Is Every Hermitian Operator Observable?

[sweet springs]

Hi.
I know that operator of observable is Hermitian. Reciprocally, any Hermitian is observable? If not, showing some examples are appreciated.
Regards.

 

[Monocles]

No, for example consider the identity operator. Certainly Hermitian, but if it were an observable then that would imply that we can directly observe the probability density.

 

[SpectraCat]

Hi.
I know that operator of observable is Hermitian. Reciprocally, any Hermitian is observable? If not, showing some examples are appreciated.
Regards.

The identity matrix is one example, so is any (real) constant times the identity matrix.

Another example is the so called "number operator" aa^{\dagger}, which is the product of lowering (annihilation) and raising (creation) operators. It is called the number operator because it returns the quantum number of the state it is applied to.

 

[sweet springs]

Hi. Thanks Monocles and SpectraCat.
Definition of observable in Dirac II-10 is "We call a real dynamical variable whose eigensates form a complete set an observable."
Identity operator satisfy eigenvalue eauation Iψ=1ψ and these {ψ} form a complete set.
Thus identity operator is observable. Am I wrong?
Regards.

 

[LukeD]

I disagree. The identity operator is 1 in the case of a particle existing and 0 in the case of it not existing. Therefore, it is an observable value. A value being observable doesn't mean that its entire distribution is observable. Only eigenvalues and average values need to be observable. In the case of the identity operator, every state has the same eigenvalue, so the operator doesn't tell you much.

The number operator is also observable since it is H/(hw)-1/2, and that is a measurable value for the quantum harmonic oscillator.

----

I'd be hard pressed to say that every Hermitian matrix in Quantum Mechanics corresponds to something measurable... But if a hermitian matrix appears as part of the Hamiltonian, then I'd say it's something observable.

Basically, if you can build a device that depends on the value of an operator, then that operator can be said to be "observable", either directly or indirectly.

 

[Truecrimson]

I agree with LukeD.

Also if A and B are hermitian, AB+BA and i(AB-BA) are also hermitian. If, say, A=\hat{x} and B=\hat{p}, It'd be hard to come up with the measuring devices, let alone the meaning of the observables.

Since you read Dirac, I have to add that the postulate that, to each observable there corresponds a Hermitian operator, is motivated because of the spectral theorem that the set of all eigenvectors of a Hermitian operator forms an orthonormal basis. (I remember Dirac didn't even mention linear algebra in the book.)

Edit: What I wanted to say in the last paragraph is that, Dirac was right to say that we need a complete set of eigenstates, but he didn't have this theorem to back it up rigorously.

 

[sweet springs]

Hi, LukeD and Truecrimson.
Now I tend to consider any hermitian is observable.
Thank you.

 

[f95toli]

Another example is the so called "number operator" aa^{\dagger}, which is the product of lowering (annihilation) and raising (creation) operators. It is called the number operator because it returns the quantum number of the state it is applied to.

The number operator is certainly an observable, there are actually quite a few ways to experimentally count the number of photons in a number state (and you don't have to do it "indirectly" by measuring the total energy of the system).

 

[SpectraCat]

I disagree. The identity operator is 1 in the case of a particle existing and 0 in the case of it not existing. Therefore, it is an observable value. A value being observable doesn't mean that its entire distribution is observable. Only eigenvalues and average values need to be observable. In the case of the identity operator, every state has the same eigenvalue, so the operator doesn't tell you much.

Ok, I guess I see what you are saying ... if we look for a particle, and it is there, by definition we have just applied the identity operator, so it is kind of by default included in any measurement we make. Is that another way to put it?

The number operator is also observable since it is H/(hw)-1/2, and that is a measurable value for the quantum harmonic oscillator.

Yeah, that was a mistake on my part. I was thinking in terms of building a device to measure just the quantum number, but of course if we measure the energy of an HO eigenstate, we immediately infer the quantum number.

So, I guess I don't know any examples of Hermitian operators that don't correspond to observables. I guess what Truecrimson said about being able to compose complicated Hermitian operators out of other Hermitian operators might work, but that seems like a practical approach that misses the OP's point.

 

[Truecrimson]

Ok, I guess what Truecrimson said about being able to compose complicated Hermitian operators out of other Hermitian operators might work, but that seems like a practical approach that misses the OP's point.

Now I think I see what you mean, reminding me of what Ballentine says in his book.

Dirac, to whom we are indebted for so much of the
modern formulation of quantum mechanics, unfortunately used the word
"observable" to refer indiscriminately to the physical dynamical variable
and to the corresponding mathematical operator. This has sometimes led
to confusion. There is in the literature a case of an argument about whether
or not the electromagnetic vector potential is an observable, one party
arguing the affirmative on the grounds that the operator satisfies all of the
required conditions, the other party arguing the negative on the grounds
that the vector potential cannot be measured.

So yes, sweet springs' conclusion that "any hermitian is observable" is right in a mathematical sense.

 

[Monocles]

Very good points - I learned something new!

