Time evolution amplitude of path integrals and the time evolution of localized states

Question

I have a conceptual question regarding the time evolution amplitude in the path integral formulation of quantum mechanics. I apologize for this long post, but I think this long text helps with stating the problem and making the actual questions clear.

Since the question is about the subtleties of the time evolution amplitude, let me introduce the "time evolution amplitude" [1], which is also called "time evolution kernel" [2], as it is laid out in some of the literature:

Definition of Kleinert (Ref. [1], page 44 (page number of PDF file: 87))
Kleinert uses in Ref. [1] the definition $$ (\mathbf{x}_b t_b | \mathbf{x}_a t_a) \equiv \langle\mathbf{x}_b | \hat{U}(t_b,t_a) | \mathbf{x}_a \rangle \qquad\qquad (1)$$ and notes that in the case of a time-independent Hamiltonian this can be written as $$ (\mathbf{x}_b t_b | \mathbf{x}_a t_a) = \langle\mathbf{x}_b | \exp(-\mathrm{i}\hat{H}(t_b-t_a)/\hbar | \mathbf{x}_a \rangle \,. \qquad\qquad (2)$$

Definition of Wipf (Ref. [2], page 10 (page number of PDF file: 3))
Wipf calls this quantity in Ref. [2] "time evolution kernel" and introduces it in the context of analyzing the time dependence of the wave function as $$K(t,q,q') = \langle q| e^{-\mathrm{i}t H / \hbar} | q'\rangle \,, \qquad\qquad (3)$$ where he denotes the position eigenstates by $|q\rangle$. Interestingly, he notes afterwards "It is the probability amplitude for the particle to propagate from $q'$ at time 0 to $q$ at time $t$ and is occasionally denoted by $$K(t,q,q') \equiv \langle q,t | q',0\rangle \quad \textrm{"} \,. \qquad\qquad (4)$$

First thoughts about this
When interpreting this, we face the little difficulty that both authors use for definitions the symbol $\equiv$ and not $:=$ or $=:$, which have in contrast to the first one the advantage of being directed. Nevertheless, in the case of Kleinert's exposition it is clear that he uses the matrix element of the time evolution operator $\hat{U}$ to define $(\mathbf{x}_b t_b | \mathbf{x}_a t_a)$. It is important to note that for some reason, which he does not tell the reader :-), he avoids using the standard Dirac brackets here and uses round brackets instead.

In the case of the script of Wipf this interpretation can be a bit more difficult: However, because of the order of Wipf's discussion it is in my opinion not far fetched to claim that he first defines the time evolution kernel via matrix elements of the time evolution operator and then proceeds to define the overlap $\langle q,t | q',0\rangle$ based on this kernel function.

In this context, it is interesting to explore what happens, if one reads the round brackets of Kleinert's exposition as Dirac brackets: Accordingly, we now want to evaluate the expression $ \langle \mathbf{x}_b t_b | \mathbf{x}_a t_a\rangle $. We note that both, Kleinert and Wipf, use the same sign convention for the time evolution of states in the Schrödinger picture: $$ | \psi_S(t) \rangle = e^{-\mathrm{i}\hat{H}t/\hbar} | \psi_S(0) \rangle \qquad\qquad (5) \,,$$ see page 10 (page 3 of the PDF file) of Ref. [2]. The analog can be found on page 34 (page 77 of the PDF file) of Ref. [1]. If we time evolve the localized state $| \mathbf{x}_a\rangle$ in the same way as a generic state $|\psi\rangle$, we obtain $$ \langle \mathbf{x}_b, t_b | \mathbf{x}_a, t_a\rangle = \langle \mathbf{x}_b, 0 | (e^{-\mathrm{i}\hat{H}t_b/\hbar})^\dagger \, e^{-\mathrm{i}\hat{H}t_a/\hbar} | \mathbf{x}_a, 0\rangle = \langle \mathbf{x}_b | e^{-\mathrm{i}\hat{H}(t_a-t_b)/\hbar}) | \mathbf{x}_a \rangle \,. \qquad\qquad (6) $$ So, with this short calculation we arrived at an expression with the opposite sign as Eq. (2) by ignoring the fact that the brackets on the left side of Eq. (2) are not Dirac brackets. I assume that this calculation is bascially the reason, why in Eq. (2) round brackets are used and why in Eq. (4) the overlap of the localized states at different times is defined via the kernel function. Reading the discussion of the time evolution amplitude in Ref. [3] strengthens this impression.

