbf{M}(t),\left[\dots,\left[\mathbf{M}(t),\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\right]\right]\dots\right]\exp\left(\mathbf{M}(t)\right), $$

where the $\text{n}^{\text{th}}$ term has $n$ commutators with $\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)$.

What you want is the following: a unitary time evolution operator $U(0,t)$ that satisfies

$$ \frac{\mathrm d}{\mathrm dt} U(0,t) = \frac{i}{\hbar} H(t) U(0,t), $$

because $U(0,t)\psi(0) = \psi(t)$ for a state $\psi$ (assuming t > 0). Substitution of $\psi(t)$ yields the TDSE. Naively solving the differential equation for $U$ gives $U(0,t) = \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right)$. However, expansion of the exponential using the aforementioned formula involves commutators of the Hamiltonian with itself at different times. Because $H$ is an operator, it is not guaranteed to commute with itself at different times. While you can write the time evolution operator in this way, the evolution operator itself will not obey the TDSE, i.e. you will have to use the full matrix expansion to obtain the correct time-dependent evolution equation for the state $\psi(t)$. The correct way to solve this differential equation--such that the unitary operator acting on $\psi$ obeys the TDSE--is recursively. The formal solution is

$$U(0,t) = 1 + \frac{i}{\hbar}\int\limits_0^{t}\mathrm dt' H(t')U(0,t').$$

The differential equation has now been converted to an integral equation. Continue to substitute the solution for $U$ into the above equation. Note that $0 < t' < t$. For the second iteration, you will find that $0 < t'' < t' < t$. This pattern continues ad infinitum. This provides a time-ordering to the expansion, which ensures that the Hamiltonians act on the state $\psi$ in the correct order in time. This yields the time-ordered exponential:

$$ \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right) = \sum\limits_{n=0}^{\infty} \left ( \frac{i}{\hbar} \right )^n \int\limits_{t''}^t \mathrm dt' \dots \int\limits_0^{t^{(n-1)}}\mathrm dt^{(n)} H(t')H(t'')\dots H(t^{(n-1)})H(t^{(n)}) $$

It is this unitary operator which a) evolves $\psi$ from an earlier time to a later time and b) obeys the TDSE on its own. Altogether:

$$ \frac{\mathrm d}{\mathrm dt}\psi(t) = \frac{\mathrm d}{\mathrm dt}U(0,t)\psi(0) = \frac{i}{\hbar}H(t)U(0,t)\psi(0) = \frac{i}{\hbar}H(t)\psi(t). $$

NB: the operator without time-ordering is a unitary operator for time-translation, but the time-translation in infinitesimal time-steps zig-zags through time instead of taking a time-ordered path. Because that operator has an expansion in an infinite number of nested commutators, its action on $\psi$ does not yield a TDSE at first glance, but I think it should be equivalent to the action of the time-ordered exponential on the state after an infinite number of algebraic manipulations (correct me if I am wrong on this point). The extra factor of $\frac{1}{n!}$ in the non-ordered exponential (c.f. the formula involving the derivative of $\mathbf{M}(t)$ above) accounts for the over-counting of equivalent paths through time.

The short answer: the unitary time-evolution operator in quantum mechanics is

$$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right), $$

where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer...

The exponential of a matrix is defined by

$$\mathbf{M} = \sum\limits_{n=0}^{\infty} \frac{\mathbf{M}^n}{n!}.$$

(Aside: The exponential of a matrix always converges for finite dimensional matrices.)

Suppose that $\mathbf{M}$ depends on time; i.e. $\mathbf{M} \equiv \mathbf{M}(t)$. Then after a little bit of algebra, one can show that

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \int\limits_0^1 \mathrm ds \exp\left(s\mathbf{M}(t)\right)\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\exp\left((1-s)\mathbf{M}(t)\right). $$

Using Baker-Campbell-Hausdorff,

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\left[\mathbf{M}(t),\left[\dots,\left[\mathbf{M}(t),\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\right]\right]\dots\right]\exp\left(\mathbf{M}(t)\right), $$

where the $\text{n}^{\text{th}}$ term has $n$ commutators with $\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)$.