 

[Fredrik]

If we define "observable" to mean "bounded self-adjoint operator", then of course every hermitian operator is an observable. We defined them to be. There is however another approach. We define our "observables" operationally, by describing the devices that are supposed to measure them. The question is then, what sort of mathematical object should we use to represent observables mathematically? This sort of thing is discussed in detail in books on the mathematics of QM, e.g. "An introduction to the mathematical structure of quantum mechanics", by F. Strocchi, and "Mathematical theory of quantum fields" by H. Araki.

I don't know this stuff myself, but I get that the basic idea is to start with a C*-algebra, define a "state" as a positive linear functional on the C*-algebra of observables, and then invoke the appropriate mathematical theorems to prove that abelian C*-algebras give us classical theories and non-abelian C*-algebras give us quantum theories. (The C*-algebra is then isomorphic to the algebra of bounded self-adjoint operators on a complex separable Hilbert space).

In this approach, it's not the case that every member of the C*-algebra corresponds to a measuring device, but I don't really have any more information on that. Perhaps someone can read those books and tell the rest of us.

 

[Truecrimson]

Thank you Fredrik!

 

[sweet springs]

Hi,
I have a new wondering.

In 9.2 of my old textbook Mathematical Methods for Physicists, George Arfken states,
------------------------------------------------------
1. The eigenvalues of an Hermite operator are real.
2. The eigen functins of an Hermite operator are orthogonal.
3. The eigen functins of an Hermite operator form a complete set.*
* This third property is not universal. It does hold for our linear, second order differential operators in Strum-Liouville (self adjoint) form.
------------------------------------------------------
The * suggests that not all Hermitian are OBSERVABLE. Can anyone suggest me some examples?
Definition of observable in Dirac II-10 is "We call a real dynamical variable whose eigensates form a complete set an observable."
Regards.

 

[Hurkyl]

Wait a moment -- isn't the identity operator supposed to correspond to a "trivial" measurement that, no matter what is going on in reality, always returns the number "1"?

 

[sweet springs]

Hi, Hurkyl

Wait a moment -- isn't the identity operator supposed to correspond to a "trivial" measurement that, no matter what is going on in reality, always returns the number "1"?

Let me confirm my understanding.
IΨ＝1Ψ
Any state is eigenvector of identity operator I with eigenvalue 1. I is both Hermitian and Observable.

I am interested in some examples if any, of operators that is Hermite but not observable in the sense that whose eigenvectors do not form a complete set.

Regards.

 

[Fredrik]

whose eigenvectors do not form a complete set.

How do you define this? Isn't it just that the identity can be expressed as \sum_a|a\rangle\langle a| where the sum is over the eigenvalues of some operator A and the eigenvector corresponding to the eigenvalue a is written as |a>? In that case, any projection operator for a closed proper subspace will do. (Its eigenvalues are 0 or 1, so any sum over its eigenvalues would only contain two terms). Such an operator is however a mathematical representation of an operationally defined observable. (It corresponds to a device that measures A and only outputs 1 if the result is in the given range, and 0 if it's not). The simplest example is |a><a|.

I agree with Hurkyl about the identity operator.

 

[kof9595995]

this is an interesting post, i'd like to keep myself updated...[

 

[sweet springs]

Hi, Fredrik

Such an operator is however a mathematical representation of an operationally defined observable. (It corresponds to a device that measures A and only outputs 1 if the result is in the given range, and 0 if it's not). The simplest example is |a><a|.

Let me ask you a question in order to check if I understand your point.

{|x>} is a complete set of eigenvectors of coordinate. ∫|x><x| dx =I. Then

a) Operator say Q=∫x>0 |x><x| dx is Hermitian but not Observable, because its eigenvector set does not contain |x> x<0. Eigenstates of Q do not form a complete set.

b) Q|x> = 0|x> for x<0. The subspace composed of {|x>|x<0 } is eigensubspace of Q with eigenvalue 0. So eigenstates of Q form a complete set.

Which one (or another one?) is the right answer?

Regards.

 

[Fredrik]

I wouldn't define an observable that way. (I don't think anyone does). Your a) is essentially correct apart from the terminology. (I also wouldn't use the position operator as an example, because it doesn't actually have any eigenvectors. It takes some fairly sophisticated mathematics to make sense of what |x> means).

Your b) is wrong, if the definition of of "complete set" is that the identity can be expressed as a sum of projection operators for 1-dimensional subspaces with exactly one term for each distinct eigenvalue.

 

[sweet springs]

Hi, Fredrik
I repeat your teachings.

An operator is OBSERVABLE whether it has degeneracy or not. For example squqre of momentum P^2 is observable while eigenspace for eigenvalue p^2 has two-fold degeneracy of |p> and |-p>.
We cannot (or must not?) find eigenspace for eiganvalue 0. [STRIKE]The idea that ALL THE NOT PROJECTED states degenerate and belong to eigenspace for eigenvalue 0 was wrong.[/STRIKE]

Thanks.

