QFT for the gifted amateur: translation of prob. density

[haushofer]

TL;DR Summary

Question about the book "QFT for the gifted amateur"

Dear all,

I was reading through the book "QFT for the gifted amateur" because I'm currently working on a popular science book about symmetries. Chapter 9 is about transformations of the wave function. On page 80 the book says

Translating the particle shouldn't change the probability density. Thus

$$< \psi(x)| \psi(x)> \, = \, < \psi(x+a)| \psi(x+a)> \, = \, < \psi(x)| U^{\dagger}(a) U(a) | \psi(x)>$$

It's the second equality that confuses me: doesn't the statement $< \psi(x)| \psi(x)> = < \psi(x+a)| \psi(x+a)>$ say that the probability density is constant everywhere (its value at x is the same as its value at x+a for arbitrary a)? What am I missing here? Or does the book misses some primes on the fields and should it read $< \psi(x)| \psi(x)> \, =\, < \psi'(x+a)| \psi'(x+a)>$?

 

[vanhees71]

I don't know, what the author means by his bracket notation. It's not the standard Dirac notation. In standard Dirac notation there are abstract vectors $|\psi \rangle$ and "wave functions", which are the "components" of the abstract vectors in the (generalized) position eigenbasis, $\psi(x)=\langle x|\psi \rangle$. Then you have
$$\langle \psi|\psi \rangle=\int_{\mathbb{R}} \mathrm{d}^3 x \langle \psi|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d}^3 x |\psi(x)|^2.$$
Thus a translation operator, defined as
$$\hat{T}_a \psi(x)=\psi(x+a),$$
is a unitary operator, because for all $\psi \in \mathbb{R}^2$ obviously
$$\langle \hat{T}_a \psi|\hat{T}_a \psi \rangle=\langle \psi|\psi \rangle.$$
Indeed with the representation of the momentum operator in position space, $\hat{p} \psi(x)=-\mathrm{i} \hbar \partial_x \psi(x)$ you get [EDIT: corrected in view of #4]
$$\hat{T}_a \psi(x) = \exp(\mathrm{i} a \hat{p}/\hbar) \psi(x)=\hat{U}(a) \psi(x)=\exp(a \partial_x) \psi(x)=(1+a \partial_x + a^2 \partial_x^2/2!+\cdots) \psi(x) = \psi(x+a).$$

 

[haushofer]

Yes, I think this is the confusing part. The inner product contains an implicit integration, and this integration is invariant under a shift x+a because you integrate from minus infinity to plus infinity.

 

[weirdoguy]

@vanhees71 is there one psi missing in the last line? Next to U.

 

[vanhees71]

Yes, indeed. I've corrected the typo in the original posting.

 

[haushofer]

Ok, this seems to me a case of confusing notation. Thanks for the help, it's clear now!

 

[romsofia]

I've seen a proof for something similar in the book "Geometry, Topology and Physics" by Mikio Nakahara, in which he uses the notation vanhees pointed out, which is... hard to get use to. But, chapter 1.2 of this book goes in depth, so maybe it'll help you. It's from the second edition, chapter 1.2, and i'll post the following proof from it:An abstract ket vector is now expressed in terms of a more concrete wavefunction $\psi(x)$ or $\psi(p)$. What about the operators? Now we write down the operators in the basis $|x\rangle$. From the defining equation $\hat{x}|x\rangle=x|x\rangle$, one obtains $\langle x| \hat{x}=\langle x| x$, which yields after multiplication by $|\psi\rangle$ from the right,
$$
\langle x|\hat{x}| \psi\rangle=x\langle x \mid \psi\rangle=x \psi(x)
$$
This is often written as $(\hat{x} \psi)(x)=x \psi(x)$.
What about the momentum operator $\hat{p}$ ? Let us consider the unitary operator
$$
\hat{U}(a)=\mathrm{e}^{-\mathrm{i} a \hat{p}}
$$
Lemma 1.1. The operator $\hat{U}(a)$ defined as before satisfies
$$
\hat{U}(a)|x\rangle=|x+a\rangle
$$
Proof. It follows from $[\hat{x}, \hat{p}]=\mathrm{i}$ that $\left[\hat{x}, \hat{p}^n\right]=\mathrm{i} n \hat{p}^{n-1}$ for $n=1,2, \ldots$ Accordingly, we have
$$
[\hat{x}, \hat{U}(a)]=\left[\hat{x}, \sum_n \frac{(-\mathrm{i} a)^n}{n !} \hat{p}^n\right]=a \hat{U}(a)
$$
which can also be written as
$$
\hat{x} \hat{U}(a)|x\rangle=\hat{U}(a)(\hat{x}+a)|x\rangle=(x+a) \hat{U}(a)|x\rangle .
$$

This shows that $\hat{U}(a)|x\rangle \propto|x+a\rangle$. Since $\hat{U}(a)$ is unitary, it preseves the norm of a vector. Thus, $\hat{U}(a)|x\rangle=|x+a\rangle$.

