1. Jun 13, 2013

There are two kinds of uncertainty in QM: one of operators (Uncertainty Principles), and one of states (superposition). Is there any direct connection between them?

2. Jun 14, 2013

### StarsRuler

Well. A toy example: if you are in a state of an operator O, with 2 possible results $o_1$ and $o_2$ , with eigenvectors $\|e_1>$ and $\|e_2>$ , and your state is $c_1|e_1>+c_2|e_2>$, and you do a measurement, with an uncertainty that comprends $c_1$and $c_2$ like possible results, then you can´t see the superposition

3. Jun 14, 2013

StarsRuler, many thanks for the enlightening example. Interesting. I presume you mean that you cannot see the superposition from a single measurement, since you could presumably see the superposition by doing an experiment on lots of identical states.
Is there any way that one of these uncertainties can be derived from the other? I see the Uncertainty Principles derived from a straightforward derivation on non-commuting Hermitian operators, but I only come across the mechanism of superposition as a postulate.

P.S. A question about your example: it uses a situation with a single operator, but uncertainty principles involve two operators, so upon reflection I am unclear as to how your example would work.

Last edited: Jun 14, 2013
4. Jun 14, 2013

### StarsRuler

No, you don´t "see" superpositions really, when you do a measurement, you obtain a result. The superposition is then in another observables you are not measuring. You represent your collapsed state by a wavefunction that is a superposition of eigenstates of another observables. Indeed, observables that don´t conmute with the measured observable

In the standard presentation of QM, superposition is a postulate ( states are represented by vectors of a Hilbert Space) . Only the uncertainty relations for observables that don´t conmute can be deduced. There are anyway tries to deduce QM from more general postulates, many of them. And strong discussions about it and the different interpretations of QM

"Uncertainty" from superposition principle is not about 2 operators. Only about the measured result of the observable in which eigenstate basis you are writing the state vector.

5. Jun 14, 2013

Thanks again, StarsRuler. On my way to a concert, so a quick question now:
Yes, precisely, but the uncertainty principles I was referring to were the kind such as the Heisenberg Uncertainty Principle, that do refer to two operators. So your example was explaining the "uncertainty" of the superposition principle, but not explaining the link between that kind of uncertainty (superposition) and the Heisenberg Uncertainty Principle (or similar ones).

6. Jun 14, 2013

### Ravi Mohan

This is how I understand it.
Consider the system with two dimensional Hilbert space, say spin 1/2. Now we have three operators
$\hat{M}_x$, $\hat{M}_y$ and $\hat{M}_z$ and consider that they dont commute. Consider the system in the eigen state of $\hat{M}_z$ and let it be $|\uparrow_z\rangle$ (spin up in $\hat{a}_z$ direction).

Now consider the relation
$\hat{M}_z$$\hat{M}_x$$\neq$$\hat{M}_x$$\hat{M}_z$

$\hat{M}_z$$\hat{M}_x$$|\uparrow_z\rangle$$\neq$$\hat{M}_x$$\hat{M}_z$$|\uparrow_z\rangle$

$\hat{M}_z$$\hat{M}_x$$|\uparrow_z\rangle$$\neq$$\lambda$$\hat{M}_x$$|\uparrow_z\rangle$ (where $\lambda$ is some number)
or

$\hat{M}_z$$\big($$\hat{M}_x$$|\uparrow_z\rangle$$\big)$ $\neq$$\lambda$$\big($$\hat{M}_x$$|\uparrow_z\rangle$$\big)$

It means $\hat{M}_x$$|\uparrow_z\rangle$ is not an eigen state of $\hat{M}_z$

Let $|\uparrow_x\rangle$ be the eigen state of
$\hat{M}_x$. Now since $\hat{M}_x$ is a measurement operator (thus self adjoint), it should be like $|\uparrow_x\rangle$$\langle\uparrow_x|$.

$\hat{M}_x$$|\uparrow_z\rangle$ = $|\uparrow_x\rangle$$\langle\uparrow_x|$$|\uparrow_z\rangle$ = $c_1$$|\uparrow_x\rangle$

Now $c_1$$|\uparrow_x\rangle$ $\neq$ $|\uparrow_z\rangle$ and this is due to the reason that $\hat{M}_z$ and$\hat{M}_x$ dont commute.

Last edited: Jun 14, 2013
7. Jun 15, 2013

Thank you very much for your reply, Ravi Mohan. The calculation is very interesting; if I may, I would like to ask a couple of questions about steps that I am not sure of.

I presume you mean "for all λ ...." Is the justification for this last step that if there did exist such a λ, then $\hat{M}_z$ = λI (I being the appropriate identity), which would lead to it being able to commute with $\hat{M}_x$, which it doesn't ?

Very interesting. Now, I am surely being dense, but I would be grateful if you could spell it out for me: why does this conclusion lead to the postulate of superposition?

8. Jun 15, 2013

### kith

The superposition principle follows from the postulate that states are elements in a certain vector space (Hilbert space).

The derivation of the uncertainty principle makes use of the Cauchy-Schwartz inequality, which holds in vector spaces with the additional structure of an inner product.

So this is probably the connection you are looking for.

9. Jun 15, 2013

Thanks, kith. Yes, I guess that is as close a connection as I'm going to get: they are both properties of Hilbert spaces. I was wondering if the connection was closer, but I guess they are only related like humans and chimps: both have a common ancestor.

10. Jun 15, 2013

### Ravi Mohan

No it doesnt mean "for all λ" at all. It depends on the definition of operator $\hat{M}_z$ and vector $|\uparrow_z\rangle$. I never said it is normalised.
Normalize the vector and you wont need this number.

Ok so I have shown that the eigenvectors of non-commuting aperators are not the same. In this case $c_1$$|\uparrow_x\rangle$ $\neq$ $|\uparrow_z\rangle$ and thus to represent $|\uparrow_z\rangle$ you will need a basis set. Now the eigenvectors of spin along x axis do form a basis. Hence you can represent $|\uparrow_z\rangle$ as superposition!

Basically I dont consider superposition as a postulate. Once you postulate that the physical system should be associated with linear vectors, superposition of staes is obvious.

Consider it the opposite way. Consider that both operators do commute. Then you can see that both have a common set of eigen vectors. We call them simultaneous eigen vectors for both operators. Hence you can have the eigen value (upto arbitrary degree of accuracy) of both the operators simultaneously. No uncertainity.
Basically it is the propertiy of linear vectors that when you use inner product structure, they follow cauchy schwarz inequality which when combined with operators give heisenberg uncertainity. Here I have demonstrated all of this "at work".

11. Jun 16, 2013