 

[Fredrik]

I don't understand that last sentece, but think of it this way: Suppose that A has a degenerate spectrum. Then we can't write \sum_a|a\rangle\langle a|=I. But we should be able to find a second operator B that commutes with A, and find state vectors |ab\rangle that are eigenvectors of both A and B. If we now have \sum_{a,b}|ab\rangle\langle ab|=I, then {A,B} is a "complete set of commuting observables". If not, then we should be able to find a third operator C that commutes with both A and B, and simultaneous eigenstates |abc\rangle. We can keep finding more and more observables until all the degeneracy is removed. Let's say that the last operator we need is Q. Then our set {A,B,C,...,Q} is a complete set of commuting observables.

I think that when this happens, there's always an observable X with non-degenerate spectrum such that all of the members in the complete set can be expressed as functions of X, e.g. A=f_A(X), B=f_B(X),\dots,Q=f_Q(X).

Avodyne's post here explains some of the details.

 

[peteratcam]

...Suppose that A has a degenerate spectrum. Then we can't write \sum_a|a\rangle\langle a|=I...[/tex]

You can't write that because |a\rangle\langle a | is not defined for a degenerate spectrum. It's not just that the lhs doesn't equal the right hand side, the left hand side doesn't mean anything.

Your b) is wrong, if the definition of of "complete set" is that the identity can be expressed as a sum of projection operators for 1-dimensional subspaces with exactly one term for each distinct eigenvalue.

That would be a bad definition of a complete set.
A set (of vectors) is complete if it spans the vector space. The set of eigenvectors of an observable is complete, whether or not there is degeneracy.

The notion of completeness with regards a complete set of mutually commuting observables is a separate one.

 

[sweet springs]

Hi. I have further questions on eigenvalue 0.

IΨ＝1Ψ
Any state is eigenvector of identity operator I with eigenvalue 1. I is both Hermitian and Observable.

Here Observable means operator whose eigenstates form a complete set.

#1
Null　Operator　O
OΨ＝0Ψ
Any state is eigenvector of identity operator O with eigenvalue 0. So O is Hermitian and Observable.

The both discussion seem to be similar but I feel the former right and the latter wrong. Both are wrong? Can anyone show me the right way?

#2
Measurement of some physical variable, for example coordinate X, leads value 0. X|0>=0|0>. |0> is a specific vector, δ(x). This eigenvalue equation looks like above OΨ＝0Ψ, Ψ is any vector. Don't we have to distinguish these two types of eigenvalue equations for eigenvalue 0 ?

Regards.

 

[Fredrik]

You can't write that because |a\rangle\langle a | is not defined for a degenerate spectrum. It's not just that the lhs doesn't equal the right hand side, the left hand side doesn't mean anything.

Yeah, I should have written something like |a\sigma\rangle instead of just |a\rangle, and written the sum in a different way too. I was a bit sloppy there.

That would be a bad definition of a complete set.
A set (of vectors) is complete if it spans the vector space. The set of eigenvectors of an observable is complete, whether or not there is degeneracy.

The notion of completeness with regards a complete set of mutually commuting observables is a separate one.

I don't think I've heard anyone define a "complete set of vectors" that way. I've heard terms like "maximal orthonormal set", but not "complete set". That's why I've been asking sweet springs to clarify what he meant. Hm, I guess it would make sense to define "complete set" as one that has a maximal orthonormal set (i.e. a basis) as a subset.

I have further questions on eigenvalue 0.

That case doesn't need to be treated separately.

Here Observable means operator whose eigenstates form a complete set.

And by "complete set" you mean a set that has a basis as a subset? I think it's much better to define observables operationally, and then say that they are represented mathematically by bounded self-adjoint operators. (The set of eigenvectors of a bounded self-adjoint operator is always "complete" in the sense defined above).

#1
Null　Operator　O
OΨ＝0Ψ
Any state is eigenvector of identity operator O with eigenvalue 0. So O is Hermitian and Observable.

The both discussion seem to be similar but I feel the former right and the latter wrong. Both are wrong? Can anyone show me the right way?

#2
Measurement of some physical variable, for example coordinate X, leads value 0. X|0>=0|0>. |0> is a specific vector, δ(x). This eigenvalue equation looks like above OΨ＝0Ψ, Ψ is any vector. Don't we have to distinguish these two types of eigenvalue equations for eigenvalue 0 ?

Regards.

Some of the symbols you're typing don't display properly for me, on either of my two computers. It's very confusing. And as I said before, the position operator isn't the greatest example, since it isn't bounded, and doesn't have eigenvectors. You have to define a "rigged Hilbert space" just to be able to define the "eigenstates" |x\rangle.

 

[peteratcam]

A set {f_i} of vectors is complete iff any vector in H can be written as a linear combination of vectors from the set {f_i}. A complete set of orthonormal vectors forms a basis.
...
[pg 149]
Another use of the word complete is in the description of a vector space. This concept is again trivial for [finite dimensional euclidean space] but non-trivial for H.
...[some stuff on the convergence of Cauchy sequences]...
All finite dimensional vector spaces are complete. That this is true for infinite dimensional Hilbert spaces is a deep theorem of analysis known as the Riesz-Fischer Theorem.

