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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?

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¿You must be assuming the HIlbert space of states is finite dimensional? See the next answer, where it is assumed the dimension is $n$. –  joseph f. johnson Dec 30 '11 at 1:40

6 Answers 6

up vote 5 down vote accepted

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.


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}

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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.

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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.

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Except if the number of dimensions is finite, which I think it has to be or determinant is not defined either. The OP seems to be assuming finite-dimensionality, in which case the power series expansion is good. –  joseph f. johnson Dec 30 '11 at 1:41
    
@joseph: The infinite dimensional determinant arises in quantum mechanics all the time, and there is no point in ignoring it. The way it becomes well defined is that you take its log and differentiate with resepct to a parameter, at which point it can become a convergent sum. It is no different in this respect from the partition function of a statistical system, which is also an infinite dimensional object which reduces to a determinant in special cases. –  Ron Maimon Dec 31 '11 at 10:25
    
Well, at this point it is incumbent on you to calculate the regularised determinant of the expression in the original query, and then we can see whether it allows you to deduce something about the kernel or eigenvalues, and I will be able to see what, if anything, goes wrong with the shift operator. (Furthermore, if you are so optimistic about the regularised determinant and willing to see it nascent in this post, what is there to prevent the exponential function or any holomorphic function from being optimistically interpreted in the same way) If you did this, it would be a valuable answer –  joseph f. johnson Jan 2 '12 at 9:43
    
You see, although it arises as a partition function, that hardly addressed the issue of eigenvalues and kernels as far as I know, so you really ought to put this thing up. –  joseph f. johnson Jan 2 '12 at 9:44

ADDED REMARK: This is an answer to Antillar Maximus' question:

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

My answer. The importance of unitary operators in QM relies upon a pair of fundamental theorems, known as Wigner's and Kadison's theorem respectively.

Consider a quantum system described in a Hilbert space ${\cal H}$. Pure states are represented by vectors $\psi \in \cal H$ with $||\psi||=1$. Actually a state is the whole set $[\psi ] := \{e^{ia} \psi \:|\: a \in R\}$ when $||\psi||=1$, because there is no physical way to fix the phase $e^{ia}$ and all physical quantities are independent form it. Notice that, with the introduced notation, one hence has $[\psi]= [\phi]$ if and only if $\psi = e^{ia} \phi$ for some $a \in R$. Henceforth $S(\cal H)$ will denote the space of pure states in $\cal H$. (It coincides to the complex projective space of $\cal H$ but it does not matter now.)

If you have two states $[\psi] \neq [\phi]$, represented by normalized vectors $\psi$ and $\phi$, as is well known $$P([\psi],[\phi])| := |\langle \phi|\psi \rangle|^2$$ is a physical quantity representing the probability transition from the former to the latter state (in both directions). Notice that, as it must be, that quantity is not affected by changes of the arbitrary phases embodied in the states.

There are very important transformations of state called (quantum) symmetries.

A symmetry, in the sense of Wigner, is a map from the space of the states to the space of the states, say $S(\cal H) \ni [\psi] \to [\psi'] \in S(\cal H)$ verifying two requirements:

(1W) the map is bijective ("into" and "onto"). In other words, with a given symmetry one can moves form a given state to any other state, and the action of the symmetry is reversible;

(2W) the map preserves the transition probabilities: $|\langle \phi|\psi \rangle|^2 = |\langle \phi'|\psi' \rangle|^2$.

There are lots of physical transformations of states verifying (1W) and (2W). For instance: time evolution, charge conjugation, parity reflection, time reversal, the action of Poincaré's group or Galileo's group, and many many others. Sometimes it turns out that a supposed symmetry actually does not exist (like parity reflection for systems subjected to weak interaction).

The natural question is therefore:

What is the most general form of a symmetry, i.e. a map from the set of the states to the same set satisfying (1W) and (2W)?

The answer was given by Wigner with a celebrated theorem commonly known as Wigner's theorem (even if Wigner gave so many contributions to the mathematical foundation of QM that it sounds very reductive to call that theorem Wigner's theorem simply).

