Heisenberg picture and Path integrals (Zee QFT)

[qft-El]

TL;DR Summary

There is a remark, which was added in the latest edition (2010) of Zee's book "Quantum field theory in a nutshell" which I do not understand. Such remark states that he's working in Heisenberg picture. I think anything there could be also interpreted in terms of Schrödinger picture, though.

Reading the introduction to path integrals given in the latest edition of Zee's "Quantum field theory in a nutshell", I have found a remark which I don't really understand. The author is evaluating the free particle propagator $K(q_f, t; q_i, 0)$
$$\langle q_f\lvert e^{-iHt}\lvert q_i \rangle\underset{\delta t:=t/N}{=}\langle q_f\lvert e^{-iH\delta t}e^{-iH\delta t}...e^{-iH\delta t}\lvert q_i \rangle=\int\prod_{j=1}^{N=1} dq_i \langle q_{f}\lvert e^{-iH\delta t}\lvert q_{N-1}\rangle...\langle q_1\lvert e^{-iH\delta t}\lvert q_i\rangle$$
Using $H=\frac{p^2}{2m}$, we can evaluate $\langle q_{j+1}\lvert e^{-i\frac{p^2}{2m}\delta t}\lvert q_j\rangle\quad\forall j=1...N-1$.
Before doing that, I want to say that this amplitude can be interpreted both in the Schrödinger picture as evolving the state $\lvert q_j\rangle$ for a time $t$ and then computing the amplitude with the position eigenbra $\langle q_{j+1}\lvert$ or as the inner product between two instaneous eigenstates of $\hat{q}(t)$ in Heisenberg picture
$$\langle q_{j+1}, t_0+\delta t\lvert q_j, t_0\rangle=\langle q_{j+1}\lvert\hat{U}(t_0+\delta t)\hat{U}^{\dagger}(t_0)q_j, t_0\rangle=\langle q_{j+1}\lvert e^{-iH\delta t}\lvert q_j\rangle$$
After all, amplitudes should not depend on the picture used. Now, if we insert the (Schrödinger picture) completeness relation of momentum eigenstates
$$1=\int\frac{dp}{2\pi}\lvert p\rangle\langle p\lvert$$
we get

$$\langle q_{j+1}\lvert e^{-i\frac{p^2}{2m}\delta t}\lvert q_j\rangle=\int\frac{dp}{2\pi}\langle q_{j+1}\lvert e^{-i\frac{p^2}{2m}}\lvert p\rangle\langle p\lvert\lvert q_j\rangle=\int\frac{dp}{2\pi}e^{-i\frac{p^2}{2m}}\langle q_{j+1}\lvert p\rangle\langle p\lvert\lvert q_j\rangle$$
Then it's just a matter of evaluating this Gaussian integral. After this the author says: "we are evidently working in the Heisenberg picture." (page 11, 2010 edition)
For the reasons above, I do not understand this remark:

1. As I said one could work in both pictures, so why does he say that? Where is he evidently using Heisenberg picture?
2. Furthermore, he seems to be using Schrödinger picture completeness relations, am I wrong?

For the sake of completeness, here is the part of the book I'm talking about.

Spoiler: Page 10-11

 

[Demystifier]

1. As I said one could work in both pictures, so why does he say that? Where is he evidently using Heisenberg picture?
2. Furthermore, he seems to be using Schrödinger picture completeness relations, am I wrong?

1. He could, but he doesn't. He writes $|q\rangle$, not $|q(t)\rangle$, which is Heisenberg, not Schrodinger, picture.
2. You are wrong. Schrodinger picture completeness would be something like
$$\int dq(t)\, |q(t)\rangle\langle q(t)|=1, \forall t$$

 

[PeterDonis]

I want to say that this amplitude can be interpreted both in the Schrödinger picture as evolving the state for a time and then computing the amplitude with the position eigenbra or as the inner product between two instaneous eigenstates of in Heisenberg picture

No; in the Heisenberg picture, what changes with time are operators, not states. The equations Zee is using are for the relationships between operators (in this case the "position" operator $q$) at different times. Once you pass to relativistic QFT, these operators will be quantum fields at different spacetime points.