Perhaps other texts avoid the word 'complete' due to the ambiguity which it creates (see, eg, this thread!)

 

[Fredrik]

That first definition is a bit strange, mostly because (I think) all the math books define a "linear combination" to have a finite number of terms. The last sentence is odd too. A Hilbert space is complete by definition, so I guess they're using a non-standard definition of that too.

 

[strangerep]

If we define "observable" to mean "bounded self-adjoint operator", then of course every hermitian operator is an observable. We defined them to be. There is however another approach. We define our "observables" operationally, by describing the devices that are supposed to measure them. The question is then, what sort of mathematical object should we use to represent observables mathematically? This sort of thing is discussed in detail in books on the mathematics of QM, e.g. "An introduction to the mathematical structure of quantum mechanics", by F. Strocchi, and "Mathematical theory of quantum fields" by H. Araki.

I don't know this stuff myself, but I get that the basic idea is to start with a C*-algebra, define a "state" as a positive linear functional on the C*-algebra of observables, and then invoke the appropriate mathematical theorems to prove that abelian C*-algebras give us classical theories and non-abelian C*-algebras give us quantum theories. (The C*-algebra is then isomorphic to the algebra of bounded self-adjoint operators on a complex separable Hilbert space).

In this approach, it's not the case that every member of the C*-algebra corresponds to a measuring device, but I don't really have any more information on that. Perhaps someone can read those books and tell the rest of us.

Here's an excerpt from Neumaier & Westra (quant-ph/08101019) that might
give insight into the algebraic approach from a different perspective.
(This is from the start of section 5.1.)

Any fundamental description of physical systems must give account of
the numerical values of quantities observable in experiments when the
system under consideration is in a specified state. Moreover, the form
and meaning of states, and of what is observable in principle, must be
clearly defined. We consider an axiomatic conceptual foundation on the
basis of quantities and their values, consistent with the conventions
adopted by the International System of Units (SI) [238], who declare:

"A quantity in the general sense is a property ascribed to
phenomena, bodies, or substances that can be quantified for, or
assigned to, a particular phenomenon, body, or substance. [...] The
value of a physical quantity is the quantitative expression of a
particular physical quantity as the product of a number and a unit, the
number being its numerical value."

They also have a footnote:

We deliberately avoid the notion of observables, since it is not clear
on a fundamental level what it means to observe something, and since
many things (such as the fine structure constant, neutrino masses,
decay rates, scattering cross sections) which can be observed in nature
are only indirectly related to what is traditionally called an
observable in quantum mechanics. [...]

IMHO, adherence to the SI term "quantity" fosters clearer thinking
about this stuff.

They go on to give a set of axioms for algebras of quantities, and
define states as linear mappings from a space of quantities to a space
of ordinary numbers. This is the basic idea of the algebraic approach.
Any class of physical systems is defined by specifying an algebra of
quantities.

As to whether any element of the algebra can be measured... that gets
tricky. Consider angular momentum J_i. To measure the "pose"
of a system relative to that of another system from the same class we
must the consider the rotation group, which is generated by angular
momentum. Although (\theta^i J_i)^2, (\theta^i J_i)^3 might be hard to imagine
measuring, we have no trouble imagining a finite rotation given by \exp(i\theta^i J_i).

In the rest of the section 5, Neumaier & Westra go on to explain how
their axioms encompass both classical and quantum systems (including
thermodynamics/statistics). The algebraic approach thus unifies
all of physics under a more coherent umbrella. In section 5.4, they
discuss limits of experimental resolution and uncertainty, and show
how the essential difference between "classical" and "quantum" boils
down to commutativity of quantities.

Ordinary Hilbert spaces become merely a convenience, appropriate to
certain cases, on which the quantity algebra can be represented as
operators.

HTH.

 

[strangerep]

That first definition is a bit strange, mostly because (I think) all the math books define a "linear combination" to have a finite number of terms. [...] A Hilbert space is complete by definition, so I guess they're using a non-standard definition of that too.

A "linear combination" in an arbitrary vector space can certainly be an infinite sum.
The case of Hilbert space is bit trickier because all vectors therein must have finite
norm. But an arbitrary infinitely-long linear combination does not necessarily have
finite norm, hence the need to place restrictions on the sum in order to have a
Hilbert space. So the two notions of "complete" are related.

 

[Fredrik]

Thanks for the reference. That one goes on my to-read list for sure. You know, it's not hard to post a direct link to the article. (I can usually search arxiv.org for the code you posted, and be taken directly to the right page, but this time it didn't work).

