Observables in Quantum Mechanics

[Tom Mattson]

This thread is an offshoot of CPL.Luke's thread imaginary numbers. That thread is about FTL travel, tachyons, etc. Since what I have to say is about plain vanilla quantum mechanics, I didn't want to clog his thread up any more than I already have, so I started this one. Here goes.

 

There seems to be a prejudice against physical observables taking on complex values. This prejudice is no doubt inspired by the statement in most QM textbooks that observables must correspond to Hermetian observables, which have real eigenvalues. But this is false! It is not the case that an observable must have real eigenvalues. All that is required of an operator to represent a physical observable is that its eigenvectors form an orthonormal basis. While Hermiticity certainly is a sufficient condition for this, it is not a necessary one. A necessary and sufficient condition for an operator to correspond to an observable is normality. A normal operator A is one which satisfies [A,A⁺]=0.

 

After all, take two observable operators: X and Y (whose eigenvalues are the x and y coordinates, respectively). What sense is there in saying that X+iY is not also an observable? Absolutely none, since both parts are observable.

[revprez]

A necessary and sufficient condition for an operator to correspond to an observable is normality. A normal operator A is one which satisfies [A,A⁺']=0.

 

Wait, I'm confused. I thought the definition of a normal operator is one that commutes with its Hermitian adjoint.

 

Rev Prez

[Tom Mattson]

Wait' date=' I'm confused. I thought the definition of a normal operator is one that commutes with its [i']Hermitian[/i] adjoint.

 

Right. That's what [A,A⁺]=0 means.

[revprez]

Right. That's what [A,A⁺']=0 means.

 

Can you just briefly explain the notation. I'm assuming you're not saying AA⁺ = A⁺A = 0.

 

BTW, is LaTeX working for you?

[Tom Mattson]

Can you just briefly explain the notation. I'm assuming you're not saying AA⁺ = A⁺A = 0.

 

No' date=' I'm not saying that. [x,y'] is the notation for the commutator. It means:

 

[x,y]=xy-yx.

 

BTW, is LaTeX working for you?

 

It's on and off, unfortunately. That's why I'm typing everything out.

 

Edit: Let's try it.

 

[math][A,A^{\dagger}]=0[/math]

 

Does that work?

 

Edit Again: No, it doesn't.

[Johnny5]

No' date=' I'm not saying that. [x,y'] is the notation for the commutator. It means:

 

[x,y]=xy-yx.

 

 

Notation for commutator, yeah.

 

So in the case where the quantities x,y commute then xy=yx whence it follows that [x,y]=0.

 

When you write [A, A⁺] does A⁺ denote the complex conjugate of A?

[Tom Mattson]

No, it's the Hermitian conjugate. You can't take the complex conjugate of an operator.

[revprez]

No' date=' I'm not saying that. [x,y'] is the notation for the commutator. It means:

 

[x,y]=xy-yx.

 

All right.

 

So what's an example of a complex-valued observable. The Hamiltonian obviously isn't.

 

It's on and off, unfortunately. That's why I'm typing everything out.

 

Edit: Let's try it.

 

[math][A,A^{\dagger}]=0[/math]

 

Does that work?

 

Edit Again: No, it doesn't.

 

Yeah, sucks doesn't it? What are they using as their converter? Texvc?

 

Rev Prez

[Tom Mattson]

So what's an example of a complex-valued observable.

 

I gave one in my opening post: X+iY.

 

Going on with that' date=' if [b']R[/b]=<X,Y,Z> is the displacement operator, and |r> are the eigenkets of X, then:

 

X|r>=x|r>

Y|r>=y|r>

 

From this it is possible to construct the normal (but not Hermitian) operator X+iY, with complex eigenvalues x+iy. The point of my opening post is to highlight the fact that X+iY is in fact an observable, and that normal, non-Hermitian operators should not be dismissed as unphysical simply because they have complex eigenvalues.

 

The Hamiltonian obviously isn't.

 

That's right, and it can't be non-Hermitian. But that's not because it's an observable, it's because H is the generator of time translations. If H is not Hermitian, it will give rise to a nonunitary time evolution operator, which will violate probability conservation.

 

Yeah, sucks doesn't it? What are they using as their converter? Texvc?

 

Don't know. But it's so jittery here that I've given up on using it.

