On 1/2 spin - which is the X axis?

[MichPod]

TL;DR

If we chose Z axis when defining electron spin states, how do we choose X and Y directions then?

For a 1/2 spin particle, every pure spin state may be represented as a superposition of two states of spin parallel to some arbitrary Z axis.

(Upd) Particularly:

$$|\uparrow_{x}>=\frac{1}{\sqrt{2}}(|\uparrow_{z}>+|\downarrow_{z}>)$$

I then wonder, if we chose the Z axis, how the X axis should be chosen to make the above equality true?

 

[Nugatory]

Summary:: If we chose Z axis when defining electron spin states, how do we choose X and Y directions then?

I then wonder, if we chose the Z axis, how the X axis should be chosen to make the above equality true?

Any direction perpendicular to the z axis will work (although once you've chosen z and x, y is determined).

 

[MichPod]

Any direction perpendicular to the z axis will work (although once you've chosen z and x, y is determined).

Well, supposedly, not all the directions for chosing the X axis are the same, as we know that for Y axis

$$|\uparrow_{y}>=\frac{1}{\sqrt{2}}(|\uparrow_{z}>+i|\downarrow_{z}>)$$

 

[PeroK]

Well, supposedly, not all the directions for chosing the X axis are the same, as we know that for Y axis

$$|\uparrow_{y}>=\frac{1}{\sqrt{2}}(|\uparrow_{z}>+i|\downarrow_{z}>)$$

You're saying that if you choose a z-axis, then the x-y axes are uniquely defined? How would that be?

 

[Nugatory]

Well, supposedly, not all the directions for chosing the X axis are the same, as we know that for Y axis

Once you've chosen the z and x axes, the y-axis is determined up to your choice of left-hand or rght-hand rule (because y has to be perpendicular to both x and z).

 

[PeroK]

Well, supposedly, not all the directions for chosing the X axis are the same, as we know that for Y axis

$$|\uparrow_{y}>=\frac{1}{\sqrt{2}}(|\uparrow_{z}>+i|\downarrow_{z}>)$$

Note that this state is determined once you have chosen the state that represents $|\uparrow_{x} \rangle$.

 

[MichPod]

You're saying that if you choose a z-axis, then the x-y axes are uniquely defined? How would that be?

Well, I am not claiming that. But it looks like this inadequate strange result follows from the math. So I wanted to see what other people think.

 

[MichPod]

Note that this state is determined once you have chosen the state that represents $|\uparrow_{x} \rangle$.

Nope. My claim is as soon as we chose Z axis and $|\uparrow_{z} \rangle$ and $|\downarrow_{z} \rangle$, both X and Y axises are automatically defined and cannot be selected arbitrarily.

 

[Vanadium 50]

That claim just isn't true.

 

[PeroK]

Well, I am not claiming that. But it looks like this inadequate strange result follows from the math. So I wanted to see what other people think.

I guess the critical point is this:

1) We set up a Cartesian coordinate system.

2) We associate two basis vectors with spin-up and spin-down in z-direction.

3) By analysing the behaviour of spin states under spatial rotations (or otherwise), we associate spin states to spin-up and spin-down in the x and y directions. There is some flexibility in this (phase factors), but otherwise they are determined.

4) The critical point is to understand the derivation of the x and y spin states, given the z basis states.

 

[PeterDonis]

My claim is as soon as we chose Z axis and and , both X and Y axises are automatically defined and cannot be selected arbitrarily.

Your claim is wrong. Once you choose the Z axis, the plane in which the X and Y axes lie is automatically defined. But defining a plane is not the same as defining a single pair of perpendicular axes within that plane. To do the latter, you need to pick one particular axis to be one of the pair; for example, if you pick a particular axis in the plane perpendicular to the Z axis to be the X axis, then the Y axis is automatically defined. But not before.

 

[PeterDonis]

The critical point is to understand the derivation of the x and y spin states, given the z basis states.

The particular directions corresponding to x and y are an arbitrary choice; the only requirement is that they both have to lie in a plane perpendicular to z. Once you pick a particular direction for, say, the x axis, the y-axis is determined.