Regarding "linear combinations", I wouldn't be surprised if different people define the term in different ways, but I think it's more appropriate to define a linear combination to only have a finite number of terms. This is why: If V is a vector space over a field F, and S is a subset of V, the "subspace generated by S" (or "spanned" by S) can be defined as any of the following

a) the smallest subspace that contains S
b) the intersection of all subspaces that contain S
c) \Big\{\sum_{i=1}^n a_i s_i|a_i\in\mathbb F, s_i\in S, n\in\mathbb N\Big\}

These definitions are all equivalent, and the fact that every member of the set defined in c) can be expressed as \sum_{i=1}^n a_i s_i, with n finite, seems like a good reason to define a "linear combination" as having only finitely many terms. That definition means that the subspace generated by S is equal to (instead of a subset of) the set of linear combinations of members of S. (Note that all of the above is true even if some subset of S is an infinite orthonormal set).

 

[sweet springs]

Hi.

Some of the symbols you're typing don't display properly for me, on either of my two computers.

Excuse me. I will restate my questions.

Definition:
OBSERVABLE is operator whose eigenvectors form a complete set (by Dirac).
OBSERVABLE is operator whose eigenspaces contain all the maximal orthonormal set, i.e. basis(Thanks to Fredrik).
Both the definition are equivalent.

Question:Are the following operators OBSERVABLE?
-Identity operator
-Null operator
-Projection to a subspace e.g. |a1><a1| for A|an>=an|an> with eigenvalues { an| a1,a2,a3,...}

I want to know how to deal with "eigenspace with eigenvalue 0".

Regards.

 

[George Jones]

A "linear combination" in an arbitrary vector space can certainly be an infinite sum.

For an arbitrary vector space, what does "infinite sum" mean? There is only one topology, the Euclidean topology, that can be given to a finite-dimensional vector space, but an infinite-dimensional vector space can be given various topologies.

 

[Fredrik]

Definition:
OBSERVABLE is operator whose eigenvectors form a complete set (by Dirac).
OBSERVABLE is operator whose eigenspaces contain all the maximal orthonormal set, i.e. basis(Thanks to Fredrik).
Both the definition are equivalent.

Question:Are the following operators OBSERVABLE?
-Identity operator
-Null operator
-Projection to a subspace e.g. |a1><a1| for A|an>=an|an> with eigenvalues { an| a1,a2,a3,...}

I want to know how to deal with "eigenspace with eigenvalue 0".

As I said, I'm not a fan of that definition, but given that definition, then all of those operators are observables. Recall that an eigenvector of a linear operator A is a non-zero vector x such that Ax=ax for some number a. Every non-zero vector is an eigenvector of the identity operator with eigenvalue 1. Every non-zero vector is an eigenvector of the null operator with eigenvalue 0. If P is a projection operator for a subspace V, then every non-zero member of V is an eigenvector of P with eigenvalue 1, and every non-zero vector that's orthogonal to all the vectors in V is an eigenvector of P with eigenvalue 0.

Eigenvalue 0 doesn't cause any additional complications at all, so it doesn't need to be handled separately.

 

[DarMM]

I don't know this stuff myself, but I get that the basic idea is to start with a C*-algebra, define a "state" as a positive linear functional on the C*-algebra of observables, and then invoke the appropriate mathematical theorems to prove that abelian C*-algebras give us classical theories and non-abelian C*-algebras give us quantum theories. (The C*-algebra is then isomorphic to the algebra of bounded self-adjoint operators on a complex separable
Hilbert space).

I just wanted to say something about this. An Abelian C*-algebra gives us a Probability theory, that is Kolmolgorov Probability. You may think of this as the kind of probability that results from our lack of knowledge. Non-Abelian gives us Quantum Theory. This is the mathematical insight of C*-algebras. Just like Non-Euclidean geometry is a generalisation of Euclidean geometry, Quantum Mechanics is a generalisation of Probability theory.

For instance the smallest and simplest Commutative C*-algebra is the Probability theory of a coin toss, 1/2 chance for tails, 1/2 chance for heads. The smallest Non-Commutative C*-algebra is the quantum theory of one particle with spin 1/2 and no other properties.

This actually provides a method of proving Bell's inequalities. If there was another theory underlying QM which gave definite values to things independant of context, then QM would simply be a result of our ignorance and the "randomness" would be due to a lack of knowledge. If this was the case the probability would be just usual probability, described by a commutative C*-algebra. However you can construct a set observables which must have correlations less than some value c in a commutative C*-algebra, but can have correlations exceeding c in Non-Commutative C*-algebras. Hence the predictions of QM are fundamentally different from such theories. Experiment supports the Non-Commutative C*-algebras and so QM is not just the result of our ignorance of an underlying theory.
By the way this proof does not assume locality.

Also the Non-Commutative C*-algebra approach provides what is in my opinion the best explanation of entanglement. As we all know, correlation is not causation. When we set up entangled particles and send them far apart to be measured we can easily see that the results are strongly correlated. However if we do the further statistical tests for causation then the correlations fail these tests. Hence the particles are not influencing each other. What then is the cause of these strange correlations? Simple, we are not dealing with old fashioned 19th/early 20th century probability. QM is a new theory of probability, one that allows stronger correlations than before. So it is not that the particles are causing effects on each other, but rather that they are more strongly correlated than is possible in probabilities that result from ignorance.

