Quantum Mechanics: Uncertainty Principles & Superposition

[nomadreid]

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?

 

[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_1and c_2 like possible results, then you can´t see the superposition

 

[nomadreid]

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.

 

[StarsRuler]

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.

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

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.

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

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.

"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.

 

[nomadreid]

Thanks again, StarsRuler. On my way to a concert, so a quick question now:

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.
"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.

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).

 

[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 don't 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 don't commute.

 

[nomadreid]

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.

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 don't 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)

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 ?

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 don't commute.

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?

 

[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.

 

[nomadreid]

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.

 

[Ravi Mohan]

I presume you mean "for all λ ..."

No it doesn't 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 won't need this number.

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?

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 don't 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".

 

[nomadreid]

Thanks for the clarification, Ravi Mohan. I believe I understand your explanation now. The fog is lifting... (but you will see more posts by me before the fog lifts completely, as I work through QM on my own).