# I Two questions about "The Physics of Virtual Particles"

#### fox26

Arnold Neumaier,

I have 2 elementary questions about your article “The Physics of Virtual
Particles”.

1. In the paragraph headed “States.” on p. 4, of 13, you talk about states of a
physical system, with a mixed state specified by a Hermitian operator ρ of trace
1 acting on the Hilbert space of the system which contains the system’s pure
state vectors. You say that this also covers the pure case with ρ = ψψ*. Do the
symbols on the right-hand side of this equation indicate the product, in the
standard product of functions sense, of ψ with its complex conjugate? If so, for a
system consisting of just one particle, ρ is a function of the arguments r and t of
ψ whose value at each (r,t) is the probability density of finding the particle at the
point r at time t in whatever space the argument r of ψ is, e.g., position space or
momentum space. This ρ doesn’t sound like a Hermitian operator of trace 1, and
doesn’t even specify a state completely; for example, if the value of ρ is the
position probability density, ρ doesn’t determine the momentum probability
things, that is, with two different meanings?

2. In the paragraph headed “Stable and unstable particles.” on p. 5, you say
that stable particles, according to the QFT formalism, must be on-shell, meaning
that their momentum p is related to the real rest mass mm (sic) (doubling due to
typo?) by the relation p2 = m2 (in units where c = 1). You clearly were not talking
just about particles that move (in vacuum) only at speed c (= 1 in your units). In
relativistic mechanics, p2 = m2v2, where m is the relativistic (moving) mass, not
the rest mass, and v2 = 1 (in your units) only for particles moving at speed c. The
v2" was omitted in that equation in 3 later paragraphs, in the first of which, for
complex p and m, after the equation “p2 = m2", you have “and v = p/m real",
implying p2 = m2v2. Is the omission of all those “v2" ‘s due to a peculiarity of QFT,
to the fact that you were thinking just of particles moving with speed c (= 1) when
you were writing this but neglected to specify that limitation, or is it just a strange
repeated typo or other similar error--or something else?
--fox26

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#### A. Neumaier

You say that this also covers the pure case with ρ = ψψ*. Do the symbols on the right-hand side of this equation indicate the product, in the standard product of functions sense, of ψ with its complex conjugate?
It is the linear operator with $\rho\phi:=\psi(\psi^*\phi)$, where the term in parentheses is the inner product. $\langle A\rangle=Tr~\rho A=\psi^*A\psi$ relates it to the usual definition. This is treated in detail in any introduction to quantum statistical mechanics.
related to the real rest mass mm (sic) (doubling due to typo?)
Yes; corrected.
the relation $p^2 = m^2$ (in units where c = 1).
This is correct. $v=p/m$ is the relativistic 4-velocity, which satisfies $v^2=c^2=1$ and is defined only for $m>0$, excluding particles moving at the speed of light (which have $p^2=0$ and no relativistic velocity vector).

• bhobba and vanhees71

#### fox26

It is the linear operator with $\rho\phi:=\psi(\psi^*\phi)$, where the term in parentheses is the inner product. $\langle A\rangle=Tr~\rho A=\psi^*A\psi$ relates it to the usual definition. This is treated in detail in any introduction to quantum statistical mechanics.

Yes; corrected.

This is correct. $v=p/m$ is the relativistic 4-velocity, which satisfies $v^2=c^2=1$ and is defined only for $m>0$, excluding particles moving at the speed of light (which have $p^2=0$ and no relativistic velocity vector).
Thank you for answering my questions. The two points in your article that I mentioned which seemed to me to be questionable are actually not so, as your answer showed, if allowance is made for your somewhat careless or lazy omission of standard parts of the symbology, such as bra or ket symbols around the two "ψ" 's to distinguish the operator from the product, and failing to use vector symbology to show you were referring to the four-momentum and the four-velocity instead of the corresponding scalar quantities. This could be confusing or misleading to someone not already familiar with the subject. One of your minor departures from standard terminology which you could easily correct is calling ħ Planck's constant. "Planck's constant" is almost universally used to name h, the constant originally used by Planck in " E = hν" when he introduced quantization. "ħ", or "h-bar", as you know, is used for h/2π.

