# What happens with non orthogonal eigenvectors

1. Apr 3, 2014

### jk22

I considered the covariance of 2 spin 1/2 as a non linear operator : $$A\otimes B-A|\Psi\rangle\langle\Psi|B$$.
The eigenvectors are but non orthogonal and I wondered what happens in that case with the probabilities : from Born"s rule it comes that the transition probability from one vector to the other is not 0. So how can we compute the probabilities of measurement for each eigenvalue ?

2. Apr 4, 2014

### Simon Bridge

If the vectors are not orthogonal, then you get the answer by resolving the vectors against a basis.
Just like you do in classical mechanics when two vectors are not 90deg to each other.

3. Apr 4, 2014

### jk22

We could do that but then they are not eigenvectors anymore.
Maybe the problem is more complicated since the operator is not linear we shall not consider linear combination of them.
In fact I got 6 eigenvectors of dimension 4 so they cannot build a basis.
2 eigevectors are for the eigenvalue 0 but isn't there a quantum rule that say we shall not consider the kern of the measurement operator since it corresponds to a non-measurement, or the zero point of the apparatus ?

4. Apr 4, 2014

### UltrafastPED

Ordinary QM is a linear theory, built on linear vector spaces.

You are bound to run into many issues when you use a non-linear operator.

For example, the concept of eigenvector comes from a linear theory, so what are these so-called eigenvectors you have found for a non-linear operator?

5. Apr 4, 2014

### jk22

I took a similar equation, if C is an operator (non-linear of nor), I write Cx=ax where x is a non null vector and a a scalar
I think the form of the non-linear operator should not come from nothing or imagination, that's why I thought the covariance operator should be reasonable enough, but I think there is no general theory for non linear stuff. In that case it seems the calculation is quite simple for the eigenproblem.

Anyhow, if I solve that equation I get manifolds as solution, namely 4 orthogonal straight line passing through 0 (those are linear) and 2 unit circle centered at 0, orthogonal and non intersecting. So subtracting the product of the averages in the covariance gives the standard eigenvalues 1 and -1 as for only the product AB, and those 2 subspaces more.

But considering that addition eigenvalue 0 leads to non-sense, for example if one use for the probability of measurement the modulus squared of the projection for the singlet state as initial state, that p(-1)=1, p(0)=1/2, so that the sum of the probabilities is bigger than 1 (because the eigenvectors are not orthogonal).

I'm trying to solve that problem. Considering that the operator is non-linear, the norm of them can be important, so that we cannot normalize them independently. I tried to put that the norm of the eigenvector for the value -1 is still unknown and to write an equation for the probabilities summing to 1.

But I faced another problem concerning the measurement problem : if a transition is possible does it happen, and how many transition can it happen during a real measurement.

As getting around this problem it seems to become inextricably unsolvable for this particular case. I thought maybe I could find here some new ideas.

6. Apr 4, 2014

7. Apr 5, 2014

### jk22

Thanks.

I noticed one thing more, the way I presented above is non coherent since it does not correspond to the average. So I tried to use the closure relationship and it gave that the result of the covariance operator can be either {1,0}, or {-1,0}.

So that studying the operator in fact gives insight to which measurement result it can be (it's in some way more precise than to look at the average).

In fact I noticed a big problem : since the operator is not symmetric, there are left and right eigenproblems. In order to get rid of that, one can symmetrize the operator, leading to

$$C=A\otimes B-\frac{1}{2}(A|\Psi\rangle\langle\Psi|B+B|\Psi\rangle\langle\Psi|A)$$

to consider the simplest case, I chose A=B=diag(1,-1). Letting Psi=(a,b,c,d), I got :

$$C=\left(\begin{array}{cccc}1-a^2& 0 & 0 & ad\\0&-1+b^2&-bc&0\\0 &-bc&-1+c^2&0\\ad&0&0&1-d^2\end{array}\right)$$

THe eigenproblem is $$C|\Psi\rangle=\lambda|\Psi\rangle$$ hence :

1) $$a-a^3+ad^2=\lambda a$$
2) $$-b+b^3-bc^2=\lambda b$$
3) $$-b^2c-c+c^3=\lambda c$$
4) $$a^2d+d-d^3=\lambda d$$

To solve this problem i supposed a,b,c,d non zero (if they are 0 this solve the equation), then it comes

1+4 implies $$\lambda=1$$
2+3 implies $$\lambda=-1$$

hence not all a,b,c,d can be 0. If i suppose b=c=0 I get a=d or a=-d
if a=d=0, then b=c or b=-c

Considering equ. 1 and 3 (hence b=d=0) I get

$$1-a^2=\lambda$$ and $$-1+c^2=\lambda$$ leading to $$a^2+c^2=2$$

and $$\lambda=1-a^2\in[-1;1]$$

The same comes out for equ. 2 and 4 (hence a=c=0)

Thus the spectrum is composed of 2 discrete eigenvalues 1 and -1 and a continuous spectrum.

We see however that the eigenvectors are not orthogonal.

Writing the closure relationship : $$\int\sum_{\lambda}|\lambda\rangle\langle\lambda|=\mathbb{1}$$ leads to the fact that the discrete eigenvalues are not considered, because of equation of the form $$p^2+q^2+\underbrace{\langle2cos(\theta)^2\rangle}_{=1}=1$$ where p and q are the component of the eigenvector corresponding to the value -1.

I get then for the average of the C operator in the singlet state :

$$\langle C\rangle=\frac{1}{\pi}\int_0^{2\pi}(1-2cos(\theta)^2)sin(\theta)^2=1/2$$

This is but different than the average computed directly since it simply gives -1.

There is already a sign difference but I don't know if it's a mistake ??

Last edited: Apr 5, 2014
8. Apr 6, 2014

### jk22

Sorry I noticed the mistake : Since the C operator is not linear we cannot exchange with the integration.

So does anyone has an idea if it is possible to calculate the density of probability of the measurement outcomes in this case ?

9. Jul 12, 2014

### jk22

Can we consider those as hidden variables ?

As it can be seen, the continuous spectrum is hidden at zero degree since the probability of getting -1 is 1, if we say the eigenvector for the variable eigenvalue of equ 1 and 3 is parametrized, we can choose it to have zero probability.

So we see the "continuous" spectrum has no influence, but it exists, can we say in some way this is a hidden variable that... does not disturb ?

I'm now trying to seek not at angle 0 but 45 degrees (like in Bell's theorem) in order to see if there are continous spectra that act like an error bar...

10. Aug 11, 2014

### jk22

Erratum : in the seek for the eigenvalues it comes that it cannot depend on a b c d hence there is no "continuous" spectrum

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