# What does the time evolution operator look like in Matrix representation?

In Matrix representation of quantum mechanics, using an energy eigenbasis, we have the state vector: $$|\psi(t)\rangle=\left(\begin{matrix} \psi_0(t)\\ \psi_1(t)\\ \psi_2(t)\\ \vdots \end{matrix}\right)$$ and $$|\psi(t)\rangle=\left(\begin{matrix} \psi_0(0)e^{iE_0t/\hbar}\\ \psi_1(0)e^{iE_1t/\hbar}\\ \psi_2(0)e^{iE_2t/\hbar}\\ \vdots \end{matrix}\right)$$ This is often represented as the state vector at time $0$ operated on by the time evolution operator $\hat U$: $$|\psi(t)\rangle=\hat U (t)|\psi(0)\rangle$$ And $\hat U (t)$ is usually represented as: $$\hat U(t)=e^{i\hat Ht/\hbar}$$ Now I know that, in the energy eigenbasis, the Hamiltonian Operator is represented as: $$\hat H=\left(\begin{matrix} E_0&0&0&\cdots\\ 0&E_1&0&\cdots\\ 0&0&E_2&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{matrix}\right)$$ How do you represent an exponential to the power of a matrix? Or is the exponential just shorthand to show $U_{ij}(t)=e^{i H_{ij}t/\hbar}$, such that $$\hat U=\left(\begin{matrix} e^{iE_0t/\hbar}&0&0&\cdots\\ 0&e^{iE_1t/\hbar}&0&\cdots\\ 0&0&e^{iE_2t/\hbar}&\cdots\\ \vdots&\vdots&\vdots&\ddots \end{matrix}\right)$$ And how does this generalize to other eigenbasis?

• Recommended (related) reading: Moler C, Van Loan C. Nineteen dubious ways to compute the exponential of a matrix, twenty-five years later. SIAM review. 2003;45(1):3-49 Feb 15, 2018 at 4:07
• I don't understand what the actual question is here. You already have $\hat U(t)$. What do you mean by "generalize to other eigenbasis"? Feb 15, 2018 at 14:16
• @jjgoings That paper is science at its best. Feb 15, 2018 at 14:51
• @EmilioPisanty I'm pretty sure the question is "what does the notation $e^A$ mean, when $A$ is a matrix?" Feb 16, 2018 at 7:54

## 1 Answer

The exponential of an operator is generally defined via Taylor series as: $$e^{\hat A} \equiv \sum_{n=0}^\infty \frac{\hat A^n}{n!}$$ Which is completely independent of the matrix representation of $\hat A$ in any basis. Now if $\hat A$ is diagonalizable, as in the case of the hamiltonian, one can show that the above Taylor series definition boils down to exponentiating the diagonal elements of the operator in its eigenbasis (in which it is diagonal), which is what you have written. For the general matrix elements of $e^{\hat A}$ in an arbitrary basis, one can first write the matrix elements in its own eigenbasis, and then use a basis transformation.Explicitly, the spectral decomposition of $e^{\hat A}$ is: $$e^{\hat A} = \sum_k e^{A_k} |A_k \rangle\langle A_k|$$ Now imagine you want the matrix elements in the basis $\{|e_i \rangle \}_i$, this can easily be calculated with the above relation: $$\langle e_i|e^{\hat A}|e_j \rangle = \sum_k e^{A_k} \langle e_i|A_k \rangle\langle A_k|e_j \rangle$$ What the above sum actually is depends on the specific operator in question. However, note the important fact that in general, $\langle e_i|e^{\hat A}|e_j \rangle \neq e^{\langle e_i|\hat A|e_j \rangle}= e^{A_{ij}}$. Matrix elements generally can not be exponentiated for non-diagonal matrices.