Spin states

[Prometheus]

I'm so confused i'm not even sure what i'm confused about, so any help appreciated.

 

So i understand the z-component of the spin state of a particle can represented by [math]|A \rangle = a_1 |\uparrow_z \rangle + a_2|\downarrow_z \rangle [/math], where [math] |\uparrow_z \rangle [/math] and [math] |\downarrow_z \rangle [/math] provide an orthonormal basis in spin space and [math]a_1, a_2 [/math] are complex numbers.

 

I'm also aware that [math]|\uparrow_z \rangle = \frac{1}{\sqrt{2}} |\uparrow_x \rangle - \frac{1}{\sqrt{2}} |\downarrow_x \rangle [/math]. But my understanding of [math]|A \rangle[/math] above is that it only spans a 2-dimensional space defined by the up and down spin of the z-component. How is it then, that the x-component of spin can be expressed as a linear combination of the z-component (and similarly the y-component)?

 

 

 

[Mordred]

To start with your Hilbert space/inner product space is a function space.

 

An inner product takes two vectors and computes a scalar. Two vectors are orthogonal if there inner product is zero

 

[latex]a\cdot b=0 [/latex]

 

https://en.m.wikipedia.org/wiki/Inner_product_space

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=https://www.math.ust.hk/~mabfchen/Math111/Week13-14.pdf&ved=0ahUKEwjBsZPp78bRAhUfHGMKHTxEA4wQFghiMBA&usg=AFQjCNF76FS1ZmkRS9-t2ozicCwdjrMPSA&sig2=lorcxcv37fWKzyg_ooOWjg

 

Spin up being orthogonal to spin down

 

The only difference being a change in sign. In this case treat up/down as equivalent to + or -

 

In spin ask yourself this question "how many dimensions are required to describe rotation?" ie a sine wave ? with an electron, the electron whether or not is spin up or spin down has Rotational symmetry. The ie its angular momentum.

 

This should provide the clues you need Particularly on Hilbert space/inner product space.

 

Here this will greatly help with complex conjugates.

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.matha.rwth-aachen.de/de/lehre/ws07/calculus/Lecture_1.pdf&ved=0ahUKEwi6y7S2xsfRAhVU0mMKHRMxCHg4ChAWCBwwAQ&usg=AFQjCNGRt0Bmjd43cIBybgNLXOY2TPWCCQ&sig2=0xMcr8MDGr9VL4y1g-wjiA

 

On this post the above is the best article and the most pertaining directly to your question.

 

A more simplified coverage here though far lacking the details of the previous link

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://www.phas.ubc.ca/~mav/p200/complex.pdf&ved=0ahUKEwia_aTVy8fRAhVQVWMKHRcDC4g4ChAWCCMwBA&usg=AFQjCNGQLX9ted9jjkd_JWx3h-Yv22lSUw&sig2=vye_DwLruXuw34TvidZIvg

Edited January 16, 2017 by Mordred

[Prometheus]

This arm has orthogonal symmetry the only difference between going up and going down is the direction itself. Everything else about that robot arm is identical (symmetric) so we can reduce this two directions to simply a change in sign. Plus or minus. For Prometheus benefict an inner product. Its range of motion is 1 dimensional. As Prometheus has posed a related question in spin I will add another type of symmetry example.

 

A mounted fan can either spin clock wise or counterclockwise. The fan blade itself doesn't change only the direction of motion "Rotational symmetry" so once again we can reduce the number of possible directions to an inner product. As rotation requires 2 dimensions to describe its range of motion inner product space is two dimensional ie spin space (I know you are having difficulty visualizing inner products Prometheus these two examples will help) Its not an attempt to call attention to you but an effort to assist you get past all the fancy mathematical terminology such as complex conjugates/ operators etc.(though these are also extremely important terms)

I'm not sure my problem here is with inner product spaces (though for sure that's a problem in general, and i don't know what i don't know).

I'm happy to accept that the z-component of spin can have two states based on the results of the Stern-Gerlach experiment, and that we can represent these two states in a 2-dimensional spin state space. But then the x-component of the spin can also be measured. So we have a separate spin state for the x-component? And also for the y-component? A total of 3 separate spin state spaces?

But the fact that one spin state can be expressed as a linear combination of another spin state suggests that in fact there is just one spin state space for the x,y and z components of spin.

I think my problem is in attempting to relate the x,y,z spatial dimensions to the state space.