THEOREM (Wigner). Let $\cal H$ be a complex separable Hilbert space and $f : [\psi] \to [\psi']$ a map transforming states to states and satisfying (1W) and (2W). Then, there exists either unitary or anti-unitary (depending on $f$) operator: $$U_f : \cal H \to \cal H\:,$$
determined by $f$ up to a constant phase (i.e. $U_f$ can only be modified to $e^{ia}U_f$ with $a\in R$, preserving its properties), such that: $$f ([\psi]) = [U_f \psi]\quad\quad \mbox{for every normalized vector $\psi \in \cal H$}$$

In particular, taking Wigner's theorem into account, we may forget of states as classes of vectors and we can safely use vectors as usual to represent states. (Life is not so easy, since an annoying and difficult problem arises as soon as one tries to represent a group of symmetries this way)

Let us come to the other, analogous, important theorem due to Kadison. Everybody knows that there exists another, more general, notion of state in QM. I mean a mixed state. A mixed state is represented by a positive, trace class operator $\rho : \cal H \to \cal H$ with unit trace $tr(\rho)=1$. The set $M(\cal H)$ of mixed states is a convex body in the real linear space of self-adjoint bounded operators on $\cal H$. This means that if $p_1,\ldots, p_n \in [0,1]$ and $\rho_1,\ldots, \rho_n \in M(\cal H)$ verify $\sum_i p_i =1$, then: $$\sum_{i=1}^n p_i \rho_i \in M({\cal H})\:. \qquad (1)$$

The expectation value of an observable $A$ respect to the state $\rho$ is

$$\langle A \rangle_\rho := tr(\rho A)\:,$$ (where some hypotheses are necessary on the domain of the self-adjoint operator $A$).

Pure states $[\psi]$ are a subcase of mixed ones, those of the form $|\psi\rangle \langle \psi|$ (notice that the arbitrary phase affecting $\psi$ does not matter here). Actually it is possible to prove that pure states are all of the states in $M(\cal H)$ such that cannot be decomposed as in (1) (for some non-trivial decomposition): They are the extremal elements of $M(\cal H)$.

Every mixed state $\rho$, in general, admits many decompositions like that in (1). One is the most natural, that obtained by the spectral decomposition: $$\rho = \sum_{k=0}^{+\infty} p_k |\psi_k \rangle \langle \psi_k|\:,\qquad (2)$$ where now the $p_k$s are the eigenvalues of $\rho$, that (by hypotheses on $\rho$) belong to $[0,1]$ and their sum is $1$. (2) has a natural interpretation as a classical mixing, with classical probabilities $p_k$, of pure quantum states $|\psi_k \rangle \langle \psi_k|$. Actually in general there are many ways to decompose $\rho$ into such a form, it is untenable separating classical probabilities (carried by the $p_n$s) and quantum probabilities (embodied in the $|\psi_k \rangle \langle \psi_k|$s).

Within this context, there is another notion of quantum symmetry due to Kadison (actually it was formulated using a dual, equivalent, approach referring to the lattice of orthogonal projectors).

A symmetry (in the sense of Kadison) is a map $f: M({\cal H})\ni \rho \mapsto \rho' \in M(\cal H)$ that verifies:

(1K) it is bijective;

(2K) it is convex linear, i.e., $$f\left( \sum_{i=1}^n p_i \rho_i \right) = \sum_{i=1}^n p_i f(\rho_i)\quad \mbox{for $p_i \in [0,1]$ with $\sum_i p_i=1$.}$$

The second conditions physically means that if we construct a mixed state $\rho$ by classically mixing some other states $\rho_i$ with respective classical probabilities $p_i$, the action of the symmetry preserves this composition (the classical probabilities) with respect to the transformed states.

Such a notion of quantum symmetry seems to be quite different from that proposed by Wigner. Nonetheless the two notions coincide in view of the celebrated (however less known than Wigner's one) Kadison's theorem.

THEOREM (Kadison). Let $\cal H$ be a complex separable Hilbert space and $f : M({\cal H}) \ni \rho \to \rho' \in M(\cal H)$ a map transforming states to states and satisfying (1K) and (2K). Then there exists either unitary or anti-unitary (depending on $f$) operator: $$U_f : \cal H \to \cal H\:,$$
determined by $f$ up to a constant phase (i.e. $U_f$ can only be modified to $e^{ia}U_f$ with $a\in R$, preserving its properties), such that: $$f (\rho) = U_f \rho U_f^\dagger \quad\quad \mbox{for every $\rho \in M(\cal H)$.}$$

It is obvious that, looking at the associated (anti) unitary operators, a Kadison symmetry defines a Wigner symmetry and vice-versa.

