# An issue with the vector model of spin and its operators

• JeremyEbert
In summary, the vector model of spin is an approximation to the true spin of an atom. It is unusual to see someone taking the model seriously, like driving a model T. Be sure to understand that its only a model. The operators S+ and S- are stepping operators that act to change the value of m up or down by one. They can only take integer or half-integer values and are not measured simultaneously. There is some controversy about all these quantum and non-quantum visualizations of spin from Stan Sykora. He maintains the high-resolution NMR spectra simulation dll in Mnova software and added some comments on http://oeis.org/A003991 that peaked my interest

#### JeremyEbert

I can't seem to reconcile a part of the vector model of spin and some of its operators.

to quote wiki just above the "Bohr model" section:

http://en.wikipedia.org/wiki/Vector_model_of_the_atom#Mathematical_background_of_angular_momenta

"2.The magnitude of the vectors must be constant (for a specified state corresponding to the quantum number),
so the two indeterminate components of each of the vectors must be confined to a circle,
in such a way that the measurable and un-measurable components (at an instant of time)
allow the magnitudes to be constructed correctly, for all possible indeterminate components."

I would assume pythagorean theorem holds true and "the two indeterminate components (S_x & S_y)
of each of the vectors must be confined to a circle" of radius r:

magnitude = sqrt(s(s+1))

r = sqrt(s(s+1) - m^2)

however the operators are defined as:

S+ = sqrt(s(s+1) - m(m+1))
S- = sqrt(s(s+1) - m(m-1))

http://en.wikipedia.org/wiki/Spin_(physics)#Spin_operator
http://en.wikipedia.org/wiki/Anti-symmetric_operator#Spin

A visual of spin 5/2 with some notes:

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Jeremy, It's unusual to see someone taking the vector model seriously - like driving a model T!
Be sure to understand that its only a model. In quantum mechanics angular momentum is not really represented this way, as precessing vectors.

If you look more closely at the definitions of S+ and S-, you'll see that they are stepping operators, that is they act to change the value of m up or down by one:

S+ |m> = √(s(s+1) - m(m+1)) |m+1>
S- |m> = √(s(s+1) - m(m-1)) |m-1>

so you can't just treat them as magnitudes.

Bill_K said:
Jeremy, It's unusual to see someone taking the vector model seriously - like driving a model T!
Be sure to understand that its only a model. In quantum mechanics angular momentum is not really represented this way, as precessing vectors.

If you look more closely at the definitions of S+ and S-, you'll see that they are stepping operators, that is they act to change the value of m up or down by one:

S+ |m> = √(s(s+1) - m(m+1)) |m+1>
S- |m> = √(s(s+1) - m(m-1)) |m-1>

so you can't just treat them as magnitudes.

Thanks Bill for you insight.

I am very new to this vector/spinor model, hence my unusual view.

I have heard that there is some controversy about all these
quantum and non-quantum visualizations of spin from Stan Sykora. He maintains the
high-resolution NMR spectra simulation dll in Mnova software and added some comments on
http://oeis.org/A003991 that peaked my interest on the subject.

I am trying to fully understand the concept of these spin operators because they are
eerily close to a root system I've been working on involving prime number distribution.

We already know a connection between number theory and quantum mechanics comes from
the discovery that the spacing's between consecutive zeros of the zeta function also appear to
behave statistically like the spacing's between consecutive eigenvalue's of large random matrices
which physicists use to obtain estimates of the average spacing between consecutive energy levels
of heavy atomic nuclei and other complex quantum systems.

So these ladder operators act to change the value of m up or down by one essentially changing the
observed value on the Z axis. Is it not correct to say that once S_z is observed, the un-measurable components
must lie somewhere along a circle of radius sqrt(s(s+1) - m^2)?

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Jeremy, In the vector model, Sx and Sy could take on any continuous value along the circumference of a circle. The modern viewpoint is different - Sx and Sy (and any other spin projection) can only take integer or half-integer values. They can't be measured simultaneously, and so for a state in which Sz has been determined there will be some probability that Sx takes on each of its allowed discrete values.

Bill_K said:
Jeremy, In the vector model, Sx and Sy could take on any continuous value along the circumference of a circle. The modern viewpoint is different - Sx and Sy (and any other spin projection) can only take integer or half-integer values. They can't be measured simultaneously, and so for a state in which Sz has been determined there will be some probability that Sx takes on each of its allowed discrete values.