 

[qft-El]

1. He could, but he doesn't. He writes $|q\rangle$, not $|q(t)\rangle$, which is Heisenberg, not Schrodinger, picture.
2. You are wrong. Schrodinger picture completeness would be something like
$$\int dq(t)\, |q(t)\rangle\langle q(t)|=1, \forall t$$

According to Modern Quantum Mechanics by J.J. Sakurai and Napolitano, what you have written is the Heisenberg completeness relation.

Spoiler: Source

Furthermore, in Heisenberg picture the position operator is time dependent, as you said. Its eigenstates are different at each instant and fixing time they satisfy such completeness relation. On the other hand, in Schrödinger picture the position operator is time independent and so are its eigenkets, in the sense they are the same at any time.

 

[Demystifier]

Furthermore, in Heisenberg picture the position operator is time dependent, as you said. Its eigenstates are different at each instant and fixing time they satisfy such completeness relation. On the other hand, in Schrödinger picture the position operator is time independent and so are its eigenkets, in the sense they are the same at any time.

OK, let me make a step back and clarify some concepts from a beginning. The dynamics in physics is described by physical quantities that depend on time. In quantum physics there are several ways to do it, e.g. Heisenberg picture where the observables such as $Q(t)$ depend on time, and Schrodinger picture where the state of the system $|\psi(t)\rangle$ depends on time. Mathematically, the state is a vector in a Hilbert space, while observable is an operator on the same Hilbert space. The Hilbert space has a complete orthogonal basis. Strictly speaking the basis is separable (countable), but in physics it is convenient to work with non-rigorous Dirac formalism in which the basis does not need to be separable. An example of non-separable basis is the position basis $\{|q\rangle\}$, which obeys
$$Q|q\rangle=q|q\rangle$$
where $Q$ is the position operator in the Schrodinger picture and $q$ is its eigenvalue. Defining
$$Q(t)\equiv U(t)QU^{\dagger}(t)$$
$$|q(t)\rangle\equiv U(t)|q\rangle$$
where $U(t)=e^{-iHt}$ is unitary operator because the Hamiltonian $H$ is hermitian, we see that
$$Q(t)|q(t)\rangle=q|q(t)\rangle$$
Note that the eigenvalue $q$ does not depend on time. The basis $\{|q\rangle\}$ is complete
$$\int dq\, |q\rangle\langle q|=1$$
which implies
$$\int dq\, |q(t')\rangle\langle q(t')|=1$$
for any $t'$. The state $|\psi(t)\rangle$ in the Schrodinger picture can be expanded in the basis $\{|q\rangle\}$ as
$$|\psi(t)\rangle=\int dq\, |q\rangle\langle q|\psi(t)\rangle$$
or alternatively
$$|\psi(t)\rangle=\int dq\, |q(t')\rangle\langle q(t')|\psi(t)\rangle$$
Note that the last two formulas are both in the Schrodinger picture, because $|\psi(t)\rangle$ depends on $t$. But they use different bases, namely different completeness relations. This means that there is no such thing as the completeness relation in Schrodinger picture or the completeness relation in Heisenberg picture. The two pictures refer to time dependence (or the lack thereof) of the observable and the state, but not of the basis.

 

[qft-El]