What you want is the following: a unitary time evolution operator $U(0,t)$ that satisfies

$$ \frac{\mathrm d}{\mathrm dt} U(0,t) = \frac{i}{\hbar} H(t) U(0,t), $$

because $U(0,t)\psi(0) = \psi(t)$ for a state $\psi$ (assuming t > 0). Substitution of $\psi(t)$ yields the TDSE. Naively solving the differential equation for $U$ gives $U(0,t) = \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right)$. However, expansion of the exponential using the aforementioned formula involves commutators of the Hamiltonian with itself at different times. Because $H$ is an operator, it is not guaranteed to commute with itself at different times. While you can write the time evolution operator in this way, the evolution operator itself will not obey the TDSE, i.e. you will have to use the full matrix expansion to obtain the correct time-dependent evolution equation for the state $\psi(t)$. The correct way to solve this differential equation--such that the unitary operator acting on $\psi$ obeys the TDSE--is recursively. The formal solution is

$$U(0,t) = 1 + \frac{i}{\hbar}\int\limits_0^{t}\mathrm dt' H(t')U(0,t').$$

The differential equation has now been converted to an integral equation. Continue to substitute the solution for $U$ into the above equation. Note that $0 < t' < t$. For the second iteration, you will find that $0 < t'' < t' < t$. This pattern continues ad infinitum. This provides a time-ordering to the expansion, which ensures that the Hamiltonians act on the state $\psi$ in the correct order in time. This yields the time-ordered exponential:

$$ \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right) = \sum\limits_{n=0}^{\infty} \left ( \frac{i}{\hbar} \right )^n \int\limits_{t''}^t \mathrm dt' \dots \int\limits_0^{t^{(n-1)}}\mathrm dt^{(n)} H(t')H(t'')\dots H(t^{(n-1)})H^{t^{(n)}} $$

It is this unitary operator which a) evolves $\psi$ from an earlier time to a later time and b) obeys the TDSE on its own. Altogether:

$$ \frac{\mathrm d}{\mathrm dt}\psi(t) = \frac{\mathrm d}{\mathrm dt}U(0,t)\psi(0) = \frac{i}{\hbar}H(t)U(0,t)\psi(0) = \frac{i}{\hbar}H(t)\psi(t). $$

NB: the operator without time-ordering is a unitary operator for time-translation, but the time-translation in infinitesimal time-steps zig-zags through time instead of taking a time-ordered path. Because that operator has an expansion in an infinite number of nested commutators, its action on $\psi$ does not yield a TDSE at first glance, but I think it should be equivalent to the action of the time-ordered exponential on the state after an infinite number of algebraic manipulations (correct me if I am wrong on this point). The extra factor of $\frac{1}{n!}$ in the non-ordered exponential (c.f. the formula involving the derivative of $\mathbf{M}(t)$ above) accounts for the over-counting of equivalent paths through time.

The short answer: the unitary time-evolution operator in quantum mechanics is

$$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right), $$

where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer...

The exponential of a matrix is defined by

$$\mathbf{M} = \sum\limits_{n=0}^{\infty} \frac{\mathbf{M}^n}{n!}.$$

(Aside: The exponential of a matrix always converges for finite dimensional matrices.)

Suppose that $\mathbf{M}$ depends on time; i.e. $\mathbf{M} \equiv \mathbf{M}(t)$. Then after a little bit of algebra, one can show that

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \int\limits_0^1 \mathrm ds \exp\left(s\mathbf{M}(t)\right)\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\exp\left((1-s)\mathbf{M}(t)\right). $$

Using Baker-Campbell-Hausdorff,

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\left[\mathbf{M}(t),\left[\dots,\left[\mathbf{M}(t),\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\right]\right]\dots\right]\exp\left(\mathbf{M}(t)\right), $$

where the $\text{n}^{\text{th}}$ term has $n$ commutators with $\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)$.