Bill,
Thank you again. I assumed the values of Sx and Sy would be quantized as well. It seems logical seeing as the Z axis is actually an arbitrary direction usually determined by an external magnetic field. Sz is just used for convention, we could substitute Sx and Sy just as easily. Is there a formula for the probability of these other allowed discrete values?

Also, I still can't understand the ladder operators fully. They have been described as the relative intensities of the states with the equivalence S+ = Sx + i Sy and S- = Sx - i Sy, but the geometry doesn't seem to fit with the eigenvalues. Can you shed some light on this for me?

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Bill_K said:
They can't be measured simultaneously, and so for a state in which Sz has been determined there will be some probability that Sx takes on each of its allowed discrete values.

Bill,

If you think of the possible spin states as a function of the magnitude of spin then:

|S| = magitude
S+-=|x +- iy| = magnitude

applying the ladder opperators on the S_x states
one could come up with something like this model.

Notice the X-axis:
One link to the fact that you cannot observe the other projections is because
they do not fall on integer or half-integer values.
Is this a proper view of the ladder operators of the un-observable states?

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JeremyEbert said:
Bill,

If you think of the possible spin states as a function of the magnitude of spin then:

|S| = magitude
S+-=|x +- iy| = magnitude

applying the ladder opperators on the S_x states
one could come up with something like this model.

Notice the X-axis:
One link to the fact that you cannot observe the other projections is because
they do not fall on integer or half-integer values.
Is this a proper view of the ladder operators of the un-observable states?

Bill,
Here is a 3D model of possible S_x and S_y states based upon the ladder operators.
You will need the Flash plugin to view. This runs through all Spin states up to s=75/2.
http://dl.dropbox.com/u/13155084/SPIN/index.html [Broken]

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Can't say as I understand your Flash model, but it sure does look cool, Jeremy. Feel free to explain it a bit more. Also, since you mentioned spin 5/2...
JeremyEbert said:
A visual of spin 5/2 with some notes:

... here is a purely numerological observation that may or may not be of interest to you in relation to Stan Sykora's comments here at http://oeis.org/A003991 (the ones that piqued your interest):

$S$=$\frac{\hbar}{2}$$\sqrt{n(n+2)}$ = $\frac{\hbar}{2}$$\sqrt{5(5+2)}$ == $\frac{\hbar}{2}$$\sqrt{5 + 8 + 9 + 8 + 5}$ = $\frac{\hbar}{2}$$\sqrt{35}$ for n = 5

In other words, for the special case of n = 5, then 35 is tetrahedral as per Sykora's example (sum of relative intensities of Spin 5/2 transition states) and also one less than a square, meaning that it follows form n(n + 2) as per the above formula for spin. The maths work out in this instance because any tetrahedral number is equal to the product of 3 consecutive integers divided by 6. Therefore, 5*6*7/6 = 35, the fifth tetrahedral number, and because the 6's cancel it is also one less than a square (=5*7).

- AC

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## 1. What is the vector model of spin and its operators?

The vector model of spin is a representation of the quantum mechanical concept of spin, which is an intrinsic property of particles. The operators in this model are mathematical operators that act on the spin states of particles to describe their behavior.

## 2. What is the significance of the vector model of spin in quantum mechanics?

The vector model of spin is important in quantum mechanics because it helps us understand the behavior of particles with spin, such as electrons. It also provides a way to describe the interactions between particles in terms of their spin states.

## 3. What is the issue with the vector model of spin and its operators?

The issue with the vector model of spin is that it does not fully capture the complexity of spin in quantum mechanics. It is a simplified model that works well for certain systems, but it may not accurately describe all types of spin interactions.

## 4. How does the vector model of spin relate to other models in quantum mechanics?

The vector model of spin is a specific representation of spin in quantum mechanics. It is related to other models, such as the spinor model and the density matrix model, which provide alternative ways of describing spin behavior in different contexts.

## 5. Are there any proposed solutions to the issue with the vector model of spin and its operators?

There have been various proposals to address the issue with the vector model of spin, such as the use of higher-dimensional representations and the incorporation of additional mathematical tools. However, there is still ongoing research and debate in the scientific community about the best way to handle the complexities of spin in quantum mechanics.