Exploring 120-Sided Dice and Eigenvalues

[mike1000]

Here is a picture of a set of 120 sided dice. Each die has 120 eigenvalues. It is easy to see that as the number of eigenvalues increases, the probability of any eigenvalue gets smaller. In the limit where the number of eigenvalues is $\infty$ the probability of anyone eigenvalue approaches zero and the shape becomes a sphere.

I suppose that in 120 dimensional space, each die would be represented by a single point. The vector describing that point would have 120 components. Each component would represent the probability that the die would be found in that particular eigenstate.

 

[DrClaude]

Here is a picture of a set of 120 sided dice. Each die has 120 eigenvalues.

How can a die have eigenvalues?

It is easy to see that as the number of eigenvalues increases, the probability of any eigenvalue gets smaller.

That's incorrect. There is not reason to assume that all outcomes are equiprobable.

I suppose that in 120 dimensional space, each die would be represented by a single point.

This I really don't get.

The vector describing that point would have 120 components. Each component would represent the probability that the die would be found in that particular eigenstate.

I get the analogy you are trying to make, but this is also incorrect. Probabilities are related to absolute values square, not components. Also, you have to differentiate the space of possible outcomes with the actual state of a given die.

 

[mike1000]

How can a die have eigenvalues?

Well a die can have eigenvalues because the result of a measurement on a die can only be one of the 120 values on the die.(I suspect there has to be some matrix operator that exists that actually has those 120 possible outcomes as real eigenvalues.)

That's incorrect. There is not reason to assume that all outcomes are equiprobable.

Yes. I should have stated in my original post that I was talking about a system where all outcomes were equally probable. The state vector would look like \begin{equation}|\psi\rangle=\frac{1}{\sqrt{120}}\begin{pmatrix}1\\1\\1\\ \vdots\\1\end{pmatrix}\end{equation}

This I really don't get.

I am assuming that each possible outcome can be considered a dimension in some probability space. Since there are 120 possible outcomes the total dimension of the space is 120. An arbitrary point in that space would be defined by a vector that has 120 components. When the die is shaking the state would be a superposition of all possible states as shown in Equation (1)

I get the analogy you are trying to make, but this is also incorrect. Probabilities are related to absolute values square, not components. Also, you have to differentiate the space of possible outcomes with the actual state of a given die.

I should have said "probability amplitudes" in my original post.

 

[Truecrimson]

I get what the OP is trying to get at, so I wouldn't say that it's wrong. Yes, you can model this with a Hilbert space and take the components to be the probability amplitudes (this is just quantum mechanics restricted to only one basis) or the probabilities themselves i.e. using 1-norm instead of 2-norm. See this lecture by Scott Aaronson: http://www.scottaaronson.com/democritus/lec9.html

 

[Swamp Thing]

I get what the OP is trying to get at, so I wouldn't say that it's wrong.

It's certainly not "not even wrong"

Maybe one can use it to motivate a lesson on density matrices on a continuum type state space.

 

[Demystifier]

Here is a picture of a set of 120 sided dice. Each die has 120 eigenvalues. It is easy to see that as the number of eigenvalues increases, the probability of any eigenvalue gets smaller. In the limit where the number of eigenvalues is $\infty$ the probability of anyone eigenvalue approaches zero and the shape becomes a sphere.

I suppose that in 120 dimensional space, each die would be represented by a single point. The vector describing that point would have 120 components. Each component would represent the probability that the die would be found in that particular eigenstate.

View attachment 133273

What's your point? What did we learn from the 120-sides case that we didn't already know from the 6-sides case?

 

[mike1000]

What's your point? What did we learn from the 120-sides case that we didn't already know from the 6-sides case?

My point was that superposition is not a real state that the die can be measured in. When the die is being shaken it is put into an indeterminate state that is described mathematically as a superposition of possible outcomes but does not correspond to an actual outcome.

Also, I realized that as the number of possible outcomes goes to infinity the shape of the die becomes a sphere. How do you determine what the outcome of a shake is when the die is a sphere? It must be the tangent point of the sphere to the table when the die hits the table.

 

[Demystifier]

My point was that superposition is not a real state that the die can be measured in. When the die is being shaken it is put into an indeterminate state that is described mathematically as a superposition of possible outcomes but does not correspond to an actual outcome.

Couldn't you do all this with 6 sides?

Also, I realized that as the number of possible outcomes goes to infinity the shape of the die becomes a sphere. How do you determine what the outcome of a shake is when the die is a sphere?

Ah, now I see why 6 is not big enough. On the sphere the spectrum is continuous, and continuos observables cannot be measured with perfect accuracy.

 

[mike1000]

Ah, now I see why 6 is not big enough. On the sphere the spectrum is continuous, and continuos observables cannot be measured with perfect accuracy.

If an observable with a continuous spectrum cannot be measured with perfect accuracy, does that mean that when we measure it we find that it actually is in a superposition of one or more states?

 

[Demystifier]

If an observable with a continuous spectrum cannot be measured with perfect accuracy, does that mean that when we measure it we find that it actually is in a superposition of one or more states?

Yes.