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Are we doing the notecards again for next week's exam?

And maybe since we have covered more topics we are allowed more equations…hint hint…?

I would honestly prefer just having an equation sheet, as long as we are given it a couple of days before the test. From the equations given on it we would be able to study more proficiently, and not be surprised by a question that we don't know how to begin on, and therefore lose many points. (No residual discontent at all here )

I would prefer the use of a note card again. Call me crazy but I bet that if an equation sheet were given, the key forms of the equations needed would be lacking. Rather we might need to derive them from basic principles. With the note card, the derivation can be skipped as you will be able to put down the more useful forms of equations.

i like making my own equations sheet better, i think in the process of doing so i usually learn better than if i have to only concentrate on the equations professors provide. in any case are we allowed an equation sheet this time too?

We are allowed an equation sheet and it's due Wednesday before the test at noon. :)

If you have a 3-dimensional square well, and you systematically push in all sides of the square well till it is extremely small… ultimately only the size of the particle - what happens to the particle? How is uncertainty preserved? Would it be that… you've enclosed the particle, so you know exactly where it is, but you have absolutely no idea about it's momentum? but… wouldn't it's momentum be known since it's confined to a box? or is it that the box itself is now so small that it now has its own uncertainty? What if while we shrink it we know exactly where it is?

My understanding is that the uncertainty in the instantaneous momentum increases(although the average is still 0) as you continue to confine the particle in x,y,z. I would guess that the energy of the particle would fluctuate so largely that it would tunnel through the box long before you were able to truly confine it.

In review of ch 3 preceeding eq 3.71 it seems as if <math><[(\hat{H}\hat{Q}])></math> would cancel as defined above eq 3.71 Would this term be zero? Why not

In answer to Dark Helmet's question about eigenstates and eigenvalues, this section of wikipedia really helped me:

“Generally, quantum mechanics does not assign definite values to observables. Instead, it makes predictions using probability distributions; that is, the probability of obtaining possible outcomes from measuring an observable. Oftentimes these results are skewed by many causes, such as dense probability clouds[23] or quantum state nuclear attraction.[24][25] Naturally, these probabilities will depend on the quantum state at the “instant” of the measurement. Hence, uncertainty is involved in the value. There are, however, certain states that are associated with a definite value of a particular observable. These are known as “eigenstates” of the observable (“eigen” can be roughly translated from German as inherent or as a characteristic[26]). In the everyday world, it is natural and intuitive to think of everything (every observable) as being in an eigenstate. Everything appears to have a definite position, a definite momentum, a definite energy, and a definite time of occurrence. However, quantum mechanics does not pinpoint the exact values of a particle for its position and momentum (since they are conjugate pairs) or its energy and time (since they too are conjugate pairs); rather, it only provides a range of probabilities of where that particle might be given its momentum and momentum probability. Therefore, it is helpful to use different words to describe states having uncertain values and states having definite values (eigenstate).

For example, consider a free particle. In quantum mechanics, there is wave-particle duality so the properties of the particle can be described as the properties of a wave. Therefore, its quantum state can be represented as a wave of arbitrary shape and extending over space as a wave function. The position and momentum of the particle are observables. The Uncertainty Principle states that both the position and the momentum cannot simultaneously be measured with full precision at the same time. However, one can measure the position alone of a moving free particle creating an eigenstate of position with a wavefunction that is very large (a Dirac delta) at a particular position x and zero everywhere else. If one performs a position measurement on such a wavefunction, the result x will be obtained with 100% probability (full certainty). This is called an eigenstate of position (mathematically more precise: a generalized position eigenstate (eigendistribution)). If the particle is in an eigenstate of position then its momentum is completely unknown. On the other hand, if the particle is in an eigenstate of momentum then its position is completely unknown. [27] In an eigenstate of momentum having a plane wave form, it can be shown that the wavelength is equal to h/p, where h is Planck's constant and p is the momentum of the eigenstate.[28]

Also: “Each eigenstate of an observable corresponds to an eigenvector of the operator, and the associated eigenvalue corresponds to the value of the observable in that eigenstate. If the operator's spectrum is discrete, the observable can only attain those discrete eigenvalues.”

Thanks John Galt, that really did help. Wikipedia to the rescue once again.

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