Definition of Herzog (Ref. [3], page 2)
Herzog is similar in the notation to Wipf and gives the relation $$ \langle x_f, t_f | x_i, t_i \rangle = \langle x_f | e^{-\mathrm{i} (t_f -t_i)/\hbar} | x_i \rangle \,, \qquad \qquad (7) $$ whereby he made some important statements regarding $|x,t\rangle$ before. He uses for the time evolution the same sign convention as the others: $|\psi(t)\rangle_S = e^{-\mathrm{i} H t /\hbar } |\psi\rangle_H$ ($S$ for Schrödinger picture and $H$ for Heisenberg picture). In the next step he uses $\langle{x} | \psi(t) \rangle_S = \psi(x,t)$ and $\psi(x,t) = \langle x | e^{-\mathrm{i} H t/\hbar} | \psi \rangle_H$ to motivate the definition $$ | x, t\rangle \equiv e^{\mathrm{i} H t/\hbar} | x\rangle \,. \qquad \qquad (8)$$ So, this discussion is very clear. The statement regarding the transition amplitude is basically the same as in the scripts of Kleinert and Wipf, however, the problem with unusual time evolution of $|x,t\rangle$ is directly visible.

Final discussion
Reading the correspondings parts of the scripts of Kleinert, Wipf, and Herzog makes it clear that there is some subtlety in the time evolution of the localized state here: It has a different sign than the time evolution of the other state, what can be seen by comparing Eq. (8) with Eq. (5). While Wipf and Herzog determine this time evolution of localized states with defintions, Kleinert circumvents this problem by using round brackets implying that $|\mathbf{x}_a t_a)$ might be not directly comparable with $|\mathbf{x}_a t_a \rangle$.

While some things might have become clear, there are also some questions:

• We have here the localized states which time evolve with the opposite sign than the normal states. But are also localized states with the "normal" time evolution, i.e., localized states obeying Eq. (5), thinkable?

  Here are some thoughts: If yes, we would distinguish $| x_a, t_a )$ evolving with a plus (combining notation from Kleinert with the more explicit equations from Herzog) from $| x_a, t_a \rangle$ evolving with a minus. $| x_a, t_a )$ would be useful for describing the time evolution amplitude, while $| x_a, t_a \rangle$ could be useful for analyzing the time evolution of a physical system with a localized state as initial state. One could calculate, e.g., $\langle n | x_0,t \rangle$, the probability amplitude for finding the system initially localized at $x_0$ after some time $t$ in the eigenstate $n$ of some operator.

• Some claim, that $ | x, t ) = e^{\mathrm{i}\hat{H} t /\hbar} | x, 0 )$ is a Heisenberg state. See, e.g., this Physics SE answer. I wonder whether this is correct. In my eyes ${}_H\langle x_f, t_f | x_i, t_i \rangle{}_H$ has basically, two problems:

  • First, Heisenberg states are states, which are not time-evolving. They are usually only defined for one reference time, e.g.. $t=0$. See for example Ref. [3]. If you need the state at a different time, evolve it with the operator or use the Schrödinger state at a fixed time, which is an argument of the state. This understanding of Heisenberg states is violated by the interpretation of these states as Heisenberg states.
  • Secondly, if one uses a Heisenberg state and evolves it to some other point in time, one has just the usual time evolution with the usual sign, see page 1 of Ref. [3].

  In some context I can make some sense of this relation to the Heisenberg picture. If I see $ | x, t )$ as part of a projection operator $P_x(t) = | x, t ) ( x, t | $, in the Heisenberg picture this operator evolves according to $P_x(t) = e^{\mathrm{i}\hat{H} t /\hbar} | x, 0 ) ( x, 0 | e^{-\mathrm{i}\hat{H} t /\hbar} = | x, t ) ( x, t | $ yielding the correct behavior for $| x, t )$.

[1] H. Kleinert, "Path Integrals - in Quantum Mechanics, Statistics, Polymer Physics, and Financial Markets", http://users.physik.fu-berlin.de/~kleinert/b5/psfiles/pi.pdf
[2] A. Wipf, "Path Integrals", URL of chapter 2: https://www.tpi.uni-jena.de/~wipf/lectures/pfad/pfad2.pdf
[3] C. Herzog, "Notes on the Path Integral - Physics 305", http://insti.physics.sunysb.edu/~cpherzog/phys405fall2017/pathintegral.pdf

Answer 1

I) OP's question (v2) seems to be mainly spurred by 2 common confusions:

1. The Heisenberg instantaneous position eigenstate $|x,t_0\rangle_H $ does not evolve in time $t$ but does depend on a time parameter $t_0$.

