How can polarisation by angle 45 degrees be represented in quantum mechanics?

[Jamma]

Hi there, mathematician learning a little QM here.

I've been watching a few of these lectures:

http://www.youtube.com/watch?v=LBFBQr_xKEM&feature=relmfu


The lectures are good, a little slow paced for me, but clear at least.

He starts talking about polarisation in the last half of the first link, and continues in the second. He introduces basis vectors for polarisation in the vertical and horizontal oritentations. I get totally all that's going on, and the maths, but then he introduces polarisation by angle 45 degrees. To do this, he adds the basis vectors for vertical and horizontal directions and normalises.

I can see where he's coming from, but I don't understand- surely this state he's created is a "mixed" state, and represents a state in which the particle has a 50% chance of being polarised vertically and 50% horizontally.

This may simplify things, and is sort of inline with the notion of 45 degrees in vector calculus, but this doesn't seem to be inline with the notation of QM, surely he should introduce a new basis vector for each new state i.e. he should have a continuum of states ranging from 0 degrees up to 90 degrees.

I can still follow what he's doing, this is just a niggling annoyance. Any help appreciated.

 

[Bill_K]

Yep, this is the fundamental idea behind quantum mechanics, and is the reason why we talk about state vectors in the first place. It's called the superposition principle. The linear combination he's constructed is both: a state in which the polarization along the 45-degree line is 100 percent, *and* a state in which there's a 50-50 chance of being polarized vertically or horizontally.

 

[Jamma]

Huh? Sorry, can you clarify? How can it be 100% certain to be polarised along a 45-degree line and also 100% certain to be polarised along one of the vertical or horizontal lines? That doesn't make sense.

How, in this scheme, would I differentiate between a polarisation that has 100% certainty of being polarised at 45 degrees and one which has 50% of being horizonatally and 50% of being vertically polarised?

 

[Jamma]

Oh, or are you saying that in this scheme, there doesn't exist a state which is polarised at angle 45 degrees with 100% certainty, one of being 50% likely to be horizontal and 50% of being vertical IS a polarisation of 45 degrees? This is very different to, say, position in which there are vectors which represent a certain place, but there are also ones where the distribution may be "smeared".

I must say, rigour seems to be left behind a tad when doing this kind of physics, I've already been wanting a better setting for the delta function!

 

[zonde]

Pure state of 100% photons being polarized at 45° is different than mixed state of 50% H photons and 50% V photons.
Simple test with polarizer will show the difference.

What QM claims is that mixed state of 50% H photons and 50% V photons is the same as mixed state of 50% +45° photons and 50% -45° photons.

 

[Jamma]

Ok, so I am correct then, what the guy is doing isn't right, because he was setting it all up as if what he'd created was a pure state of 100% photons being polarised at 45°, when he should have been saying that he was setting up one with 50% V and 50% H?

Am I right in saying that? (for brevity, in the lecture, he puts V as |x> or (1,0) and H as |y> or (0,1) and puts the 45° one as (|x>+|y>)/(root(2)) or (root(2),root(2)).

"What QM claims is that mixed state of 50% H photons and 50% V photons is the same as mixed state of 50% +45° photons and 50% -45° photons. "
I assume here that this applies to any angle. Is this because no observable could ever distinguish between the two states?

 

[netheril96]

I've got confused by all your guys. Polarization of 45 degrees is the superposed state of [tex]\frac{1}{\sqrt{2}}\left(|H\rangle+|V\rangle\right)[/tex], not a mixed state.

 

[Jamma]

Isn't that what a mixed state is, a linear combination of the pure basis states? If it is not a mixed state, then how do you express the mixed state with 50% prob H and 50% V?

 

[netheril96]

Isn't that what a mixed state is, a linear combination of the pure basis states? If it is not a mixed state, then how do you express the mixed state with 50% prob H and 50% V?

Mixed state is described in terms of density matrix.

 

[ThomasT]

About mixed state and pure state

 

[Jamma]

Hmm, ok, thanks. I think that I just had my definitions of mixed and pure states wrong, I assumed that mixed states were states which were a linear combination of different basis vectors.

I think my reason for this was to do with getting probabilities out of them by multiplying by conjugates, yadda yadda.

So mixed states are actually operators, you put in a ket vector and it gives you another ket vector? How does this set up work? So if I want to find the probability that it is in a state "x", do I simply let it act on "x" and see if it is an eigenvector of it, and if it is, the eigenvalue corresponds to the probability? I see how everything else follows through, act with an observable and you get another mixed state etc. etc.

 

[strangerep]

So mixed states are actually operators, you put in a ket vector and it gives you another ket vector? How does this set up work? So if I want to find the probability that it is in a state "x", do I simply let it act on "x" and see if it is an eigenvector of it, and if it is, the eigenvalue corresponds to the probability?

There's a bit more to it than that. Take a look at this Wiki page:

http://en.wikipedia.org/wiki/Density_operator

Better still, find a copy of Ballentine's textbook "QM - A Modern Development".

I must say, rigour seems to be left behind a tad when doing this kind of physics,
I've already been wanting a better setting for the delta function!

You said you're a mathematician so I guess you've heard of Schwarz space,
its closure as a Hilbert space, and its dual as the space of tempered distributions
a.k.a "Gel'fand triple"? More often known in QM as "rigged Hilbert space"?
That's what's really going on here. Ballentine gives quite readable introduction
to these points.

HTH.

 

[Jamma]

Cool, might have a look for that book in my library, thanks. Tbh, I've never heard of Schwartz space before, seems like quite an isolated concept, but will check it out.