In this approach, it's not the case that every member of the C*-algebra corresponds to a measuring device, but I don't really have any more information on that. Perhaps someone can read those books and tell the rest of us.

Basically the Hermitian subalgebra of the whole C*-algebra is meant to correspond to measuring devices.

 

[Hurkyl]

Basically the Hermitian subalgebra of the whole C*-algebra is meant to correspond to measuring devices.

This goes back to my earlier gripe -- there's nothing physically stopping me from making a measuring device that outputs complex numbers. Singling out those elements whose anti-Hermetian part is zero as being more "real" is just an extension of the old bias that the complex numbers with zero imaginary part are somehow more real than the rest of them.

 

[DarMM]

This goes back to my earlier gripe -- there's nothing physically stopping me from making a measuring device that outputs complex numbers. Singling out those elements whose anti-Hermetian part is zero as being more "real" is just an extension of the old bias that the complex numbers with zero imaginary part are somehow more real than the rest of them.

I wouldn't so much see the "reality" of the measured output as being the reason for requiring Hermiticity. Of course if I take an observable A and an observable B, I can measure A + iB by just getting their values and put them into a complex number. Rather it has more to do with Unitarity. If an observable is not Hermitian then the transformation associated with it is not Unitary and it does not represent a good quantum number or even allow sensible quantum evolution. For example if H, the Hamiltonian wasn't Hermitian then time evolution wouldn't be Unitary, which would make the theory collapse. Similarly for momentum, linear and angular, rotations and translations wouldn't be unitary.
Hence Hermitian operators represent our observables, because only they represent good quantum numbers. For example only then will we be sure that when we obtain A = a that we are in a specific state by the spectral theorem.
Another example would be that measuring A + iB only really makes sense if A and B are compatible observables. So if the Hamiltonian, Linear Momentum and Angular Momentum have to be Hermitian, functions of them essentially exhaust all operators.
There are other reasons for Hermiticity, which I can go into if you want.

 

[Hurkyl]

It's clear why translation and other automorphisms should be unitary: they have to preserve the C* structure. And it's clear why the inner automorphisms -- those of the form T \mapsto U^* T U -- require U to be unitary element.


But (a priori, anyways) that has absolutely nothing to do with whether an element of the C*-algebra should correspond to a measuring device.



Actually, you bring up an interesting example. If you have a one-parameter family of unitary transformations U(t) -- such as time translation -- the corresponding infinitessimal element U'(0) is anti-Hermitian, not Hermitian. That we divide out by i to get a Hermitian element appears to me to be for no reason deeper than "people like Hermitian elements".

 

[sweet springs]

Hi.

Eigenvalue 0 doesn't cause any additional complications at all, so it doesn't need to be handled separately.

Thanks a lot, Fredrik.　Now I am fine with this eigenvalue 0 concern.

Now going back to my original question

Hi,
In 9.2 of my old textbook Mathematical Methods for Physicists, George Arfken states,
------------------------------------------------------
1. The eigenvalues of an Hermite operator are real.
2. The eigen functins of an Hermite operator are orthogonal.
3. The eigen functins of an Hermite operator form a complete set.*
* This third property is not universal. It does hold for our linear, second order differential operators in Strum-Liouville (self adjoint) form.
------------------------------------------------------

Advice on * of 3. , showing some "not forming a complete set" examples, are open and appreciated.

Regards.

 

[DarMM]

It's clear why translation and other automorphisms should be unitary: they have to preserve the C* structure. And it's clear why the inner automorphisms -- those of the form T \mapsto U^* T U -- require U to be unitary element.


But (a priori, anyways) that has absolutely nothing to do with whether an element of the C*-algebra should correspond to a measuring device.



Actually, you bring up an interesting example. If you have a one-parameter family of unitary transformations U(t) -- such as time translation -- the corresponding infinitessimal element U'(0) is anti-Hermitian, not Hermitian. That we divide out by i to get a Hermitian element appears to me to be for no reason deeper than "people like Hermitian elements".

Well let's concentrate on just the Hamiltonian. As you said we could work with A = iH, but we choose to work with A = iH. However this isn't really a very interesting case, it's similar to the case I described before in terms of sticking an i in front. So for example A + iB is fine as an observable, get a machine that measures both and add them together inside the machine and the machine will have measured a + ib. Nothing wrong with that. However these are trivial complex observables, formed from Hermitian observables anyway. What about a genuine complex observable like a non-Hermitian Hamiltonian which can produce eigenstates like E + i\Gamma? The problem is that such eigenvalues are actually describe decaying non-observable particles and there won't be a conservation of probability.
My basic idea is that while we can have things like iH, it is difficult to justify an arbitrary non-Hermitian operator. In the case of the Hamiltonian its because of the loss of conservation of probability, in the case of other operators it's usually because non-Hermitian observables don't form an orthogonal basis and hence aren't good quantum numbers.