 

[Demystifier]

Quite generally, the book "QFT for the gifted amateur" is excellent for developing conceptual intuitive understanding of the main ideas, but not very reliable as a reference for technical mathematical aspects of QFT. The authors themselves say in the preface that they are experimentalists.

 

[AndreasC]

I don't know, what the author means by his bracket notation. It's not the standard Dirac notation. In standard Dirac notation there are abstract vectors $|\psi \rangle$ and "wave functions", which are the "components" of the abstract vectors in the (generalized) position eigenbasis, $\psi(x)=\langle x|\psi \rangle$. Then you have
$$\langle \psi|\psi \rangle=\int_{\mathbb{R}} \mathrm{d}^3 x \langle \psi|x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \mathrm{d}^3 x |\psi(x)|^2.$$
Thus a translation operator, defined as
$$\hat{T}_a \psi(x)=\psi(x+a),$$
is a unitary operator, because for all $\psi \in \mathbb{R}^2$ obviously
$$\langle \hat{T}_a \psi|\hat{T}_a \psi \rangle=\langle \psi|\psi \rangle.$$
Indeed with the representation of the momentum operator in position space, $\hat{p} \psi(x)=-\mathrm{i} \hbar \partial_x \psi(x)$ you get [EDIT: corrected in view of #4]
$$\hat{T}_a \psi(x) = \exp(\mathrm{i} a \hat{p}/\hbar) \psi(x)=\hat{U}(a) \psi(x)=\exp(a \partial_x) \psi(x)=(1+a \partial_x + a^2 \partial_x^2/2!+\cdots) \psi(x) = \psi(x+a).$$

I think it is because in QFT instead of time as a parameter, you have spacetime position, which might be what is symbolized here as x. In that case, the notation makes sense, and position is not really an operator any more, like time is not an operator in non relativistic QM (in Schrodinger picture).

EDIT: By not an operator I meant the X operator used in non relativistic QM is not an observable any more.

 

[vanhees71]

In QFT the spacetime arguments come through the field operators, out of which you build local observables. The state usually has no spacetime argument.

 

[AndreasC]

In QFT the spacetime arguments come through the field operators, out of which you build local observables. The state usually has no spacetime argument.

Right, because we normally use Heisenberg picture. I got a bit confused because I was thinking Schrodinger picture in non relativistic QM and momentarily forgot that in QFT the Schrodinger picture states again only have time as label. Bit annoying how that works.

 

[vanhees71]

True, but there's no problem to formulate QFT in the Schrödinger picture although it's pretty inconvenient. See, e.g.,

B. Hatfield, Quantum Field Theory of Point Particles and
Strings, Addison-Wesley, Reading, Massachusetts, 10 edn.
(1992).

 

[AndreasC]

True, but there's no problem to formulate QFT in the Schrödinger picture although it's pretty inconvenient. See, e.g.,

B. Hatfield, Quantum Field Theory of Point Particles and
Strings, Addison-Wesley, Reading, Massachusetts, 10 edn.
(1992).

What is really annoying to me is that the Schrodinger picture of QFT only moves time evolution to the states, and they still don't have space dependence. That's no way to treat space and time equally!

I'm thinking now if it would be possible to also move space arguments to the states, in some kind of "Schrodinger+" picture.

I'm also realizing that there are some things in QFT I don't understand as well as I thought, because I didn't really reexamine my prior knowledge I transferred from QM. For instance, given that X is not really an observable in the same sense, then it doesn't really make sense to take its eigenstates to insert a complete basis, right? In that case the correspondence between the Hilbert space and L^2 brakes down, and now I realized I no longer know how I am supposed to take inner products between states. Interesting.

I'm guessing in general it is not even possible right now to know that, since the right Hilbert space for interacting QFT is not known. So maybe I should be content with at least knowing how to take inner products in Fock space, which is not something I know. Oof.

In the course of writing this, I realized that I do actually know that for the Fock space, since I know Fock states form a complete set. But yeah I hadn't really realized I can not just integrate over all space or whatever to get an inner product.

 

[vanhees71]

The canonical formalism is anyway usually asymmetric in time and space. Also states have no a-priori connection to position.

 

[AndreasC]

Also states have no a-priori connection to position.