2. The dependence of the time parameter in the Heisenberg instantaneous position eigenstates $$ | x, t_f \rangle_{H}~=~ e^{i\hat{H}\Delta t/\hbar} | x, t_i \rangle_{H}, \qquad\Delta t~:=~t_f-t_i,\tag{1} $$ is opposite the time-dependence in the Schrödinger picture: $$ | \psi (t_f) \rangle_{S}~=~ e^{-i\hat{H}\Delta t/\hbar} | \psi (t_i) \rangle_{S}.\tag{2} $$ (Here we have for simplicity assumed that the Hamiltonian $\hat{H}$ has no explicit time dependence.)

See also e.g. this, this & this related Phys.SE posts.

Answer 2

I think I have found an answer to my question and also to my discussion with Qmechanic regarding the nature of Heisenberg states. The book "Modern Quantum Mechanics" (second edition) by Sakurai and Napolitano discusses in its chapter 2.2 the Schrödinger picture (SP) and the Heisenberg picture (HP) in much more detail than some of the other literature on QM and offers by that an explanation. Pages 82 and 88 (numbers of the second edition) are especially helpful.

Heisenberg states and Schrödinger states
First, the book introduces Heisenberg states and Schrödinger states in a very clear manner by emphasizing the distinction between the "reference time" (my own nomenclature) $t_0$ and the time parameter $t$. The "Heisenberg-picture state ket" is given by $$|\alpha,t_0;t \rangle_H = | \alpha, t_0\rangle\,. \qquad \qquad (1)$$ It is frozen at $t_0$ and independent of $t$. The "Schrödinger-picture state ket" is given by $$|\alpha,t_0;t \rangle_S = \hat{U}(t,t_0) (| \alpha, t_0\rangle = e^{-\mathrm{i}\hat{H}(t-t_0)/\hbar} | \alpha, t_0\rangle\,\,, \qquad \qquad (2)$$ where I assumed a time-independent $\hat{H}$.

That implies that these state have the following coincidence property: $$ \forall t: \quad |\alpha,t_0;t \rangle_H = |\alpha,t_0 \rangle = |\alpha,t_0;t_0 \rangle_S \,. \qquad \qquad (3)$$

Heisenberg states with different reference times
Additionally, we can derive from Eqs. (1), (2), and (3) how "Heisenberg-picture state kets" with different reference times are related: $$ |\alpha,t_1;t \rangle_H = |\alpha,t_1;t_1 \rangle_S = \hat{U}(t_1,t_1) |\alpha,t_1\rangle = \hat{U}(t_1,t_0) |\alpha,t_0\rangle \\ = \hat{U}(t_1,t_0) \hat{U}^\dagger(t_0,t_0) |\alpha,t_0;t_0\rangle_S = \hat{U}(t_1,t_0) |\alpha,t_0;t_0\rangle_S = \hat{U}(t_1,t_0) |\alpha,t_0,;t\rangle_H \,. \qquad (4) $$ I calculated this in a bit lengthy way, in order to use directly and explicitly Eqs. (1) and (2). This calculation shows that "Heisenberg-picture state kets" differing in the reference times, are related by the "ordinary" time evolution operator for ket states with the standard minus sign.

( We can also derive the equation $|\alpha,t_0;t>_H = e^{+i\hat{H}(t'-t_0)/\hbar} |\alpha,t_0;t'>_S$, which can be found (in the more common notation without $t_0$) in some literature. Here we have a different sign, but this is also a different kind of relation. )

Transition amplitudes in the Heisenberg picture and the Schrödinger picture
An important step to avoid possible confusion is to discuss what this implies for transition amplitudes (and analogously for wave functions). I try to briefly summarize the statements of Sakurai and Napolitano:

A transition amplitude is an overlap of a bra and ket, whereby the ket is "state ket" and the bra is a "base bra" (or vice versa "base ket" and "state bra"). In the Schrödinger picture the "state ket" has the usual time evolution, while the "base bra" is just a time-independent eigentstate of some operator. Assuming a common reference time, which we omit here, we have in the Schrödinger picture: $$ \mathcal{A}(t) = \langle b | \cdot | a; t \rangle_S = \langle b | \cdot U(t) | a \rangle \qquad \qquad (5) $$ In the Heisenberg picture one uses a time-independent "Heisenberg-picture state ket", but has a base bra, which is time-dependent with the for a bra unusual sign. Thereby, the transition amplitude is independent of the picture in use, as it has to be. $$ \mathcal{A}(t) = (\langle b | U(t) ) \cdot | a; t \rangle_H = (\langle b | U(t) ) \cdot | a \rangle = \langle b | \cdot U(t) | a \rangle \qquad \qquad (6) $$ So a "Heisenberg-picture base ket" time evolves with the opposite sign than a "Schrödinger-picture state ket". (The "Heisenberg-picture state bra" and the "Schrödinger-picture base bra" have no time dependency.)