The correspondence between the choice of axes and the math is that which direction you pick for the x axis, for example, determines which direction you would orient a Stern-Gerlach apparatus in order for its operation on a qubit prepared in the spin-z up state to be correctly described by the way the spin-z up state is written in the spin-x basis. So different choices of x-axis in the math correspond to different ways you have to orient the apparatus for its operation on a spin-z up qubit to correspond with the math in the spin-x basis.

 

[MichPod]

4) The critical point is to understand the derivation of the x and y spin states, given the z basis states.

BTW, aside of the topic of my question, I probably understand how such derivation may be done for one-spin particle, but I am not sure if I ever saw such a derivation for half-spin particle in the QM course (I don't know QFT though). Can it be derived? Based on what idea?

 

[PeroK]

BTW, aside of the topic of my question, I probably understand how such derivation may be done for one-spin particle, but I am not sure if I ever saw such a derivation for half-spin particle in the QM course (I don't know QFT though). Can it be derived? Based on what idea?

It should be covered in all introductory QM textbooks.

 

[MichPod]

It should be covered in all introductory QM textbooks.

Could you bring in just one book name which does it acceptably for you?

 

[PeroK]

Could you bring in just one book name which does it acceptably for you?

Sakurai Modern QM. The very beginning.

 

[MichPod]

Sakurai Modern QM. The very beginning.

Thank you,

 

[MichPod]

I think I finally got what may be the answer to my question.
Of course, when we choose the Z axis direction, this alone cannot predefine the X axis direction.

But choosing of $|\uparrow_{z}>$ and $|\downarrow_{z}>$ does do it, surprisingly.

There is a degree of freedom with how we choose the spin-up and spin-down along the Z axis states because each state vector may be multiplied by a $e^{i\varphi}$ coefficient (upd: with different $\varphi$ for $|\uparrow_{z}>$ and $|\downarrow_{z}>$ i.e. $e^{i\varphi_{1}}|\uparrow_{z}>$ and $e^{i\varphi_{2}}|\downarrow_{z}>$) without the corresponding physical state being changed. It is this degree of freedom with choosing $|\uparrow_{z}>$ and $|\downarrow_{z}>$ which defines the direction of axis X (and Y as well).

 

[PeroK]

I think I finally got what may be the answer to my question.
Of course, when we choose the Z axis direction, this alone cannot predefine the X axis direction.

But choosing of $|\uparrow_{z}>$ and $|\downarrow_{z}>$ does do it, surprisingly.

There is a degree of freedom with how we choose the spin-up and spin-down along the Z axis states because each state vector may be multiplied by a $e^{i\varphi}$ coefficient without the corresponding physical state being changed. It is this degree of freedom with choosing $|\uparrow_{z}>$ and $|\downarrow_{z}>$ which defines the direction of axis X (and Y as well).

That cannot possibly be the case.

 

[MichPod]

That cannot possibly be the case.

Just for a case, made a small update/clarification above:

with different $\varphi$ for $|\uparrow_{z}>$ and $|\downarrow_{z}>$ i.e. $e^{i\varphi_{1}}|\uparrow_{z}>$ and $e^{i\varphi_{2}}|\downarrow_{z}>$

 

[PeterDonis]

There is a degree of freedom with how we choose the spin-up and spin-down along the Z axis

No, there isn't. What you are describing is a purely mathematical "degree of freedom", not a physical one. And it has nothing whatever to do with the physical relationship between spin-z up and down states and spin-x up and down states. The mathematical manipulations you are making would affect the mathematical representations of all states the same way, so they would not change the relationship between different states at all.