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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.

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In infinite dimensions, I think one can still do something, and that is look for the poles of the power series (these are bounded operators so it makes sense) for $1\over I-U(t)e^{itE}$ regarded as a series in $E$, a real variable. A pole leads to a lack of invertibility here, too. –  joseph f. johnson Dec 30 '11 at 2:55
    
This is a repeat of the fine answer by Adam Zalcman. There is no need for duplicate answers in general, but I won't downvote you, because you probably didn't know. –  Ron Maimon Dec 30 '11 at 11:42
    
There is a substantive difference with great pedagogical significance: Mr. Zalcman includes an unnecessary step and leaves out a necessary one: he decided it was important to explain the homogeneity degree of det, but unimportant to explain why the eigenvalues are the roots of det(tI-A). To me, this seemed to misjudge the level of the question. Anyway, it was an omission. –  joseph f. johnson Dec 30 '11 at 17:37
    
His answer is superior to yours pedagogically, precisely because he doesn't shy away from the implication of the homogeneity degree! He explains that scaling an operator by a constant C introduces a factor of C^N where N is the dimension of the space. There is no point in separating the finite and infinite dimensional cases, because you need to define the determinants arising in quantum mechanics by a limiting procedure from the finite dimensional case anyway. This limiting process is best left explicit, and you check at the end if there is no dependence on the regulator. –  Ron Maimon Dec 31 '11 at 10:17
    
Perhaps I am wrong, but when you extend the definition of det. that way, it is not clear that you can still conclude to the non-invertibility of the linear transformation of which it is the det, so the rest of the argument, which depends on their being a kernel, is left hanging. So using the power series is more direct and allows one to immediately conclude there is a non-trivial kernel. Furthermore, if one says « implications of the ....», well, that means « implications ». There was nothing in the other answer about the many things the homogeneity degree can lead to in regularisation..... –  joseph f. johnson Dec 31 '11 at 16:54

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!

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You are totally mistaken about notation, operators, real analysis, and the exponential. You might as well withdraw the answer. An operator on a normed space can certainly be plugged into a formal power series and the topic of spectral analysis studies when it converges to another operator and when it doesn't. This is simple mathematics, see e.g. Serge Lang Real Analysis. The point is, $A$ an operator, $A-A^2+A^3-\dots$ has partial sums which only rely on the algebra of transformations: « multiplication » means composition of transformations, which, since they are maps, is well-defined. –  joseph f. johnson Dec 30 '11 at 17:40
    
Since each partial sum is a transformation, you can ask whether it has a norm or not (since it acts on a normed space). The notion of convergent sequence can be defined as usual using these norms of the operators. That's strong convergence. Point-wise convergence of operators can be defined even more simply, and weak convergence can be defined in a more complicated way. The exponential converges strongly and pointwise for any bounded operator, and unitary operators are certainly bounded. Amazingly, it converges even for many interesting unbounded operators.... –  joseph f. johnson Dec 30 '11 at 17:48
    
I don't wish to argue, but the moment you say "on a normed space" are you not working with a representation? –  Antillar Maximus Dec 30 '11 at 19:23
    
I am happy to answer questions. I could have said that more clearly, but: no. Unless you wish to say that as soon as one says « operator » or « transformation » one is implicilty working in a representation, which would be a valid point of view, I suppose. By definition, a transformation is a map on some space. Here it is a linear normed space. The norm is certainly independent of basis or coordinates, so if by « representation » you mean « matrix representation » (and your post is very confused on this issue) then no, again, the operator norm is independent of the matrix representation. –  joseph f. johnson Dec 30 '11 at 19:37
    
Furthermore, if one works in abstract « operator algebras » where the operator actually isn't an operator, just an object satisfying certain axioms, then if the algebra is a normed algebra, all the same power series, analytic continuation, convergence goes through for exponential and other meromorphic functions. –  joseph f. johnson Dec 30 '11 at 19:39

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