Observables in Quantum Mechanics

[BlackHole]

I think that commutators can be solved with the methods of algebraic topology. It's just that general relativity uses differential geometry instead. The second choice is better.

[Johnny5]

I think that commutators[/url'] can be solved with the methods of algebraic topology. It's just that general relativity uses differential geometry instead. The second choice is better.

 

Ok thanks.

[Severian]

I was studying GR about a month ago' date=' and tensors came up.

 

Is the commutator algebra here, related to tensors?

[/quote']

 

They are not directly linked. A commutator is a realation between operators and operators can be tensors, but they doen't have to be (for example, the momentum operator is a vector). Similarly tensors don't have to be operators (the tensors in GR are not operators because it is a classical theory).

 

(Of course, some maths purist will come along and point out that a vector is a rank 1 tensor and a scalar is a rank 0 tensor, which is true, but normally in physics (and we are in the physics section) the word tensor is reseverd for rank 2 and above.)

[Johnny5]

(Of course' date=' some maths purist will come along and point out that a vector is a rank 1 tensor and a scalar is a rank 0 tensor, which is true, but normally in physics (and we are in the physics section) the word tensor is reseverd for rank 2 and above.)[/quote']

 

Ok. You know that actually helps.

 

Regards

[Tannin]

Thanks to all for opening this interesting subject of complex observables!

The previous discussion referred to the non-Hermitian operator X+iY. It's eigenvalues were supposed to be x+iy and then the meaning of vector in 2D plane was attached to the above complex number.

I used to know, that when trying to build two-dimensional operator (i.e. something operating in two-dimensional space and returning position vector), the concept of direct product is used. Then X and Y are residing in different vector spaces and X+iY is just some formal mathematical construction - an object defined as association of something from X-space and Y-space. I can't see how we can move from this point towards the vector in 2D plane...

Can we still find some physical observable having complex values?

When discussing light propagation in matter, the refractive index is a complex quantity, the real part is responsible for refraction, and the imaginary part for absoprtion and emission. But I'm not sure that the refractive index is actually the physical observable, because I've never seen the expression for refractive index operator.

[Tom Mattson]

I used to know' date=' that when trying to build two-dimensional operator (i.e. something operating in two-dimensional space and returning position vector), the concept of direct product is used.

[/quote']

 

You could use that. We don't even need to consider the issue of complex eigenvalues here. We can just look at the operator [imath]\hat{R}=<\hat{X},\hat{Y}>[/imath] whose eigenvalues are vectors [imath]\vec{r}=<x,y>[/imath] in [imath]\mathbb{R}^2[/imath]:

 

[math]\hat{R}|\vec{r}>=\vec{r}|\vec{r}>[/math]

 

The notion of the direct product arises naturally from the fact that our eigenvalues can be construed as ordered pairs.

 

Then X and Y are residing in different vector spaces and X+iY is just some formal mathematical construction - an object defined as association of something from X-space and Y-space.

 

I'm not sure I follow you.

 

In one sense [imath]\hat{X}[/imath] and [imath]\hat{Y}[/imath] both operate on the same vector space, in that they both operate on the same type of vector space, namely [imath]\mathbb{R}[/imath].

 

In another sense the two operators act on different vector spaces, because you could say that they operate on their own private, disjoint copy of [imath]\mathbb{R}[/imath].

 

I can't see how we can move from this point towards the vector in 2D plane...

 

I think you already said it yourself: We take a formal combination of the components. What is it that you think is missing?

 

Can we still find some physical observable having complex values?

 

This thread is about the definition of the word "observable".

 

That is, do we define "observable" to be "that which corresponds to a Hermitian operator? Or do we define it as "that which corresponds to a normal operator?

 

I would say that the former is acceptable, but too restrictive and that the latter is to be preferred.

 

When discussing light propagation in matter, the refractive index is a complex quantity, the real part is responsible for refraction, and the imaginary part for absoprtion and emission. But I'm not sure that the refractive index is actually the physical observable, because I've never seen the expression for refractive index operator.

 

I haven't seen it either, nor would I expect to. Loosely speaking, the index of refraction is a macroscopic property of a material, applicable when you can ignore the details of its structure. If you are considering the (quantum mechanical) dynamics of a particle through a medium, then I cannot imagine that you would ever find the notion of a refractive index useful. Of course, I could be wrong, as I don't know much about solid state physics.

[Severian]

I think this discussion is rather semantic at present, since no-one can think of how a 'complex observable' would be manifest.

 

When you make a measurement you are given a real number of a series of real numbers. Even if somehow you managed to measure a 'complex observable' the apparatus is not going to suddenly spit out a great big 'i' indicating that it is complex - you will either get one real number (the imaginary part of the complex number) or you will get 2 (the real and imaginary parts) (more likely you would get a magnitude and a phase).

 

Actually we already have this in QM - we can measure the size of a cross-section and the phase between different processes. But while this may look like we are measuring a complex number, we are not - we are separately measuring the magnitude and phase, which are both real numbers.

 

Furthermore, in QFT (or QM) one does not make the statement that all measurable quantities must be represented by Hermitian operators. One instead says that observables that we know how to measure such as momentum, energy, production rates, must be real and therefore represented by Hermitian operators. Since we can measure these things. they are undisputably real, so this must be true.

 

I challenge you to come up with a 'complex observable' and I will show you that it is not a complex observable, but simply two real observables.

[Tannin]

I admit I didn't check this, so please correct me...

May be the real and complex observables may be distinguished by their properties under coordinate transformations. Real quantity behaves like a scalar, i.e. remains invariant, while the complex number should transform like a vector in two dimensions.

Thus if the measurement outcome is a 2D-vector (and not just series of real numbers transforming independently), maybe it may be termed a complex observable.

[Tom Mattson]

When you make a measurement you are given a real number of a series of real numbers. Even if somehow you managed to measure a 'complex observable' the apparatus is not going to suddenly spit out a great big 'i' indicating that it is complex - you will either get one real number (the imaginary part of the complex number) or you will get 2 (the real and imaginary parts) (more likely you would get a magnitude and a phase).

 

The main point of this thread (esp. in Post #20) is not that you can get a measurement apparatus to spit out an 'i'' date=' but rather that observables of the form [imath']z=x+iy[/imath] aren't any more dubious than observables of the form [imath]\vec{r}=x\hat{i}+y\hat{j}[/imath]. For some reason people reject the former as observables while readily accepting the latter. But, as you noted, in both cases one is just measuring 2 real components and putting them together in a formal linear combination.

 

Furthermore, in QFT (or QM) one does not make the statement that all measurable quantities must be represented by Hermitian operators.

 

But some people do make that statement.

 

As an example, take the entry from Math World entitled Hermitian Operator. It says:

 

Every observable must therefore have a corresponding Hermitian operator.

 

This thread is in response to remarks such as that.