Recent content by MichPod

I see and I value your opinion on this seriously. Factually, Gao's works about "wave function measurement" are following works of Aharonov and Vaidman (I am saying it while understanding there is a risk that Gao adds some of his own ingridients). Do you have any opinion/impression and can you...

Some solutions of time-dependent Schrodinger equation may in fact be eigenvalue solutions of time-independent Schrodinger equation as well, if you allow such an informality as to substitute ##\Psi(r,t)## into an equation which normally expects ##\psi(r)##. Others are not, but may be expressed...

After six years, bumping this topic up in hope I have some luck and somebody would like / would be able to comment.

I did not study/read it all, but I was very positively impressed from the first paragraphs. A.I Lvovsky - Quantum Physics, an introduction based on photons Of course it is not only about photons.

Thank you. Thank you for confirming my question has some sense.

Yep. And they also have another version (quite a different one) of this course+videos given by another professor: https://ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/

Agree. And I did not mean anything opposite. The mattress has some physical particles connected with springs, then phonons correspond to the modes of the vibration of the mattress as a whole.

There is no any good book on QM unless you already know QM.

Providing the Hermitian conjugation of ##\hat{a}## is ##\hat{a}^\dagger##, this operator is Hermitian (self-adjoint)

To continue, as a discrete approximation we can model a K-G field with some mattress of discrete particles connected with springs. Acting with creation operators on the vacuum (ground) state of that mattress would correspond to adding phonons to the "lattice" of mattress. So I think that asking...

I expected there is an obvious analogy between the x coordinate of an oscillator and the value of a quantum field at particular point. Why was I wrong? BTW, how can I see that the field operator is not Hermitian, as you said? I though it is for the case of K-G equation.

For a harmonic oscillator? Per Schrodinger picture, it is just 'x'. How can I see that the field operator is not Hermitian? I though it is for the case of K-G equation.

Hmmm... interesting. From your impression, is Ballentine more advanced than Sakurai? I understand they both are ranked as graduate level textbooks.

Why the field is not observable? (I expected the value of the field is observable because it is observable in a classical case). Just to be on the same page, we can observe a coordinate of a quantum harmonic oscillator at a specific moment of time, right? By analogy, I though, we can observe an...