Conclusions & answers

• Now, we have a method for evaluating the overlap, i.e., the propagation amplitude, in different pictures. The results are, as it should be, the same for the different variants and pictures: The transition amplitude is given by a "SP base bra" at $x_f$ (-> reference time $t_f$) and a "SP state ket" at $x_i,t_i$: $$ \langle x_f | x_i, t_f; t_i\rangle_S = \langle x_f| \hat{U}(t_i,t_f) | x_i, t_f \rangle = \langle x_f| \hat{U}(t_i,t_f) | x_i\rangle = \langle x_f| e^{-i\hat{H}(t_i-t_f)/\hbar} | x_i\rangle\,, \qquad\qquad (7)$$ where Eq. (2) and the value of the reference time $t_f$ were used. As one would expect, this calculation in the SP can be also carried out with a "SP state bra" and a "SP base ket": $$ {}_S\langle x_f, t_i; t_f| x_i\rangle = (\hat{U}(t_f,t_i) | x_f,t_i\rangle)^\dagger |x_i\rangle = \langle x_f | \hat{U}^\dagger(t_f,t_i) | x_i\rangle = \langle x_f | \hat{U}(t_i,t_f) | x_i\rangle \qquad\qquad (8) $$ In the Heisenberg picture we can use a "HP state ket" together with a "HP base bra" and $t_i$ as reference time: $$ (\langle x_f | \hat{U}(t_i,t_f) )| x_i, t_i;t\rangle_H = \langle x_f | \hat{U}(t_i,t_f) | x_i, t_i\rangle = \langle x_f | \hat{U}(t_i,t_f) | x_i \rangle \qquad\qquad (9) $$ The sign is the same as in Eq. (6) of the question post, but opposite to Eq. (7), which is used for path integrals. From this I would answer my second question that the time evolution with the opposite sign is the one of "HP base kets" but not the transformation behavior (regarding reference times) of "HP state kets". The latter is what I would call a Heisenberg state. Therefore, I personally would not call $|x,t)$ a Heisenberg state, "HP base ket" should be a good classification.
• As it is explained in chapter 2.5 of the book by Sakurai and Napolitano the path integral formulation can be obtained by evaluating $\psi(x,t) = <x|\psi,t>_S = <x| U(t) |\psi> = (x,t|\psi>$, where I set the reference times to 0. Hereby, $<x| U(t) = (x,t|$ is my notation for the "HP base bra". After inserting identies at different times consisting of "HP base bra/kets", one obtains the the following overlaps used for the path integral formulation: $ (x_i,t_i|x_{i+1},t_{i+1})$.
• This answers my first question, whether localized states with the "normal" time evolution are thinkable. Localized states with the "normal" time evolution are "Schrödinger-picture state kets", while the ones with the opposite time evolution are "Heisenberg-picture base kets". (The transformation between the time-independent "Heisenberg-picture state kets" at different reference times has the "normal" sign, see Eq. (4).)
• Regarding the answer by Qmechanic, I have a few comments: The "Heisenberg instantaneous position eigenstate" of Qmechanic seems to be the "HP base bra/ket" in the nomenclature of Sakurai and Napolitano. I agree with Qmechanic regarding the time evolution of the state and also the name for it is plausible for me, as it can be also understood as the eigenstate of $\hat{x}^{(H)}(t) = \hat{U}^\dagger(t) \hat{x} \hat{U}(t)$ with the same eigenvalue for all $t$ as it is described in this post by Qmechanic. In contrast to Qmechanic, I think that this state can be described as time evolving, i.e., a function of time (not only reference time). That is what Sakurai and Napolitano say about "HP base kets/bras". In contrast to this type of Hilbert space vector, "HP state bras/kets" depend only on some reference time parameter, but are constant for all times, see Eq. (1). Regarding this aspect I find the answer of Qmechanic and its notation a bit misleading. I tend to associate bras/kets with an index $H$ as "HP state bras/kets", since often only these states are discussed and denoted with this index. (When discussing the calculation of expectation values in SP and HP, there is usually no need to discuss the "base ket/bras". Only "state ket/bras" and operators are used.)