Arguing that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without taylor expansion [closed]

Question

I would like to argue that the time derivative of $\exp(-iHt)$ is $-iH\exp(-iHt)$ without Taylor expansion. $H$ is the Hamiltonian and it is hermitian. Thus it can be diagonalized. But I cannot see how I can ignore the eigenvectors as if they do not exist and use chain rule, when I attempt to bring down the $H$ from the exponential.

Answer 1

Exponential of a matrix is defined as Taylor expansion e.g.

https://en.wikipedia.org/wiki/Matrix_exponential

So, it is hard not to use Taylor expansion due to the definition.

Answer 2

I'm not sure about your question but I'll try to clarify you a little bit about function of operators. Since they can be undestood as matrices, note that it's impossible to define something like $\exp(A)$. What one can do with operators is just multiply them. Hence, when one writes down a function of an operator $f(A)$ is indeed talking about the taylor serie of this function which will be:

$\sum_{n} C_n A^n$

where $A^n$ is an operation that takes matematically sense. Therefore, you cannot treat this kind of functions as usually functions and in order to omit any confusion, think always in terms of Taylor series.

Answer 3

Since the Hamiltonian is a self-adjoint operator, you can use the spectral theorem to write $$H = \int_\mathbb{R} \lambda \, dP_\lambda.$$ Then, as the exponential is a measurable function, $$ \exp(-iHt) = \int_\mathbb{R} \exp(-it\lambda) \, dP_\lambda. $$ Now, by the Leibniz integral rule, we have that $$\frac{d}{dt}\exp(-iHt) = \int_\mathbb{R} \bigg(\frac{\partial}{\partial t}\exp(-it\lambda)\bigg) \, dP_\lambda = \int_\mathbb{R} -i\lambda\exp(-it\lambda) \, dP_\lambda = -iH\exp(-iHt).$$

Answer 4

If $H$ is a hermitian matrix with eigenvectors $$H|n\rangle=\epsilon_n|n\rangle,$$ (note that the eigenvectors and eigenvalues do not depend on time) then we could write $$ H= \sum_n|n\rangle\langle n|\epsilon_n,\\ e^{-iHt}=\sum_n|n\rangle\langle n|e^{-i\epsilon_n t},\\ \partial_t \left[e^{-iHt}\right]=\partial_t\left[\sum_n|n\rangle\langle n|e^{-i\epsilon_n t}\right]=\sum_n|n\rangle\langle n|(-i\epsilon_n)e^{-i\epsilon_n t}=\\ -i\sum_m|m\rangle\langle m|\epsilon_m\sum_n|n\rangle\langle n|e^{-i\epsilon_n t}=-iHe^{-iHt} $$

Answer 5

But I cannot see how I can ignore the eigenvectors as if they do not exist and use chain rule

Do you want to prove it without using the spectrum or Taylor expansion?

Let $U(t) = e^{-iHt}$. By definition

$$ {dU\over dt} =\lim_{\delta \rightarrow 0} \frac {1}{\delta } (U(t+\delta) - U(t) )= \lim_{\delta \rightarrow 0}\frac {1}{\delta } \left ( e^{-i\delta H} - \mathbb I\right )U(t) \equiv OU(t) $$

Where I used the fact that $e^{-iH(t+\delta)}=e^{-iHt}e^{-iH\delta}$. From now, I'll assume that the above limit exist. We know that

$$ O^\dagger = lim_{\delta \rightarrow 0}\frac {1}{\delta } \left (e^{i\delta H} - \mathbb I\right) = \lim_{\delta\rightarrow 0} e^{i\delta H}\lim_{\delta\rightarrow 0} \frac {1}{\delta } \left(\mathbb I - e^{-i\delta H}\right )= -O $$

Where I used the fact that all the limits exist and $H$ is hermitian. Now, since $O$ is anti-hermitian, it could be written as $O= -iM$ for some hermitian operator $M$. We have

$$ d_t U (t) = -iM U(t) $$

For some hermitian operator $M$. To show it is in fact $H$, we use this formula over some state $|\psi(t)\rangle = U(t)|\psi\rangle$ to get

$$ {d|\psi\rangle\over dt} = -iM|\psi(t)\rangle $$

But we know the same equation is true if we replace $M$ by $H$, so

$$ (M-H)|\psi(t)\rangle = 0 $$

For all state $|\psi\rangle$, which completes the proof.