Hmm, what exactly do you mean by that?

 

[vanhees71]

In QT the states are described by the statistical operator $\hat{\rho}$. There's nothing in its definition related to position. Position comes in when you want to calculate the probability distribution for finding a particle at a given position, $P(\vec{x})=\langle \vec{x}|\hat{\rho} \vec{x} \rangle$.

In relativistic QFT you start with the unitary representations of the Poincare group, and the position operators must be constructed from the representations. That's possible in the strict sense for massive particles of any spin and for massless particles only for $S=0$ and $S=1/2$. For a photon (massless with spin 1) there's no position observable. What can be unambigously defined are the probabilities for detecting a photon at a given position, where this position is determined by the position of the detector. It's given by the energy density of the electromagnetic field (properly normalized).

 

[AndreasC]

In QT the states are described by the statistical operator ρ^.

That's one formalism. The other is treating the Hilbert space vectors as states, that's what I had in mind.

What I meant was that in that formalism, the Schrodinger picture moves the time dependence from the operator over to the state vector, so now the vectors have time arguments, and the operators do not.

What I was wondering was if it would be possible to move also the space arguments over to the state vectors in a similar way (I'm guessing not, but I have to think about why).

 

[vanhees71]

Pure states are described by rays in Hilbert space or equivalently by statistical operators of the form $\hat{\rho}=|\psi \rangle \langle \psi|$ with $\langle \psi|\psi \rangle=1$.

 

[AndreasC]

Pure states are described by rays in Hilbert space or equivalently by statistical operators of the form $\hat{\rho}=|\psi \rangle \langle \psi|$ with $\langle \psi|\psi \rangle=1$.

Right, and that is what I'm talking about. These pure states in the Schrodinger picture are (vector valued) functions of time, and observables are functions of space, unlike the Heisenberg picture where observables are functions of spacetime, and the states are not parametrized by other time or space. What I was thinking about was if it is possible to come up with a reasonable "Schrodinger+" picture where you move both the time AND space dependence over to the states. It's not really related to the original post, just something I was thinking about.

 

[vanhees71]

But why should a generic state have anything to do with position?

 

[AndreasC]

But why should a generic state have anything to do with position?

Why should it have to do with time?

 

[vanhees71]

That's true too. The state describes the preparation of the system at the initial time. In this sense the "most natural picture of time evolution" is the Heisenberg picture.

 

[AndreasC]

That's true too. The state describes the preparation of the system at the initial time. In this sense the "most natural picture of time evolution" is the Heisenberg picture.

Of course. I'm just wondering if you could do what Schrodinger does for Heisenberg, but also for space.

 

[vanhees71]

How should this be possible? We should start with the simple case first, where we have position as an observable, e.g., for a massive particle as an electron. In contradistinction to position time is never an observable, but a parameter in QT. The reason is simply that otherwise $\hat{t}$ would be the conjugate observable to $\hat{H}$, the Hamiltonian (usually representing energy), and then both $\hat{t}$ and $\hat{H}$ would have both $\mathbb{R}$ as their spectra. That's, because from the commutator relations between two observables (from here on I use the sloppy language and identify observables with their representing self-adjoint operators) $[\hat{A},\hat{B}]=\mathrm{i} \hbar \hat{1}$, you can proove that then both $\hat{A}$ and $\hat{B}$ have the entire real numbers as spectra.

For time and energy that's in clear contradiction to observations for at least two reasons:

(a) our world is pretty stable, i.e., the the energy is bounded from below. If this were wrong, any system could get to ever lower energy states, and nothing would be stable. That was Pauli's argument against time being an observable in his famous Handbuch article about wave mechanics.

(b) one of the key successes of early QM was the explanation for the spectral lines of many atoms beyond hydrogen, i.e., it solved the very puzzle, which was among the strongest motivations for physicists to look for a "new quantum mechanics", going beyond the Bohr-Sommerfeld quantization prescriptions, which worked for hydrogen (and the harmonic oscillator or rotator), but for nearly nothing else.

So space and time are distinct on a very fundamental level. Time is an parameter, and the Hamiltonian describes the dynamics of the system. According to the formalism of quantum theory, what's observable in principle are probabilities for the outcome of measurements on ensembles of equally prepared quantum systems: For a complete set of compatible observables $\{\hat{A}_1,\ldots,\hat{A}_n \}$ there are 1D eigenspaces, i.e., the possible outcomes when measuring these compatible observables are $(a_1,\ldots,a_n)$ with the $a_j$ in the spectra of $\hat{A}_j$. Then let $|t,a_1,\ldots,a_n \rangle$ be a arbitrary set of common normalized eigenvectors. These are all determined up to a phase factor. The same holds for the state vector $|\psi(t) \rangle$, which describes the initial preparation of the system.