 

[MichPod]

The physical states are represented mathematically with the rays in the Hilbert space or, equivalently, with the elements of the projective Hilbert space. And yet, we are using just vectors of the Hilbert space (in the best case, normalized to 1 but still with a phase degree of freedom) for our actual calculations. These vectors are sometimes arbitrary chosen representors of the equivalence class in the Hilbert space. I think it is easy (obvious) to see that depending on which phase shift we assign to such representors $|\uparrow_{z}>$ and $|\downarrow_{z}>$, we may get different actual physical states as a result of their superposition when we add them with the same coefficient $\frac{1}{\sqrt{2}}$. I should agree that all this has no physical significance, on the other side, it clarifies how the math formalism works.

 

[Vanadium 50]

Of course, when we choose the Z axis direction, this alone cannot predefine the X axis direction.

But choosing of (z-up) and (z-down)does do it, surprisingly.

You've said that before. It wasn't true then and it's not true now.

Suppose a professor sets up two grad students in separate labs, and tells them "z is North, go figure out what x and y are. Do any experiment you want." Do you really think both will come up with the same answer?

 

[PeterDonis]

I think it is easy to see that depending on which phase shift we assign to the representors and , we may get different actual physical states as a result of their superposition when we add them with same coefficient .

That's because, if you are superposing two different states, the relative phase shift between them does have physical meaning. Your statement that we can arbitrarily choose the phase of the Hilbert space vector we use to represent the state is only true for the overall state of the entire system; it is not true for individual terms in a superposition.

 

[MichPod]

You've said that before. It wasn't true then and it's not true now.

Suppose a professor sets up two grad students in separate labs, and tells them "z is North, go figure out what x and y are. Do any experiment you want." Do you really think both will come up with the same answer?

I think you are missing my point. Choosing Z axis direction by itself does not define X direction. Yet choosing the particular $|\uparrow_{z}>$ and $|\downarrow_{z}>$ vectors in the Hilbert space to represent spin-up and spin-down (along the Z-axis) physical states (we have a degree of freedom of choosing these vectors) does define X and Y axises.

 

[MichPod]

Your statement that we can arbitrarily choose the phase of the Hilbert space vector we use to represent the state is only true for the overall state of the entire system; it is not true for individual terms in a superposition.

100% agree here.

It's interesting to notice that when we write the superposition, we are using Hilbert space vectors for individual terms, so we write a superposition not as a weighted sum of physical states but as a weighted sum of vectors. On the other side, when we are presented with a particular Hilbert vector, like $|\uparrow_{z}>$, we are used to think about it as about the state, not as about the state representor only.

 

[PeroK]

I think you are missing my point. Choosing Z axis direction by itself does not define X direction. Yet choosing the particular $|\uparrow_{z}>$ and $|\downarrow_{z}>$ vectors in the Hilbert space to represent spin-up and spin-down (along the Z-axis) physical states (we have a degree of freedom of choosing these vectors) does define X and Y axises.

So, if you are in the lab, the $z$ direction is up, z-spin-up is the vector $(1,0)$ and z-spin-down is the vector $(0, 1)$ (in your Hilbert space of spin states), which direction is $x$?

 

[PeroK]

In fact, even more to the point, once you have decided on the entire definition of $x, y$ and $z$ spins states in you Hilbert space and that $z$ is up, which way is $x$?

 

[MichPod]

So, if you are in the lab, the $z$ direction is up, z-spin-up is the vector $(1,0)$ and z-spin-down is the vector $(0, 1)$ (in your Hilbert space of spin states), which direction is $x$?

This representation obviously would not help me to answer this question. And I do not know of any other representation for 1/2-spin particles which would help. For orbital momentum of an electron in H-atom with principal number n=2 and angular number l=1 state (spin-1 case, right?) there exist a wave function (defined over a sphere surface) representing the different vertical momentum (-1, 0,1) so there it is easily done.

 

[PeroK]

This representation obviously would not help me to answer this question. And I do not know of any other representation for 1/2-spin particles which would help. For orbital momentum of an electron in H-atom with principal number n=2 and angular number l=1 state (spin-1 case, right?) there exist a wave function (defined over a sphere surface) representing the different vertical momentum (-1, 0,1) so there it is easily done.

Okay, so you have a hydrogen atom in the lab and it's in the state $\Psi_{211}$. Which direction is $x$ in the lab?