However I can see that what I've said is basically an argument as to why non-Hermitian operators would be bad things to measure. What I haven't explained is why they actually can't be measured physically. I'll explain that in my next post since it takes a bit of work to set up.

 

[Hurkyl]

However these are trivial complex observables, formed from Hermitian observables anyway.

For the record, by this definition of "trivial", all elements of a C*-algebra are trivial: you can compute the real and imaginary parts just like an ordinary scalar:

X = (Z + Z*) / 2
Y = (Z - Z*) / (2i)​

giving

X* = X
Y* = Y
Z = X + iY​

 

[DarMM]

For the record, by this definition of "trivial", all elements of a C*-algebra are trivial: you can compute the real and imaginary parts just like an ordinary scalar:

X = (Z + Z*) / 2
Y = (Z - Z*) / (2i)​

giving

X* = X
Y* = Y
Z = X + iY​

Yeah, true. Bad example on my part, hopefully I can explain why we restrict ourselves to Hermitian observables in the next post. Once I have outlined the idea from the algebraic point of view hopefully we can have a more fruitful discussion.
Also I should say that in the relativistic context not all Hermitian operators are observables.

 

[DarMM]

Okay, when we make a measurement, a quantum mechanical object in the state \psi interacts with a classical measuring apparatus to record a value of some quantity \mathbb{A}. Mathematically this quantity is represented by an operator A. All the statistics for the observable such as the expectation, standard deviation, uncertainty e.t.c. can be worked out from the state and the observable. Let's take the expectation value, in your opinion should the expectation be represented as
\langle \psi, A\psi\rangle
or
\langle A\psi, \psi\rangle
Which one of these should represent an experiment to measure A?

 

[Hurkyl]

Let's take the expectation value, in your opinion should the expectation be represented as
\langle \psi, A\psi\rangle
or
\langle A\psi, \psi\rangle
Which one of these should represent an experiment to measure A?

Well, it would depend on how we chose to use the Hilbert space to represent states.

I'm going to go with the former, though. If \psi is a ket corresponding to the expectation functional \rho, then I prefer to have \rho(A) = \psi^* A \psi, which corresponds to the convention relating duals to inner products I assume we're using.

 

[DarMM]

Well, it would depend on how we chose to use the Hilbert space to represent states.

I'm going to go with the former, though. If \psi is a ket corresponding to the expectation functional \rho, then I prefer to have \rho(A) = \psi^* A \psi, which corresponds to the convention relating duals to inner products I assume we're using.

Funnily enough I should say, before I go on, that some people use my example above to argue why observables should not be Hermitian. That is they feel that physics should give the same answers regardless of which choice you use \langle \psi, A\psi\rangle or \langle A\psi, \psi\rangle. Or to put it in loose language "experiments cannot test the inner product".

Anyway on to the more important fact. It is an observed consequence of atomic measurement that if we measure a physical quantity and then measure that quantity again with no other quantities measured in between, then the chance of us obtaining the same answer is 100%. Given that this is fact of measurement how can we model it? Well if we obtained the value a we are in the state |a\rangle. Then since we know that we have no chance of measuring another value b for the same observable we want the probability to vanish for transition from |a\rangle to |b\rangle. That is we want \langle b,a\rangle = 0. So in order to match experiment observables must be represented by operators whose eigenvectors are orthogonal. Would you agree?
(Please tell me if something is incorrect.)

 

[Hurkyl]

I ground through some calculations, and I'm pretty sure that transition amplitude from \psi to b ought to be

the coefficient of |b\rangle in the representation of |\psi \rangle relative to the eigenbasis​

and not

the inner product of b with \psi.​

Of course, if you have an orthonormal eigenbasis, they are the same.

 

[DarMM]

I ground through some calculations, and I'm pretty sure that transition amplitude from \psi to b ought to be

the coefficient of |b\rangle in the representation of |\psi \rangle relative to the eigenbasis​

and not

the inner product of b with \psi.​

Of course, if you have an orthonormal eigenbasis, they are the same.

Really, why do you say? Perhaps I'm missing something, but I thought the usual definition of the transition probability was \langle b,a\rangle. How did you calculate what the transition probability was? Maybe I'm just being silly though!

 

[Hurkyl]

Well, the heuristic calculation I went through was as follows:

First, I want to make a toy example of a unitary operator that collapses the state in question. I chose the following one for no particular reason other than it was simple:

T|a,e_b> = |a,e_{b+a}>​

The Hilbert state space here is the tensor product of the state space we are interested in, with a basis labeled by the eigenvalues of whatever operator we're interested in, and another state space representing a toy environment, with basis states labeled by complex numbers. ("e" for "environment")

I chose a generic pure state in ket form:

|\psi\rangle = \sum_a c(a) |a\rangle​

computed the density matrix of the state:

T(|\psi \rangle \otimes |0 \rangle)​

and took the partial trace to get the resulting density matrix:

\sum_{a,b} c(b)^* c(a) \langle e_a | e_b \rangle\, |a\rangle\langle b|​

Since this evolution was supposed to collapse into a mixture of the eigenstates, I convinced myself that implies the environment states do need to be orthogonal, giving the density matrix:

\sum_{a} |c(a)|^2 |a\rangle\langle a|​

which is the statistical mixture that has probability |c(a)|^2 of appearing in state |a\rangle.