What you want is the following: a unitary time evolution operator $U(0,t)$ that satisfies

$$ \frac{\mathrm d}{\mathrm dt} U(0,t) = \frac{i}{\hbar} H(t) U(0,t), $$

because $U(0,t)\psi(0) = \psi(t)$ for a state $\psi$ (assuming t > 0). Substitution of $\psi(t)$ yields the TDSE. Naively solving the differential equation for $U$ gives $U(0,t) = \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right)$. However, expansion of the exponential using the aforementioned formula involves commutators of the Hamiltonian with itself at different times. Because $H$ is an operator, it is not guaranteed to commute with itself at different times. While you can write the time evolution operator in this way, the evolution operator itself will not obey the TDSE, i.e. you will have to use the full matrix expansion to obtain the correct time-dependent evolution equation for the state $\psi(t)$. The correct way to solve this differential equation--such that the unitary operator acting on $\psi$ obeys the TDSE--is recursively. The formal solution is

$$U(0,t) = 1 + \frac{i}{\hbar}\int\limits_0^{t}\mathrm dt' H(t')U(0,t').$$

The differential equation has now been converted to an integral equation. Continue to substitute the solution for $U$ into the above equation. Note that $0 < t' < t$. For the second iteration, you will find that $0 < t'' < t' < t$. This pattern continues ad infinitum. This provides a time-ordering to the expansion, which ensures that the Hamiltonians act on the state $\psi$ in the correct order in time. This yields the time-ordered exponential:

$$ \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right) = \sum\limits_{n=0}^{\infty} \left ( \frac{i}{\hbar} \right )^n \int\limits_{t''}^t \mathrm dt' \dots \int\limits_0^{t^{(n-1)}}\mathrm dt^{(n)} H(t')H(t'')\dots H(t^{(n-1)})H(t^{(n)}) $$

It is this unitary operator which a) evolves $\psi$ from an earlier time to a later time and b) obeys the TDSE on its own. Altogether:

$$ \frac{\mathrm d}{\mathrm dt}\psi(t) = \frac{\mathrm d}{\mathrm dt}U(0,t)\psi(0) = \frac{i}{\hbar}H(t)U(0,t)\psi(0) = \frac{i}{\hbar}H(t)\psi(t). $$

NB: the operator without time-ordering is a unitary operator for time-translation, but the time-translation in infinitesimal time-steps zig-zags through time instead of taking a time-ordered path. Because that operator has an expansion in an infinite number of nested commutators, its action on $\psi$ does not yield a TDSE at first glance, but I think it should be equivalent to the action of the time-ordered exponential on the state after an infinite number of algebraic manipulations (correct me if I am wrong on this point). The extra factor of $\frac{1}{n!}$ in the non-ordered exponential (c.f. the formula involving the derivative of $\mathbf{M}(t)$ above) accounts for the over-counting of equivalent paths through time.

The short answer: the unitary time-evolution operator in quantum mechanics is

$$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t dt' H(t')\right), $$

where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer...

The exponential of a matrix is defined by

$$\mathbf{M} = \sum\limits_{n=0}^{\infty} \frac{\mathbf{M}^n}{n!}.$$

(Aside: The exponential of a matrix always converges for finite dimensional matrices.)