#### weirdoguy

"ħ", or "h-bar", as you know, is used for h/2π
And it's also universally called "Planck's constant". There is no departure from standard terminology.

#### A. Neumaier

omission of standard parts of the symbology, such as bra or ket symbols around the two "ψ" '
$\psi=|\psi\rangle$ and $\psi^*=\langle\psi|$. The product is the standard operator product, familiar from matrix algebra.
failing to use vector symbology to show you were referring to the four-momentum and the four-velocity
4-vectors are never bold in relativity; the bold is for the 3-dimensional case only.

#### vanhees71

Gold Member
Well, it's dangerous. I'd not have come to the idea that, given a vector $\psi$, $\psi^*$ is its dual, which I'd label as $\psi^{\dagger}$. Why not using Dirac's notation? I know that some people don't like it (including Weinberg, but I guess that's because he doesn't like Dirac at all for some reason, given the somewhat negative attitude towards him in the otherwise great historical intro chapters of his textbooks on QFT and QM), but this I never understood, because it's so intuitive, and mathematicians tend to have sparse notation which is way less save concerning "calculational safety", i.e., you could never mess up $|\psi \rangle$ with its dual $\langle \psi|$, and never a scalar product $\langle \psi|\phi \rangle$ with the operator $|\psi \rangle \langle \phi$ or $|\psi \rangle |\phi \rangle:=|\psi \rangle \otimes |\phi \rangle$.

Already in my introductory linear-algebra course, I realized this. It was a very good two-semester course, which I've been very thankful for forever, because I got a very good basis for all physics from it (if there's a single important math subject for physics it's linear algebra), but I struggled a lot when it came to basis changes, transformations between representations, representing matrices of linear mappings, and so on. I always scratched my head about questions, where is $\hat{T}$ and where is $\hat{T}^{-1}$ when transforming the representing matrices, i.e., is it $\hat{T} \hat{A} \hat{T}^{-1}$ or the other way $\hat{T}^{-1} \hat{A} \hat{T}$. Using Ricci calculus (which were totally forbidden in our math department) and always writing the invariant vectors as a whole including the various basis vectors and using upper and lower indices (when not dealing with Cartesian bases in Euclidean vector spaces), resolves this trouble forever, because it's easy to derive, how it must be. It's almost failsave just from the notation!

I was very proud of the physicists' careful notation when I once attended a pretty abstract course on topological vector spaces in the math department. In the exercise sessions, by just the intuitive way of writing mathematical objects as done in our physics lectures, I was much faster in doing calculations. Of course, the mathematicians were in great advantage concerning to make rigorous watertight proofs out of the intuitive physics notations :-).

The short message: don't underestimate the mnemotechnical advantages of physicists's math notation!

• bhobba

#### A. Neumaier

Well, it's dangerous. I'd not have come to the idea that, given a vector $\psi$, $\psi^*$ is its dual
It is standard notation for working witch complex vectors in linear algebra. $A^*$ is the complex conjugate transpose, and becomes the adjoint in infinite dimension. If the matrix or operator $A$ has a single column only it gives the corresponding row (linear functional).

mathematicians tend to have sparse notation which is way less safe concerning "calculational safety", i.e., you could never mess up $|\psi \rangle$ with its dual $\langle \psi|$, and never a scalar product $\langle \psi|\phi \rangle$ with the operator $|\psi \rangle \langle \phi$ or $|\psi \rangle |\phi \rangle:=|\psi \rangle \otimes |\phi \rangle$.
In standard linear algebra a column vector (ket) is always $\psi$ and a row vector (bra) always $\psi^*$ or $\psi^T$, a scalar product is necessarily $\psi^*\phi$ or $\psi^T\phi$, and a dyadic product (rank 1 matrix) necessarily $\psi\phi^*$. This is completely safe, one cannot interpret it wrongly - the dimensions of the vectors would not match to make a meaningful multiplication. The formulas are just freed from clutter - especially here on PF, where one needs to write many characters to display a pointed bracket.