The time evolution of the observables $\hat{A}_j(t)$ and $|\psi(t) \rangle$ is pretty arbitrary, i.e., there are two unitary operators $\hat{A}(t)$ and $\hat{C}(t)$ describing the time evolution,
$$\hat{A}_j(t)=\hat{A}(t) \hat{A}_j(0) \hat{A}^{\dagger}(t), \quad |\psi(t) \rangle=\hat{C}(t) |\psi(0) \rangle.$$

Note that for the eigenvectors the time evolution is
$$|a_1,\ldots,a_n;t \rangle=\hat{A}(t) |a_1,\ldots,a_n; 0 \rangle,$$
so that for all $t$
$$\hat{A}_j(t) |a_1,\ldots,a_n;t \rangle = a_1 |a_1,\ldots,a_n; t \rangle.$$

The unitary operators $\hat{A}(t)$ and $\hat{C}(t)$ fulfill the equations of motion
$$\dot{\hat{A}}(t)=\frac{\mathrm{i}}{\hbar} \hat{X}(t) \hat{A}(t), \quad \dot{\hat{C}}=-\frac{\mathrm{i}}{\hbar} \hat{Y}(t) \hat{C}(t),$$
and the only thing necessary for the self-adjoint operators $\hat{X}(t)$ and $\hat{Y}(t)$ is that
$$\hat{X}(t) + \hat{Y}(t)=\hat{H}(t).$$
The Schrödinger picture is defined by $\hat{X}(t)=0$ and $\hat{Y}(t)=\hat{H}$, the Heisenberg picture in the opposite way $\hat{X}(t)=\hat{H}$ and $\hat{Y}(t)=0$.

Then there are many "intermediate pictures", which split $\hat{H}$ in any arbitrary, convenient way. E.g., to do "perturbation theory" you choose $\hat{X}=\hat{H}_0$, where you can solve the equation of motion for $\hat{A}$ exactly and then $\hat{Y}=\hat{H}_1$ to do a perturbative expansion in powers of $\hat{H}_1$. For scattering theory it's often convenient to choose $\hat{H}_0$ as the Hamiltonian of free particles (or fields) and $\hat{H}_1$ as the part that describes interactions.

No matter, how you choose the picture of time evolution, the observable (probabilistic) facts about the system come out the same (at least as long as you solve everything exactly), i.e., the probabilities,
$$P(t,a_1,\ldots,a_n)=|\langle a_1,\ldots,a_n;t|\psi(t) \rangle|^2,$$
are always the same no matter how you split $\hat{H}$ in $\hat{X}(t)$ and $\hat{Y}(t)$. So are all other derived quantities like expectation values,
$$\langle O(t) \rangle=\langle \psi(t)|\hat{O}(t) \psi(t).$$
If you take as a complete set of observables for a single (spin-0) particle the three components of the position vector you get the wave function,
$$\psi(t,\vec{x})=\langle x(t)|\psi(t) \rangle,$$
which is also invariant under a change of the picture of time evolution, modulo an overall phase factor, which never occurs in any calculation of observable quantities like probabilities etc.

The reason is that the transformation from one picture to another is a unitary transformation $\hat{B}_{12}$ for the states and the observables, i.e., if you have two splits,
$$\hat{H}=\hat{X}_1(t) +\hat{Y}_1(t)=\hat{X}_2(t) +\hat{Y}_2(t),$$
then the states and observables transform from one to the other picture with
$$|\psi_2(t)=\hat{B}_{12}(t) |\psi_1(t) \rangle, \quad \hat{O}_{2}(t)=\hat{B}_{12}(t) \hat{O}_{1}(t) \hat{B}_{12}^{\dagger}(t),$$
where
$$\hat{B}_{12}(t)=\hat{C}_2(t) \hat{C}_1^{\dagger}(t) =\hat{A}_2(t) \hat{A}_1^{\dagger}.$$
The latter equality follows from the fact that
$$\hat{U}_1(t)=\hat{A}_1^{\dagger}(t) \hat{C}_1(t)$$
and
$$\hat{U}_2(t)=\hat{A}_2^{\dagger}(t) \hat{C}_2(t)$$
both fulfill the same equation of motion,
$$\partial_t \hat{U}=-\frac{\mathrm{i}}{\hbar} \hat{H} \hat{U},$$
and the same initial condition $\hat{U}(0)=\hat{1}$, such that they are equal.