OK, let me make a step back and clarify some concepts from a beginning. The dynamics in physics is described by physical quantities that depend on time. In quantum physics there are several ways to do it, e.g. Heisenberg picture where the observables such as $Q(t)$ depend on time, and Schrodinger picture where the state of the system $|\psi(t)\rangle$ depends on time. Mathematically, the state is a vector in a Hilbert space, while observable is an operator on the same Hilbert space. The Hilbert space has a complete orthogonal basis. Strictly speaking the basis is separable (countable), but in physics it is convenient to work with non-rigorous Dirac formalism in which the basis does not need to be separable. An example of non-separable basis is the position basis $\{|q\rangle\}$, which obeys
$$Q|q\rangle=q|q\rangle$$
where $Q$ is the position operator in the Schrodinger picture and $q$ is its eigenvalue. Defining
$$Q(t)\equiv U(t)QU^{\dagger}(t)$$
$$|q(t)\rangle\equiv U(t)|q\rangle$$
where $U(t)=e^{-iHt}$ is unitary operator because the Hamiltonian $H$ is hermitian, we see that
$$Q(t)|q(t)\rangle=q|q(t)\rangle$$
Note that the eigenvalue $q$ does not depend on time. The basis $\{|q\rangle\}$ is complete
$$\int dq\, |q\rangle\langle q|=1$$
which implies
$$\int dq\, |q(t')\rangle\langle q(t')|=1$$
for any $t'$. The state $|\psi(t)\rangle$ in the Schrodinger picture can be expanded in the basis $\{|q\rangle\}$ as
$$|\psi(t)\rangle=\int dq\, |q\rangle\langle q|\psi(t)\rangle$$
or alternatively
$$|\psi(t)\rangle=\int dq\, |q(t')\rangle\langle q(t')|\psi(t)\rangle$$
Note that the last two formulas are both in the Schrodinger picture, because $|\psi(t)\rangle$ depends on $t$. But they use different bases, namely different completeness relations. This means that there is no such thing as the completeness relation in Schrodinger picture or the completeness relation in Heisenberg picture. The two pictures refer to time dependence (or the lack thereof) of the observable and the state, but not of the basis.

Yes. "Heisenberg/Schrödinger picture completeness relations" were a shorthand for "completeness relations respectively of eigenstates of instantaneous position operator in Heisenberg picture or eigenstates of position operator in Schrödinger picture." The distinction between the two pictures couldn't be more clear to me. The core of my question is: what is he doing that makes evident the use of Heisenberg picture? Again, we have that the instantaneous eigenstates
$$\hat{q}(t)\lvert q, t\rangle=q\lvert q, t\rangle$$
satisfy
$$\lvert q, t\rangle=e^{iHt}\underbrace{\lvert q, 0\rangle}_{=\lvert q\rangle\text{ Schrödinger picture eigenket}}$$
For this reason I'm saying that starting from the beginning every amplitude he writes has that meaning

Before doing that, I want to say that this amplitude can be interpreted both in the Schrödinger picture as evolving the state $\lvert q_j\rangle$ for a time $t$ and then computing the amplitude with the position eigenbra $\langle q_{j+1}\lvert$ or as the inner product between two instaneous eigenstates of $\hat{q}(t)$ in Heisenberg picture
$$\langle q_{j+1}, t_0+\delta t\lvert q_j, t_0\rangle=\langle q_{j+1}\lvert\hat{U}(t_0+\delta t)\hat{U}^{\dagger}(t_0)q_j, t_0\rangle=\langle q_{j+1}\lvert e^{-iH\delta t}\lvert q_j\rangle$$
After all, amplitudes should not depend on the picture used.

 

[Demystifier]

The core of my question is: what is he doing that makes evident the use of Heisenberg picture?

Good question! In my opinion, nothing. Zee is rather sloppy at this point. In fact, in this book, Zee is sloppy at many points. The Zee's book may be good for getting a big picture and intuitive ideas, but it's definitely not good for those who care about details and precision.

 

[weirdoguy]

In fact, in this book, Zee is sloppy at many points.

In all of his books

 

[qft-El]

Good question! In my opinion, nothing. Zee is rather sloppy at this point. In fact, in this book, Zee is sloppy at many points. The Zee's book may be good for getting a big picture and intuitive ideas, but it's definitely not good for those who care about details and precision.

To be honest what really puzzles me is that I checked the previous edition and it turns out this remark was added in the latest one...

 

[Demystifier]

To be honest what really puzzles me is that I checked the previous edition and it turns out this remark was added in the latest one...

Interesting! I also had impression that the first edition was better, for other reasons.