Suppose that $\mathbf{M}$ depends on time; i.e. $\mathbf{M} \equiv \mathbf{M}(t)$. Then after a little bit of algebra, one can show that

$$ \frac{d}{dt}\exp\left(\mathbf{M}(t)\right) = \int\limits_0^1 ds \exp\left(s\mathbf{M}(t)\right)\frac{d}{dt}\mathbf{M}(t)\exp\left((1-s)\mathbf{M}(t)\right). $$

Using Baker-Campbell-Hausdorff,

$$ \frac{d}{dt}\exp\left(\mathbf{M}(t)\right) = \sum\limits_{n=0}^{\infty} \frac{1}{n!}[\mathbf{M}(t),[\dots,[\mathbf{M}(t),\frac{d}{dt}\mathbf{M}(t)]]\dots]\exp\left(\mathbf{M}(t)\right), $$

where the $\text{n}^{\text{th}}$ term has $n$ commutators with $\frac{d}{dt}\mathbf{M}(t)$.

What you want is the following: a unitary time evolution operator $U(0,t)$ that satisfies

$$ \frac{d}{dt} U(0,t) = \frac{i}{\hbar} H(t) U(0,t), $$

because $U(0,t)\psi(0) = \psi(t)$ for a state $\psi$ (assuming t > 0). Substitution of $\psi(t)$ yields the TDSE. Naively solving the differential equation for $U$ gives $U(0,t) = \exp\left(\frac{i}{\hbar}\int \limits_0^t dt' H(t')\right)$. However, expansion of the exponential using the aforementioned formula involves commutators of the Hamiltonian with itself at different times. Because $H$ is an operator, it is not guaranteed to commute with itself at different times. While you can write the time evolution operator in this way, the evolution operator itself will not obey the TDSE, i.e. you will have to use the full matrix expansion to obtain the correct time-dependent evolution equation for the state $\psi(t)$. The correct way to solve this differential equation--such that the unitary operator acting on $\psi$ obeys the TDSE--is recursively. The formal solution is

$$U(0,t) = 1 + \frac{i}{\hbar}\int\limits_0^{t}dt' H(t')U(0,t').$$

The differential equation has now been converted to an integral equation. Continue to substitute the solution for $U$ into the above equation. Note that $0 < t' < t$. For the second iteration, you will find that $0 < t'' < t' < t$. This pattern continues ad infinitum. This provides a time-ordering to the expansion, which ensures that the Hamiltonians act on the state $\psi$ in the correct order in time. This yields the time-ordered exponential:

$$ \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t dt' H(t')\right) = \sum\limits_{n=0}^{\infty} \left ( \frac{i}{\hbar} \right )^n \int\limits_{t''}^t dt' \dots \int\limits_0^{t^{(n-1)}} dt^{(n)} H(t')H(t'')\dots H(t^{(n-1)}H^{t^{(n)}} $$

It is this unitary operator which a) evolves $\psi$ from an earlier time to a later time and b) obeys the TDSE on its own. Altogether:

$$ \frac{d}{dt}\psi(t) = \frac{d}{dt}U(0,t)\psi(0) = \frac{i}{\hbar}H(t)U(0,t)\psi(0) = \frac{i}{\hbar}H(t)\psi(t). $$

NB: the operator without time-ordering is a unitary operator for time-translation, but the time-translation in infinitesimal time-steps zig-zags through time instead of taking a time-ordered path. Because that operator has an expansion in an infinite number of nested commutators, its action on $\psi$ does not yield a TDSE at first glance, but I think it should be equivalent to the action of the time-ordered exponential on the state after an infinite number of algebraic manipulations (correct me if I am wrong on this point). The extra factor of $\frac{1}{n!}$ in the non-ordered exponential (c.f. the formula involving the derivative of $\mathbf{M}(t)$ above) accounts for the overcounting of equivalent paths through time.

The short answer: the unitary time-evolution operator in quantum mechanics is

$$ U(0,t) = \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right), $$

where $\hat{T}$ denotes time ordering. This is the unitary operator that yields the correct TDSE. The longer answer...

The exponential of a matrix is defined by

$$\mathbf{M} = \sum\limits_{n=0}^{\infty} \frac{\mathbf{M}^n}{n!}.$$

(Aside: The exponential of a matrix always converges for finite dimensional matrices.)

