Recent content by gaiussheh

Thank you indeed for helping, and apologize if I've been annoying. It is obviously up to you whether you wish to further this. My point is that it is not true to say ##\displaystyle \lim_{n\to\infty} \langle \psi_n| \hat{x}| \psi_n\rangle =+\infty##, at least for the case I constructed. My...

Not sure if I have understood this correctly, however if I expend my wavefunction using the eigenstates of harmonic oscillator, the result is compatible with the integration result (=0). In this sense I have constructed an extension of the operator ##x## to my wavefunction ##\psi## such that the...

Does it make sense to "define" the "inner product" as ##\langle \psi | x \rangle \psi \rangle = \lim_{n\to\infty} \langle \psi | x | \psi_n \rangle##, where ##\psi_n## is in the Schwartz space as it is dense? This way, one can handle ##x | \psi \rangle ## even if it is not in the Hilbert space...

Thanks for these suggestions. It was not my first course in QM but my first TA of it. I got quite good scores back in my undergraduate and didn't find this course to be that much of abstract formulation. I was just amazed by how much math I have neglected! P.S. I asked the lecturer of this...

Can you not construct a ##\mathrm{e}^{x^2}## potential and measure it?

Well, that is actually where my confusion starts. I take an operator to be a map from a function to a function. It may not be closure in the Hilbert space, so people usually restrict its domain to, for example, the Schartz space. I just do not understand the rationale behind this. As long as...

My concern on this is that even if you have a wavefunction like ##\mathrm{e}^{-x^2}##, you can still construct ##\hat{O}=\mathrm{e}^{x^2}##, and again your wave function got kicked outside by this operator. Once you start to constrain the behaviour of operators, it becomes a rabbit hole, and you...

Not sure if I am confusing you! I am not talking about the form of the operator but the domain. I mean, QM always wants to pick a subspace of the Hilbert space such that an operator always maps a square-integrable function to a square-integrable function. Why is this necessary? I can always...

I think PVM is actually not stopping me from talking about position measurement outcomes (and associated probabilities) of my (strange) example ##\langle\psi|E_{\Omega}|\psi\rangle##? However, the operator refuses to act on it.

I wonder why one should exclude this wavefunction simply for this reason - It seems I can talk about ##\langle x | \psi \rangle^2## as the probability (density) of measurement outcome ##x##, I can get ##\langle x \rangle## out of ##\langle x | x|\psi \rangle##, I can measure it energy ##\langle...

I came up with a more natural example that avoids the discussion of functions in a square well. See https://www.physicsforums.com/threads/what-happens-when-an-operator-maps-a-vector-out-of-the-hilbert-space.1078949/

This question is closely related to my previous thread mentioning that a linear operator can map a ket out of the original Hilbert space. That example was about infinite squares well, so it may be seen as an artificial example. More recently, I came up with a more "natural" example that does not...

Well, you don't deduce the boundary from the potential. It is, in fact, the opposite. You know that you are dealing with a confined particle such that it has zero probability of appearing outside the well. You know that the wave function must be continuous. Thus you construct something like...

I think my question is more like "Okay you can say this problem is overly-simplified, it doesn't exist in the real world, blah blah blah, but really, how do you know that when you talk about abstract vector spaces representing quantum states, you can always write down ##\hat{O}|\psi\rangle##...

In quantum mechanics, is the result of a linear operator acting on a ket vector always a ket vector? I’ve seen many textbooks state this, but in the case of an infinite square well, when the momentum operator acts on the ground state wave function ##\mathrm{sin}(πx/l)##, the resulting cosine...