This toy seems reasonable since it gives the statistical mixture I was expecting, and eigenstates (e.g. |a\rangle) remain fixed, so the mixture generally remains stable.

So, if transition probabilities make sense at all, the transition probability from |\psi\rangle to |a\rangle has to be |c(a)|^2 -- in other words, the right computation for transition amplitude is the "coefficient of |a\rangle" function, rather than the "inner product with |a\rangle" function.

 

[strangerep]

A "linear combination" in an arbitrary vector space can certainly be an
infinite sum.

For an arbitrary vector space, what does "infinite sum" mean? There is only one topology, the Euclidean topology, that can be given to a finite-dimensional vector space, but an infinite-dimensional vector space can be given various topologies.

That's why I tried to distinguish such an arbitrary vector space
from a Hilbert space in my post.

I should possibly have said formal linear combination. I was thinking of the
"universal" space mentioned in Ballentine section 1.4.

Certainly, one can't do very much useful stuff in such an arbitrary vector space
before equipping it with a topology.

 

[strangerep]

[...] I think it's more appropriate to define a linear combination to only
have a finite number of terms.

Consider the usual kind of inf-dim Hilbert space which has an
orthonormal basis consisting of an infinite number of vectors.
In general, an arbitrary vector in that space can be expressed
as an infinite sum over the basis vectors. Surely such a sum
qualifies as a linear combination?

This is why: If V is a vector space over a field F, and S is a subset of V, the "subspace generated by S" (or "spanned" by S) can be defined as any of the following

a) the smallest subspace that contains S
b) the intersection of all subspaces that contain S
c) \Big\{\sum_{i=1}^n a_i s_i|a_i\in\mathbb F, s_i\in S, n\in\mathbb N\Big\}

These definitions are all equivalent, and the fact that every member of the set defined in c) can be expressed as \sum_{i=1}^n a_i s_i, with n finite, seems like a good reason to define a "linear combination" as having only finitely many terms. [...]

One can also find inf-dim subspaces in general, in which case those arguments about
finite sums don't apply.

 

[Fredrik]

...an infinite sum over the basis vectors. Surely such a sum
qualifies as a linear combination?

The theorem I mentioned is valid for arbitrary vector spaces. (I'm including the proof below). It implies that if we define "linear combination" your way, the following statement is false:

The subspace generated (=spanned) by S is equal to the set of linear combinations of members of S.

So let's think about what you said in the text I quoted. The definition of "subspace generated by" and the theorem imply that a vector expressed as

x=\sum_{n=1}^\infty \langle e_n,x\rangle e_n

with infinitely many non-zero terms does not belong to the subspace generated by the orthonormal basis. That's odd. I didn't expect that.

I think the explanation is that terms like "linear combination" and "subspace generated by" were invented to be useful when we're dealing with arbitrary vector spaces, where infinite sums may not even be defined. And then we stick to the same terminology when we're dealing with Hilbert spaces.

I haven't tried to prove it, but I'm guessing that the subspace spanned by an orthonormal basis is dense in the Hilbert space, and also that it isn't complete. But vectors like the x mentioned above can be reached (I assume) as a limit of a sequence of members of the subspace generated by the basis. (A convergent sum of the kind that appears on the right above is of course a special case of that).

One can also find inf-dim subspaces in general, in which case those arguments about
finite sums don't apply.

The theorem holds for infinite-dimensional vector spaces too. The proof is very easy. Let V be an arbitrary vector space, and let S be an arbitrary subset. Define \bigvee S to be the intersection of all subspaces W_\alpha such that S\subset W_\alpha. I'll write this intersection as

\bigvee S=\bigcap_\alpha W_\alpha

Define W to be the set of all linear combinations of members of S. (Both here and below, when I say "linear combination", I mean something with a finite number of terms).

W=\Big\{\sum_{i=1}^n a_i s_i|a_i\in\mathbb F, s_i\in S, n\in\mathbb N\Big\}

I want to show that W=\bigvee S. First we prove that W\subset\bigvee S.

Let x be an arbitrary member of W. x is a linear combination of members of S, but S is a subset of every W_\alpha. So x is a linear combination of members of W_\alpha for every \alpha. The W_\alpha are subspaces, so that implies that x is a member of every W_\alpha. Therefore x\in\bigvee S[/tex], and that implies W\subset\bigvee S.<br /> <br /> Then we prove that \bigvee S\subset W. It&#039;s obvious from the definition of W that it&#039;s closed under linear combinations. That means that it&#039;s a subspace. So it&#039;s one of the terms on the right in<br /> <br /> \bigvee S=\bigcap_\alpha W_\alpha<br /> <br /> That implies that \bigvee S\subset W.