Suppose that $\mathbf{M}$ depends on time; i.e. $\mathbf{M} \equiv \mathbf{M}(t)$. Then after a little bit of algebra, one can show that

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \int\limits_0^1 \mathrm ds \exp\left(s\mathbf{M}(t)\right)\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\exp\left((1-s)\mathbf{M}(t)\right). $$

Using Baker-Campbell-Hausdorff,

$$ \frac{\mathrm d}{\mathrm dt}\exp\left(\mathbf{M}(t)\right) = \sum\limits_{n=0}^{\infty} \frac{1}{n!}\left[\mathbf{M}(t),\left[\dots,\left[\mathbf{M}(t),\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)\right]\right]\dots\right]\exp\left(\mathbf{M}(t)\right), $$

where the $\text{n}^{\text{th}}$ term has $n$ commutators with $\frac{\mathrm d}{\mathrm dt}\mathbf{M}(t)$.

What you want is the following: a unitary time evolution operator $U(0,t)$ that satisfies

$$ \frac{\mathrm d}{\mathrm dt} U(0,t) = \frac{i}{\hbar} H(t) U(0,t), $$

because $U(0,t)\psi(0) = \psi(t)$ for a state $\psi$ (assuming t > 0). Substitution of $\psi(t)$ yields the TDSE. Naively solving the differential equation for $U$ gives $U(0,t) = \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right)$. However, expansion of the exponential using the aforementioned formula involves commutators of the Hamiltonian with itself at different times. Because $H$ is an operator, it is not guaranteed to commute with itself at different times. While you can write the time evolution operator in this way, the evolution operator itself will not obey the TDSE, i.e. you will have to use the full matrix expansion to obtain the correct time-dependent evolution equation for the state $\psi(t)$. The correct way to solve this differential equation--such that the unitary operator acting on $\psi$ obeys the TDSE--is recursively. The formal solution is

$$U(0,t) = 1 + \frac{i}{\hbar}\int\limits_0^{t}\mathrm dt' H(t')U(0,t').$$

The differential equation has now been converted to an integral equation. Continue to substitute the solution for $U$ into the above equation. Note that $0 < t' < t$. For the second iteration, you will find that $0 < t'' < t' < t$. This pattern continues ad infinitum. This provides a time-ordering to the expansion, which ensures that the Hamiltonians act on the state $\psi$ in the correct order in time. This yields the time-ordered exponential:

$$ \hat{T} \exp\left(\frac{i}{\hbar}\int \limits_0^t \mathrm dt' H(t')\right) = \sum\limits_{n=0}^{\infty} \left ( \frac{i}{\hbar} \right )^n \int\limits_{t''}^t \mathrm dt' \dots \int\limits_0^{t^{(n-1)}}\mathrm dt^{(n)} H(t')H(t'')\dots H(t^{(n-1)})H^{t^{(n)}} $$

It is this unitary operator which a) evolves $\psi$ from an earlier time to a later time and b) obeys the TDSE on its own. Altogether:

$$ \frac{\mathrm d}{\mathrm dt}\psi(t) = \frac{\mathrm d}{\mathrm dt}U(0,t)\psi(0) = \frac{i}{\hbar}H(t)U(0,t)\psi(0) = \frac{i}{\hbar}H(t)\psi(t). $$

NB: the operator without time-ordering is a unitary operator for time-translation, but the time-translation in infinitesimal time-steps zig-zags through time instead of taking a time-ordered path. Because that operator has an expansion in an infinite number of nested commutators, its action on $\psi$ does not yield a TDSE at first glance, but I think it should be equivalent to the action of the time-ordered exponential on the state after an infinite number of algebraic manipulations (correct me if I am wrong on this point). The extra factor of $\frac{1}{n!}$ in the non-ordered exponential (c.f. the formula involving the derivative of $\mathbf{M}(t)$ above) accounts for the over-counting of equivalent paths through time.