For:

[latex] |\uparrow_z \rangle = \frac{1}{\sqrt{2}} |\uparrow_x \rangle - \frac{1}{\sqrt{2}} |\downarrow_x \rangle [/latex]

Is the interpretation that the measurement of the z-component spin will be up with certainty and the measurement of the x-component will be up with probability 0.5 and down with probability 0.5 (which would be a manifestation of the generalised uncertainty principle). I hope so,because it'll mean i'm starting to get it...

Edited January 18, 2017 by Prometheus

[Mordred]

 

I'm happy to accept that the z-component of spin can have two states based on the results of the Stern-Gerlach experiment, and that we can represent these two states in a 2-dimensional spin state space. But then the x-component of the spin can also be measured. So we have a separate spin state for the x-component? And also for the y-component? A total of 3 separate spin state spaces?

 

But the fact that one spin state can be expressed as a linear combination of another spin state suggests that in fact there is just one spin state space

Exactly you have two states that can be determined+1 superimposed state that is indetermined. Total of three.

 

Your getting incredibly close to understanding Dirac notation So close in fact I'm trying figure out the final ingredient to give you that final

 

Ah ha moment. I believe this may be the symmetry relations between these different states themself. I had asked in another thread whether or not you have worked with matrices and tensors, which you replied no.

 

However it may be more revealing to show a brief example of Dirac with matrices and revisit vector associations.

 

As I'm more used to relativity applications of Dirac I think I will use an example from SR.

 

Not to confuse you but to give a different angle of approach. That final Ah ha moment might be just knowing Dirac is used in all branches of physics not just QM.

 

This example here is inner space products

 

[latex] |\uparrow_z \rangle = \frac{1}{\sqrt{2}} |\uparrow_x \rangle - \frac{1}{\sqrt{2}} |\downarrow_x \rangle [/latex]

 

A different notation to represent the above being the eugenvalues can represented by base vectors

 

[latex]|+\rangle=\dbinom{1}{0}[/latex] and [latex]|-\rangle=\dbinom{0}{1}[/latex]

 

expanding on the above we can introduce the following matrices for the base vectors

[latex]|n_1\rangle=\begin{pmatrix}1\\0\\0\\0\end{pmatrix}|n_2\rangle=\begin{pmatrix}0\\1\\0\\0\end{pmatrix}|n_3\rangle= [\begin{pmatrix}0\\0\\1\\0\end{pmatrix}...|n_m\rangle\begin{pmatrix}0\\0\\0\\0\\.\\.\\.\\.\\n\end{pmatrix}[/latex]

 

your Ket notation forms your column base vectors. your Bra notation [latex]\langle n^m|[/latex] fills the row base vectors in a similar manner above. Your [latex]|n^m|[/latex] being absolute value

 

Keep in mind for simplicity above I used normalized units [latex]c=g=\hbar=1 [/latex]

 

 

Now apply the above to say for example

 

[latex]N_{ij}[/latex] the first character in your subscript being the row. N can be anything from a single value, a matrix, a tensor or even a spinor,manifolds, Euclidean space,Pauli matrices,Hilbert space etc

 

 

Does that help?

 

side note there is 1 other Golden rule to remember (well several) but in terms of subscripts vs superscripts

 

if I have

[latex]N^m[/latex] m becomes [latex]\frac{1}{m}[/latex]

 

so for example I have a matrix A the inverse of this matrix is [latex]A^{-1}[/latex] so [latex]|x\rangle=A^{-1}|\acute{x}\rangle[/latex]

 

lets see how that works I will use matrix

[latex]g_{ij}=\begin{pmatrix}1&0&0\\0&r^2&0\\0&0&r^2 sin^2\theta\end{pmatrix}[/latex] clockwise rotation

 

[latex]g^{ij}=\begin{pmatrix}1&0&0\\0&\frac{1}{r^2}&0\\0&0&\frac{1}{r^2sin^2\theta}\end{pmatrix}[/latex] anticlockwise rotation

 

I will stop there before I start to confuse you

 

What I posted is to help you understand the notation itself. Hopefully it will help you understand the two equations in your OP. The proofs behind those two equations I will leave to you though you already answered your own question

Edited January 19, 2017 by Mordred

[Prometheus]

Cheers. I think i'm fine with the base vectors notation, not sure about the other stuff though. Going to ask some question on entanglement that might put it to the test.