 

[vanhees71]

I've not followed the entire thread, but I think Zee is very confusing (if this is really what's in his book).

Let's take the example of first-quantized non-relativistic QM. Then the description of time evolution of wave functions in the position representation are as follows. For simplicity I describe the motion of a single particle in a single spatial dimension.

Schödinger Picture
The entire time evolution is in the states. A pure state is represented by a time-dependent "state ket", $|\psi(t) \rangle$. The time evolution is generated by the Hamiltonian $\hat{H}(\hat{x},\hat{p})$, and I thus assume for simplicity that $\hat{H}$ is not explicitly time-dependent. By definition the "fundamental operators" $\hat{x}$ and $\hat{p}$ and thus all functions dependent on them are time independent.

Since the time dependence is thus entirely on the state kets, it obeys the equation of motion
$$\mathrm{i} \hbar \mathrm{d}_t |\psi(t) \rangle=\hat{H} |\psi(t) \rangle,$$
which obviously is (at least formally) solved by the operator-exponential function,
$$\hat{U}(t,t_0)=\exp \left [-\frac{\mathrm{i}}{\hbar} \hat{H} (t-t_0) \right ].$$
So given the state at $t_0$, $|psi_0 \rangle$ ("prepartion of the system") the state ket at the later time, $t$ is given by
$$|\psi(t) \rangle=\hat{U}(t,t_0) |\psi_0 \rangle.$$
It's also clear that the unitary time-evolution operator fulfills the equation of motion
$$\mathrm{i} \hbar \partial_t \hat{U}(t,t_0) = \hat{H} \hat{U}(t,t_0), \quad \hat{U}(t,t_0)=\hat{1}.$$
The physics is now in the wave function $\psi(t,x)=\langle x|\psi(t) \rangle.$ Here $|x \rangle$ is the (generalized) position eigenvector with eigenvalue $x$: $\hat{x} |x \rangle=x |x \rangle$. Since $\hat{x}$ is independent of time, so is $|x \rangle$.

The wave function, which is independent of the picture of time evolution (modulo perhaps an irrelevant phase factor) thus obeys the equation of motion
$$\mathrm{i} \hbar \partial_t \psi(t,x) = \langle x| \mathrm{i} \hbar \mathrm{d}_t |\psi(t) \rangle=\langle x |\hat{H}(\hat{x},\hat{p})|\psi(t) \rangle=\hat{H}(x,-\mathrm{i} \partial_x) \psi(t,x).$$
Note that $\hat{H}(\hat{x},\hat{p})$ acts in the abstract Hilbert space of kets, while $\hat{H}(x,-\mathrm{i} \partial_x)$ acts in the dense subset of the Hilbertspace $\mathrm{L}^2(\mathbb{R})$ of square Lebesgue-integrable functions, where $\hat{H}(x,-\mathrm{i} \hbar \partial_x)$ is well defined, and where this operator is self-adjoint.

The solution can be formally written as
$$\psi(t,x)=\langle x|\psi(t) \rangle=\langle x|\hat{U}(t,t_0)|\psi_0 \rangle = \int_{\mathbb{R}} \mathrm{d} x' \langle x|\hat{U}(t,t_0|\rangle x' \rangle \langle x' |\psi_0 \rangle=\int_{\mathbb{R}} \mathrm{d} x' U(t,x;t_0,x') \psi_0(x'),$$
and
$$\hat{U}(t,t_0)(t,x;t_0,x')=\langle x|\exp[-\mathrm{i} \hat{H} (t-t_0)/\hbar]|x' \rangle.$$

