What does a unitary transformation mean in the context of an evolution equation?

Question

Let be the unitary evolution operator of a quantum system be $U(t)=\exp(itH)$ for $t >0$.

Then what is the meaning of the equation

$$\det\bigl(I-U(t)e^{itE}\bigr)=0$$

where $E$ is a real variable?

Answer 1

Multiply both sides of your equation

\begin{equation} \det(I-U(t)e^{itE})=0 \end{equation}

by $e^{-intE}$ where $n$ is the number of dimensions of the state vector space. We obtain

\begin{equation} \det(e^{-itE}I-U(t))=0 \end{equation}

(See below for how this works.) This is a special case of the following equation

\begin{equation} \det(\lambda I - A) = 0 \end{equation}

whose solutions $λ_k$ are precisely the eigenvalues of operator $A$.

Hence, the meaning of your equation is:

Each $e^{-itE}$ satisfying your equation is an eigenvalue of the unitary operator $U(t)$.

Note that all eigenvalues of unitary operators are complex numbers with absolute value 1.

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Above we used the following property of determinant: for any scalar $b$

\begin{equation} b^n\det(A) = \det(bA) \end{equation}

Determinant of operator $A$ is defined by Leibniz formula as

\begin{equation} \det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{k=1}^{n}a_{\sigma(k),k} \end{equation}

which implies that for scalar $b$

\begin{equation} b^n\det(A) = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma)\prod_{k=1}^{n}ba_{\sigma(k),k}=\det(bA) \end{equation}

Answer 2

I would like to elaborate Adam Zalcman's answer, from a physical angle. What Adam actually showed is that all the eigenvalues of the time-evolution operator are of the form $e^{i\phi}$ where $\phi$ is real. The mathematical implication is that $U$ does not change the norm of states.

Let's look at the systems eigen-states, $\{|n\rangle\}$, which are defined by $$H|n\rangle=\epsilon_n|n\rangle$$ These states span the whole Hilbert space, so knowing how $U$ acts on them tells you everything you need to know about time evolution of an arbitrary state. Note that these states are also eigenvectors of $U$, because $$\begin{align} U(t)|n\rangle&=e^{\frac{i}{\hbar} H t}|n\rangle = \sum_k \frac{\left(\frac{i}{\hbar}Ht\right)^k}{k!}|n\rangle\\ & = \sum_k \frac{\left(\frac{i}{\hbar}\epsilon_n t\right) ^k }{k!}|n\rangle = e^{\frac{i}{\hbar} \epsilon_n t}|n\rangle \end{align}$$ and, indeed, each eigenvalue is of the form $e^{i\phi}$. This means physically that each eigenstate evolve in a very simple way - simply by changing its phase. An arbitrary state is of the form $$|\psi\rangle=\sum_n c_n |n\rangle$$ and its norm is $$\sqrt{\langle\psi|\psi\rangle}=\sqrt{\sum_{n,m}c_m^*c_n \langle m|n\rangle}=\sqrt{\sum_{nm}c_m^*c_n\delta_{mn}}=\sqrt{\sum_n |c_n|^2}$$ Since application of $U$ changes each $c_n$ only by its phase, it does not change the norm of $|\psi\rangle$.

BTW, since the absolute phase is not measurable, this implies that if the system is in a pure an eigenstate, it does not evolve in time. However, if the system is in a superposition of eigenstates, each eigenstate evolves with a different phase, according to their different energies, their relative phase changes and this is what causes stuff to change. Here's a nice applet that allows you to play with that.

Answer 3

Just a mathematical note in response to the previous answer:

$e^{i \hat{H} t /\hbar}$ is not defined as the exp-series, although it is common to define it so in physics textbooks. But it is not possible to do this as the series is generally not converging (in the operator norm). One has to use the spectral calculus, in which the "calculation" $$ e^{i \hat{H} t /\hbar} |n\rangle = e^{i E_n t /\hbar} |n\rangle $$ becomes (some kind of) a definition.

Answer 4

If the dimension of the state space is finite, say $n$, then your question makes sense since the determinant makes sense.

Now suppose that $E$ is a real number such that (for all $t$) $$\det (I-U(t)e^{itE} ) =0.$$ Your equation implies that $I-U(t)e^{itE}$ is not invertible (if it were invertible, its determinant would be non-zero). This implies that there exists a non-zero vector $v$, a state, for which $$v=U(t)e^{itE}v$$ and hence $e^{-itH}v=e^{itE}v$ and hence $$ Hv = -Ev$$ so $v$ is an eigenvector with eignevalue $-E$, i.e., $-E$ is an energy level of the system and when it is in the state $v$ it will, if measured, produce the result of energy$ = -E$ with probability one.

But in infinite dimensions you can't be so direct.

Answer 5

A more general question would be, why is a unitary transformation useful?

A unitary transformation preserves the norm, i.e the norm is invariant under basis transformations (as stated by others above). But why is this useful? This is quite useful because in many cases, we cannot make measurements in the space of interest, but we can always transform to an accessible isomorphic space. From the time evolution expression you have, it is important to identify that the Hamiltonian is the infinitesimal generator of time evolution.

The exponential notation is just that... a notation for a particular combination of operators. I disagree with the claim about convergence etc being applicable to an operator. As far as I know, no such formalism exists for operators. If you want to talk about real analysis, you can only do so with respect to a representation of the operator, not the operator itself. This critical distinction is often ignored in introductory QM (and fatally so!).

• Operators = abstract objects/transformations.
• Representation = What we get when an Operator acts on a basis set.

+1 for a very good question that anybody studying intro QM ought to have!