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To calculate the wave equation of a time-independent Hamiltonian we use: $$ \Psi_{i}(r,t)=e^{-iH^{0}t}\psi_{i}(r,0). $$

We also know that the time-independent Hamiltonian $H^{0}=T+V$ is given to the sum of kinetic and potential energies, in an isolated atom. However, we cannot write $e^{-iH^{0}t}$ as the product of the operator, $e^{-iTt}e^{-iVt}$. Why is that?

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3 Answers 3

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This is because of the BCH formula $$\begin{align}e^Z~=~&e^Xe^Y \cr\Downarrow~&\cr Z~=~&X+Y+\frac{1}{2}[X,Y]+{\cal O}(X^2Y,XY^2),\end{align}$$ or equivalently, the Zassenhaus formula. But check out the Trotter product formula.

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Let $t$ be a dummy variable to keep track of orders in a series expansion: $$e^{tA} e^{tB} = \Big(1 + tA + \frac{1}{2} A^2 t^2 + \mathcal{O}\left(t^3\right) \Big) \Big(1 + tB + \frac{1}{2} B^2 t^2 + \mathcal{O}\left(t^3\right) \Big) = 1 + (A + B)t + \frac{1}{2} (A^2 + 2 AB + B^2) t^2 + \mathcal{O}\left(t^3\right) \, . $$ Now look at the series expansion for $$e^{t(A+B)} = 1 + (A+B)t + \frac{1}{2} (A + B)^2 t^2 + \mathcal{O}\left( t^3 \right) = 1 + (A+B)t + \frac{1}{2} \left(A^2 + AB + BA + B^2\right) t^2 + \mathcal{O}\left( t^3 \right) \, .$$ If $A$ and $B$ are numbers, then $AB + BA = 2AB$ and the series expansions agree. But in quantum mechanics the kinetic and potential energies are operators. Then $$AB + BA = 2AB + [B, A]$$ and generally the commutator $[B, A]$ is nonzero. You can check that $$\exp (tA) \exp (tB) = \exp t\Big(A + B + \frac{1}{2}[A, B] \Big) + \mathcal{O}\left(t^3 \right) $$ which is the start of the BCH formula mentioned in the other answer. With some more algebra and combinatorics, you can work out the higher order terms, too.

Because all the terms in the BCH formula except $A + B$ involve commutators, the usual formula for the exponential of numbers is recovered when all commutators vanish. As a generalisation of that, a good principle to remember is that the only thing that can "go wrong" when trying to derive an operator version of a formula that is true for numbers, is that the operators can fail to commute. So for commuting operators formulas you are used to for numbers still hold.

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When we put the Hamiltonian $H$ in an exponential like this, we are not assuming the $H$ represents a scalar value. The exponential notation used in this case is just shorthand for the Taylor series equivalent for the exponential function:

$e^{-iHt} \equiv 1 + (-iHt) + \frac{(-iHt)^2}{2!} + \frac{(-iHt)^3}{3!} + \cdots$

In many contexts we treat $H$ as a matrix, which means we are then dealing with powers of the matrix $H$.

If you think of $H$ as the sum of two matrices, one each for $T$ and $V$, and then substitute them into the power series shown above, you will have to compute powers of $T+V$. When you do this, you have to remember that matrices don't commute in general, so the order of the terms matters. Because they don't commute in general, you cannot simply assume that every term in the Taylor series can be condensed down to some complex number multiplied by powers of $T$ and $V$ like $T^nV^m$. Terms might look more like $TVT^2V$, etc. Until you are familiar with it, it's tempting to try and use this shorthand the way we did in calculus for real/complex areguments. However, since the exponent may involve quantities whose products don't commute, we cannot say that the exponential involving $H$ is equal to the product of an exponential involving $T$ and another exponential involving $V$.

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