Heisenberg Picture
Here the time evolusion is entirely on the operators that represent observables, and these observables obey the equation of motion ("canonical quantization": classical-mechanics Poisson brackets are represented by the corresponding commutators of operators divided by $\mathrm{i} \hbar$):
$$\mathrm{d}_t \hat{O}(t)= \frac{1}{\mathrm{i} \hbar} [\hat{O},\hat{H}].$$
This is formally solved again by the time-evolution operator,
$$\hat{U}(t,t_0)=\exp \left [-\frac{\mathrm{i}}{\hbar} \hat{H} (t-t_0) \right ],$$
as
$$\hat{O}(t)=\hat{U}^{\dagger}(t,t_0) \hat{O}(t_0) \hat{U}(t,t_0).$$
This you can check by simply take the time derivative of this equation and
$$\partial_t \hat{U} = -\mathrm{i}/\hbar \hat{H} \hat{U}.$$
The state of the system in the Heisenberg Picture is then described by the time-independent state ket at initial time $t_0$, i.e., by $|\psi \rangle=|\psi_0 \rangle=\text{const}$.

Now in the wave function the entire time-dependence comes from the position eigenkets,
$$\hat{x}(t) |x(t) \rangle=x |x(t) \rangle$$
Obviously because of the solution of the equation of motion for $\hat{x}$ we have
$$|x(t) \rangle = \hat{U}^{\dagger}(t,t_0) |x(t_0) \rangle.$$
We can without loss of generality assume that at $t_0$ both pictures use the same kets and operators. Then $|x(t_0) \rangle_{\text{H}}=|x \rangle_{\text{S}}$, $|\psi(t_0) \rangle_{\text{S}}=|\psi_0 \rangle_{\text{H}}$, where the subscripts H and S stand for Heisenberg and Schrödinger picture, respectively.

The wave function is of course the same in both pictures: Indeed in the Heisenberg picture we have
$$\psi_{\text{H}}(t,x)=\langle x(t) | \psi \rangle=\langle x(t)|\psi_0 \rangle= \langle \hat{U}^{\dagger} x(t_0)|\psi_0 \rangle = \langle x(t_0)|\hat{U} \psi_0 \rangle={}_{\text{S}}\langle x|\psi(t) \rangle_{\text{S}}=\psi_{\text{S}}(t,x).$$
In the Heisenberg picture you have
$$U_{\mathrm{H}}(t,x;t_0,x')=\langle x(t)|x'(t_0) \rangle=\langle x(t_0) | \hat{U}|x'(t_0) \rangle={}_{\text{S}} \langle x|\hat{U}|x' \rangle_{\text{S}}= U_{\mathrm{S}}(t,x;t_0,x').$$
So the notation in the OP is using the Schrödinger rather than the Heisenberg picture to derive this "full propagator" in the path-integral representation.

 

[qft-El]

I've not followed the entire thread, but I think Zee is very confusing (if this is really what's in his book).

Regarding the book itself, I attached the pages of the book in the spoiler at the end of the first post, so you can check that's exactly what it's written there.

So the notation in the OP is using the Schrödinger rather than the Heisenberg picture to derive this "full propagator" in the path-integral representation.

Yes, that's my point. The only thing that bugs me now is:

that I checked the previous edition and it turns out this remark was added in the latest one...

which makes the author's point even more mysterious!

 

[vanhees71]

I never understood, why Zee's textbook on QFT is considered to be a good one, but that's of course a matter of personal taste. If you want a witty but really deep and very clear textbook on QFT, I'd recommend Coleman's QFT lecture notes:

S. Coleman, Lectures of Sidney Coleman on Quantum Field
Theory, World Scientific Publishing Co. Pte. Ltd., Hackensack
(2018), https://doi.org/10.1142/9371

 

[qft-El]

Oh, I have those notes, there is also a free version on arXiv if I recall correctly. Since you take out the topic, the books I use are:

• A Modern Introduction to Quantum Field Theory, Michele Maggiore.
• An Introduction to Quantum Field Theory, Peskin&Schroeder
• The Quantum Theory of Fields, S.Weinberg

I happened to read Zee because I heard about it a lot. I haven't read enough to say what I think about the book, though.

 

[vanhees71]

Yes, there's part of the Coleman lectures on arXiv:

https://arxiv.org/abs/1110.5013

 

[LittleSchwinger]

I've always loved Itzykson and Zuber. Especially their treatment of LSZ reduction.