[revprez]

Notation for commutator' date=' yeah.

 

So in the case where the quantities x,y commute then xy=yx whence it follows that [x,y']=0.

 

When you write [A, A⁺] does A⁺ denote the complex conjugate of A?

 

Just to add on Tom's explanation, a Hermitian operator is its own complex conjugate.

[Tom Mattson]

A⁺ isn't the complex conjugate of A. It's the Hermitian conjugate. Just to highlight the difference: take the following matrix representation of some operator (doesn't matter what it represents):

 

   [1 0 1]
A=  [2 2 0]
   [1 0 2]

 

Now if you took the complex conjuate A^* of A, you would just get A again (because A is real). So A^*=A.

 

But if you took the Hermitian conjugate A⁺ of A, you would not get A back. That's because A⁺=A^*T, where T denotes "transpose". So you would get for A⁺:

 

  [1 2 1]
A[sup]+[/sup]=[0 2 0]
  [1 0 2]

 

So A⁺ does not equal A.

[revprez]

A⁺ isn't the complex conjugate of A. It's the Hermitian[/i'] conjugate. Just to highlight the difference: take the following matrix representation of some operator (doesn't matter what it represents):

 

Yeah, for a Hermitian operator A⁺ is just A^T.

 

So X + iY or (<x,y>)e^iwt is some rotation operator, correct?

 

Rev Prez

[Tom Mattson]

So X + iY or (<x' date='y>)e[sup']iwt[/sup] is some rotation operator, correct?

 

No, not quite. First, x+iy does not equal (<x,y>e^iwt). And second, it doesn't give a rotation.

 

What the operator X+iY does is maps coordinates x and y into vectors (x,y) in the complex plane. In other words it's a linear transformation T:R²-->C. I can't cite a practical application of it at this time but that's OK for the purpose of this thread, which is to discuss why the inference from "complex eigenvalue" to "unphysical, unobservable" is invalid.

[revprez]

No, not quite. First, x+iy does not equal (<x,y>e^iwt[/sup']).

 

^{But it does equal <x,y>ei(theta). That's just Euler's formula.}

 

^{And second, it doesn't give a rotation.}

 

^{Well, it doesn't have to, but of X and Y are parameterized by some t then it should, right?}

 

^{What the operator X+iY does is maps coordinates x and y into vectors (x,y) in the complex plane. In other words it's a linear transformation T:R2-->C. I can't cite a practical application of it at this time but that's OK for the purpose of this thread, which is to discuss why the inference from "complex eigenvalue" to "unphysical, unobservable" is invalid.}

 

^{Can't the Hamiltonian decompose into complex kinetic and potential energies, and in fact isn't that what we'd expect in a quantum harmonic oscillator?}

 

^{Rev Prez}

[Tom Mattson]

But it does equal <x,y>e^{i(theta)[/sup']. That's just Euler's formula.}

 

^{By <x,y>, do you mean (x2+y2)1/2?}

 

^{If so, then I agree that x+iy=<x,y>eiq, and that it is the Euler formula.}

 

^{I originally disagreed because when I see <x,y> I normally think of the R2 vector xi+yj. And of course, x+iy does not equal (xi+yj)eiwt.}

 

^{Well, it doesn't have to, but of X and Y are parameterized by some t then it should, right?}

 

^{Hmmm...Maybe I'm not seeing what you mean, because I still don't think so. If I make a row vector [x y], and do a transformation on it to get [x iy], then I am struggling to see how that could be called a "rotation". About what axis and through what angle was it rotated?}

 

^{Can't the Hamiltonian decompose into complex kinetic and potential energies,}

 

^{Only if those complex quantities add up to a real quantity. If complex energy eigenvalues are allowed, then probability is not conserved.}

 

^{and in fact isn't that what we'd expect in a quantum harmonic oscillator?}

 

^{*I* wouldn't expect it. In fact I've never seen it treated that way. Can you elaborate?}

[revprez]

By <x,y>, do you mean (x²+y²)^1/2[/sup']?

