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1

I think what you are missing is that these energies are eigenvalues of the time-independent Hamiltonian. i.e. They correspond to stationary states that do not change in time. The scenario you describe is not time-independent - therefore the difference between the energy levels will carry some uncertainty corresponding to the lifetime of the excited state.


3

The total energy of the universe is a vexed issue since different commentators have different views about what the concept means. See the question Total energy of the Universe for a sampling of the various viewpoints. If you Google for zero energy universe you'll find several papers purporting to show that the total energy is zero. However since their ...


3

Like everything of importance in physics it's an experimentally testable fact. If it wasn't, we wouldn't be talking about it. Secondly it is, of course, built into the theory, otherwise the theory wouldn't be correct. What it is NOT, is a measurement "problem".


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It has obviously been experimentally tested many times, but it's also derivable theoretically from the mathematics of quantum mechanics. Two basic ideas are important here--the first is the notion of a Fourier series, which allows you to represent any arbitrary periodic function (a square wave, for example) as a potentially infinite sum of different sine and ...


1

It is actually a mathematical law, for conjugate operators. For any two operators the following applies: $$ \langle (\Delta A)^2 \rangle \langle (\Delta B)^2 \rangle \geq \frac{1}{4}|\langle[A,B]\rangle| $$ So for conjugate coordinates this means the Heisenberg uncertainty principle - for more detail - see Sakurai Chap. (supposed to be a remark but ...


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In essence you are mixing the notion of electron as a point particle, and quantum physics. While classical mech have us believe that electron are point-wise particles, Quantum mech actually forbids it, by the same Uncertainty principle you are invoking. Thus the electron has some dimensionality in the z axis as well as the y axis, I'll go ahead ad ...


0

Regarding your third question. If you want to see the state vector $|\psi\rangle$ you considered explicitly, you must choose a basis to expand it. For example, the position eigenkets $|x\rangle$. Then you can expand your state kets with respect to $|x\rangle$. That is $|\psi\rangle=\int dx |x\rangle\langle x|\psi\rangle$. Where $\langle ...


2

Yes, this is possible - but only for states with zero total angular momentum. To see why, the first step is seeing that if $\Delta\sigma_x^2=\Delta\sigma_y^2=0$ on state $\psi$, then $\psi$ is an eigenstate of both $\hat\sigma_x$ and $\hat\sigma_y$: $$ ...


1

When you measure the system's position you actually push it out of the ground state, so your "trick" of using the initial total energy to compute the momentum after the position measurement doesn't work. To be specific, when you measure the position, suppose you get $x_0$ as the result. Then the particle's wave function is now a reasonably sharp peak at ...


1

You are on the right track. Pushing one more step to the final answer may leave you disappointed: $\sigma_x \sigma_K$ can equal zero! To see this, I find it more helpful to think just in terms of $x$ and $K$ as linear operators satisfying certain commutation relations, rather than thinking explicitly in terms of integrals of wavefunctions. Specifically, we ...


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As an example, http://en.wikipedia.org/wiki/Particle_decay, you can regard $\Delta t$ as the lifetime of the particle decayed.


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I will give a counter-example. Consider the potential below, $V$ is considered to be very large but not infinity. The ground state of the system should be approximate as the ground state of an infinite potential well. The uncertainty is greater than $\hbar/2$. While when the particle is confined in the parabolic potential well, the local "ground state" ...



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