How to obtain state vector for polarised light

[etotheipi]

If I'm using the basis vectors |u> and |r> for two polarisation states which are orthogonal in state space, I've seen the representation of a general state oriented at angle theta to the horizontal written as $$\lvert\theta\rangle = \cos(\theta) \lvert r \rangle + \sin(\theta) \lvert u \rangle$$This representation makes some sense since we could substitute in values of 0 and 90 degrees for theta and see that it reduces to the basis states, however I don't know how it is derived in the first place.

I asked my teacher and he just said to consider the horizontal and vertical components of light at some angle, however this didn't convince me very much since I've learned that basis states are abstract and their behaviour doesn't necessarily mirror what they look like in a geometrical sense. For example, although "up" and "down" spin states are not orthogonal in a physical sense, they are orthogonal in state space. Consequently, it doesn't seem good enough to just state that since the component of light in the x direction is cos(θ), the |r> component of the abstract state vector must also take this value.

Instead, I have learned that we obtain other state vectors by seeing what values make the probabilities correct. This state vector indeed satisfies these requirements, which we can easily "show" by taking the square of the inner product with one of the basis states and testing different values of theta.

However, is there a way of actually deriving this result instead of just checking that it works? Thank you!

 

[tnich]

Consequently, it doesn't seem good enough to just state that since the component of light in the x direction is cos(θ), the |r> component of the abstract state vector must also take this value.

In the case of polarization, though, the $|r\rangle$ state does correspond to a physical direction. You can demonstrate that with a laser pointer and a couple of polarizing filters.

 

[etotheipi]

In the case of polarization, though, the $|r\rangle$ state does correspond to a physical direction. You can demonstrate that with a laser pointer and a couple of polarizing filters.

My confusion is that, in the case a spin, we can choose the two basis vectors to be |u> and |d> - these definitely correspond to physical directions, as we can see through e.g. the Stern-Gerlach experiment - however we can also use them to construct the state vector for any other direction (e.g. |r>) even though there's no way of physically obtaining that direction with only up and down vectors. Instead, they relate to other state vectors corresponding to different spatial directions only really in terms of probabilities.

Even though in the case of polarisation |u> and |r> happen to be orthogonal in both state space and physical space, shouldn't it be only the relationship in state space which matters?

 

[tnich]

My confusion is that, in the case a spin, we can choose the two basis vectors to be |u> and |d> - these definitely correspond to physical directions, as we can see through e.g. the Stern-Gerlach experiment - however we can also use them to construct the state vector for any other direction (e.g. |r>) even though there's no way of physically obtaining that direction with only up and down vectors. Instead, they relate to other state vectors corresponding to different spatial directions only really in terms of probabilities.

Even though in the case of polarisation |u> and |r> happen to be orthogonal in both state space and physical space, shouldn't it be only the relationship in state space which matters?

If you use a polarizing filter to set the direction of linear polarization of a photon to $|\psi\rangle$, and then measure its polarization against an arbitrary pair of orthogonal linear polarizations $|r\rangle$ and $|u\rangle$ with the direction of $|r\rangle$ differing from that of $|\psi\rangle$ by angle $\theta$, then you would expect the probability of measuring the polarization as $|r\rangle$ to be $\cos^2\theta$ and the probability of measuring it as $|u\rangle$ to be $\sin^2\theta$. So I would argue that in this case, the angle $\theta$ does have a physical meaning.

 

[strangerep]

If I'm using the basis vectors |u> and |r> for two polarisation states which are orthogonal in state space, I've seen the representation of a general state oriented at angle theta to the horizontal written as $$\lvert\theta\rangle = \cos(\theta) \lvert r \rangle + \sin(\theta) \lvert u \rangle$$
[...] is there a way of actually deriving this result instead of just checking that it works?

You have 2 orthogonal state vectors $|u\rangle$ and $|r\rangle$, thus you are dealing with a 2D Hilbert space $H$ necessarily spanned by these 2 state vectors (because they are orthogonal). I.e., any vector in $|\psi\rangle \in H$ can be expressed as a linear combination: $$|\psi\rangle ~=~ a |u\rangle ~+~ b |r\rangle ~,$$ where $a,b$ are complex constants. If you want $|\psi\rangle$ to be normalized, it must satisfy the constraint: $$1 ~=~ \langle \psi | \psi \rangle ~.$$ Now, (exercise!) what constraint does this normalization condition impose between $a$ and $b$?

 

[etotheipi]

You have 2 orthogonal state vectors $|u\rangle$ and $|r\rangle$, thus you are dealing with a 2D Hilbert space $H$ necessarily spanned by these 2 state vectors (because they are orthogonal). I.e., any vector in $|\psi\rangle \in H$ can be expressed as a linear combination: $$|\psi\rangle ~=~ a |u\rangle ~+~ b |r\rangle ~,$$ where $a,b$ are complex constants. If you want $|\psi\rangle$ to be normalized, it must satisfy the constraint: $$1 ~=~ \langle \psi | \psi \rangle ~.$$ Now, (exercise!) what constraint does this normalization condition impose between $a$ and $b$?

Would the constraint be $$|a|^{2} + |b|^{2} = 1$$which is indeed satisfied by cos(θ) and sin(θ)?

 

[strangerep]

Would the constraint be $$|a|^{2} + |b|^{2} = 1$$which is indeed satisfied by cos(θ) and sin(θ)?

Yes.