 

^{Yeah, my bad. Should be |X + iY|}

 

^{Hmmm...Maybe I'm not seeing what you mean, because I still don't think so. If I make a row vector [x y], and do a transformation on it to get [x iy], then I am struggling to see how that could be called a "rotation". About what axis and through what angle was it rotated?}

 

^{Isn't the distance preserved in a transformation between [x y] and [x iy]?}

 

^{Only if those complex quantities add up to a real quantity.}

 

^{Right, got that.}

 

^{If complex energy eigenvalues are allowed, then probability is not conserved.}

 

^{And the total energy has to be real anyway.}

 

^{*I* wouldn't expect it. In fact I've never seen it treated that way. Can you elaborate?}

 

^{I'll get back to you on that.}

 

^{Rev Prez}

[Tom Mattson]

Isn't the distance preserved in a transformation between [x y] and [x iy]?

 

Yes, but parity and translation also preserve the norm of a vector, and neither is a rotation.

 

The Moral: Preservation of distance is not enough to guarantee that an operation is a rotation.

[Severian]

since you can reformulate all of physics in a way which doesn't use complex numbers at all (although it doesn't look nice) I don't see how there can be anything 'non-physical' about them being used in physics. Just think of every 'number' as being a 2d vector in some vector-space. The first entry is the 'real' part while the second is the 'imaginary' part, and the magnitude of the vector is the modulus. This should tell you that the imaginary entry is just as physical as the real one.

[Dave]

Fresh start. Please don't get off-topic this time.

 

Cheers.

[Tom Mattson]

Just think of every 'number' as being a 2d vector in some vector-space. The first entry is the 'real' part while the second is the 'imaginary' part' date=' and the magnitude of the vector is the modulus. This should tell you that the imaginary entry is just as physical as the real one.[/quote']

 

Your mention of vectors hits on something that I think is important. I am trying to find the article, but someone (on Wikipedia, I believe) said that it is only natural that we require observables to correspond to Hermitian operators because there is no measurement apparatus that has complex numbers as its output. I would object to this on two points:

 

1. There is also no measurement apparatus that has R³ vectors as its output, but I'm sure the same person would happly consider, say, the position operator X an observable.

 

2. There is no reason a measurement apparatus could not be constructed to return complex numbers. To measure X+iY, simply have it measure X and Y, and return X+iY on the display.

 

I will try to find that article because now I'm curious to know who wrote it.

[Johnny5]

Wait' date=' I'm confused. I thought the definition of a normal operator is one that commutes with its [i']Hermitian[/i] adjoint.

 

Rev Prez

 

Revprez says the above in post 2, and in post 7

 

Tom says

No' date=' it's the Hermitian conjugate. You can't take the complex conjugate of an operator.[/quote']

 

So my question is, does Hermitian adjoint mean the same thing as Hermitian conjugate? And is this the thing with the dagger in it? Is that + sign above the A supposed to represent the dagger symbol?

 

I mean the + thing here:

 

You have [A,A⁺] being the notation for the commutator of A (A is an operator) and its Hermitian adjoint/Hermitian conjugate. And the commutator is general defined as: [c,d]=cd-dc

 

So in this case we have:

 

[A,A⁺] = AA⁺- A⁺A

 

And it will commute with its Hermitian adjoint provided

 

[A,A⁺] = AA⁺- A⁺A=0

[Tom Mattson]

(stuff)

 

"Yes" on all counts.

[revprez]

since you can reformulate all of physics in a way which doesn't use complex numbers at all (although it doesn't look nice) I don't see how there can be anything 'non-physical' about them being used in physics. Just think of every 'number' as being a 2d vector in some vector-space. The first entry is the 'real' part while the second is the 'imaginary' part, and the magnitude of the vector is the modulus. This should tell you that the imaginary entry is just as physical as the real one.

 

Correct me if I'm wrong, but are you and Tom proposing that measurement itself is a map from the complex numbers to the reals?

[Severian]

Correct me if I'm wrong, but are you and Tom proposing that measurement itself is a map from the complex numbers to the reals?

 

That is not really what I was saying, but yes that is correct. (I was going a step further and saying that there is really nothing 'unreal' about complex numbers in the first place.)

[Johnny5]

I am reading this now at Wolfram, on the commutator

 

Wolfram on commutator

 

They mention tensors if you look down a little.

 

I was studying GR about a month ago, and tensors came up.

 

Is the commutator algebra here, related to tensors?

 

 

Specifially I am wondering about equation 9 at the wolfram site.