#### vanhees71

Gold Member
Well, a ket is no column vector in standard notation but an abstract vector in a abstract Hilbert space, but it's indeed semantics, and unfortunately when reading papers across different scientific communities you have to learn their specific language and notation.

#### fox26

And it's also universally called "Planck's constant". There is no departure from standard terminology.
The most authoritative source for the correct use of this terminology that I know of is NIST (the National Institute of Science and Technology). In all their publications I've looked at which list the value of, or otherwise mention, Planck's constant, it is symbolized only by "h", never by "ħ", including in the latest (2014) CODATA Recommended Values of the Fundamental Physical Constants, where h is listed under "Symbol" on the same line as the name "Planck's constant", which is listed under "Quantity" (whose 2014 CODATA value and 1 SD uncertainty is listed as 6.626070040(81) x 10-34 Js), while ħ is listed on a separate line, and is of course given as h/2π. All references to Planck's constant which I've recently looked at--several--including those in extensive online text's for QM, symbolize it as h, while ħ is called "h-bar" or "the reduced Planck's constant", never just "Planck's constant". This was also the case in the textbooks, one of which was the well-known one by Messiah, used by the courses in elementary and second year QM which I took decades ago, and also books by Roger Penrose, in one of which, Shadows of the Mind, on p. 271, he refers to ħ as Dirac's symbol for Planck's constant, divided by 2π. Nowhere have I seen ħ called "Planck's constant" except in Neumaier's article, and by you. What papers, textbooks, or standard physical constants sources call ħ "Planck's constant", other than that article, or others by Neumaier?

• vanhees71

#### fox26

$\psi=|\psi\rangle$ and $\psi^*=\langle\psi|$. The product is the standard operator product, familiar from matrix algebra.

4-vectors are never bold in relativity; the bold is for the 3-dimensional case only.
Not to harp on this, but ψψ* standardly indicates operator product if ψ and ψ* represent operators, but since the context was QM, ψ and ψ* would normally be assumed to be wavefunctions, complex-valued functions with domain in some Euclidean space or manifold. Of course, someone familiar with the subject would realize that the function product sense isn't appropriate for ρ = ψψ* to be an operator, which you had called ρ shortly before, and maybe realize that ψ should be considered as enclosed in ket symbols, and ψ* be consider to represent ψ enclosed in bra symbols, so the product would be the operator product, rather than the function product. However, someone just learning what you were discussing might have trouble with this, and even an experienced physicist might well not be completely certain what you meant.

Also, in your reply to me, you symbolize the inner product of ψ* and Φ by "(ψ*Φ)". In my linear algebra course, and in everything I've seen inner product used in ever since, such an inner product would have been symbolized by "(ψ*,Φ)" or "ψ*⋅Φ". You do seem to me to often omit symbols standardly used to indicate meanings where such omission by you leads to uncertainty on the reader's part about what you mean, and unnecessary time and effort spent by him or her in puzzling out what that meaning is.

I gave a reply shown above to weirdoguy's defense of your calling ħ "Planck's constant".

Finally, Einstein himself used for 4-vectors, I think, capital letters, e.g. "A", in his General Relativity papers, not lower case letters such as the "v" you used. Also, in your reply you had "v2 = c2 = 1", leading those unfamiliar with your non-standard notational conventions to think that "v" denoted the same type of thing as "c" did, i.e., a scalar, which in discussions of Relativity it almost always does. You can't square a 4-vector. To indicate its norm or magnitude, which can be squared, " | | " or "|| ||" is standardly used.

"Two questions about "The Physics of Virtual Particles""

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