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friend
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What is <x'|x> in quantum mechanics? I've seen it, but I don't know what it's suppose to mean in physical terms. It this the amplitude for an unspecified particle to go from x to x' ?
A. Neumaier said:It is the inner product of two unnormalized position states, and has the value ##\delta(x'-x)##. It is moot to try to interpret it in terms of transitions..
One can force an interpretation upon it, but I don't know a natural setting where this would appear as a transition amplitude.friend said:How does it compare to <x'|e-iHt|x> ? I think this does have a physical interpretation, right?
This is consistent with my statementfriend said:OK, so it is a probability amplitude.
You can give it a name, and that's it. Some people like to play with names, just because it makes formulas appear less abstract.A. Neumaier said:One can force an interpretation upon it, but I don't know a natural setting where this would appear as a transition amplitude.
Thanks, that makes a lot of sense.A. Neumaier said:But the propagators that are actually used in quantum mechanics are all between momentum states, not between position states. Because momentum states can be prepared (beams) while position states cannot.
It's much simpler than that.friend said:So would <x'|x> simply be a propagator as above with H=0 ? Does that have any physical meaning? For example, could that be a propagator for a virtual particle, that exists at each position in space but does not have any permanent energy?
In addition, the transition amplitude interpretation of a propagator is still ill-conceived. The absolute square of a probability amplitude is a probability, a number between 0 and 1, while the absolute square of a propagator value can be any number, including infinity, and hence cannot have a probability interpretation.A. Neumaier said:But the propagators that are actually used in quantum mechanics are all between momentum states
No. The integrals are sums over histories (in QM literally, in QFT in a vague sense, as everything diverges). It is at the origin of the intuition that a quantum particle travels all possible paths. But one shouldn't take this intuition too seriously - it is just a feeble attempt to make the path integral less abstract than it is, and becomes nonsense if taken too real.friend said:Are these virtual particles?
It must be stressed that you cannot prepare a particle in a state represented by ##|x \rangle##, because it's not a Hilbert-space vector. It belongs to a larger space, namely the dual of the dense subspace of Hilbert space, where the position operator is defined (i.e., the domain of the position operator).Nugatory said:It's much simpler than that.
If you're going to think of ##\langle{x}'|x\rangle## as a special case of ##\langle{x}'|e^{-iHt/\hbar}|x\rangle##, it's the ##t=0## case, not the ##H=0## case. ##|x\rangle## is the position eigenstate with eigenvalue ##x##, and if that's the state of the particle at time ##t=0## then ##e^{iHt/\hbar}|x\rangle## will be its state at all times ##t\ge{0}##. In general that state will be a superposition of eigenstates, and ##\langle{x}'|e^{-iHt/\hbar}|x\rangle## picks out the amplitude of the ##|x'\rangle## components in that superposition. Not surprising, it is equal to ##\delta(x'-x)## at ##t=0## when the state is ##|x\rangle## with no other position eigenstates contributing.
This notation only makes sense when the whole system consists of a single, spinless particle. H is the energy, nt the charge.friend said:Does ##\exp(-\mathrm{i} \hat{H} t/ħ) ## identify which particle the ##|\ x \rangle## refers to? I think ## \hat{H} ## specify the charge of the particle referred to by ##|\ x \rangle##. But does ## \hat{H} ## specify the spin statistics of ##|\ x \rangle## ?
Right. We would need something like |x1, x2> in order to starting talking about whether it is symmetric or antisymmetric under a permutation of the subscripts. As I recall, this is related to is spin statistics. (It's been a while since I looked at this). So is the way |x1, x2> behaves in permutation something we assume? Or can that be derived by some operator such as the Hamiltonian. I'm reminded that the Hamiltonian has things in it like a mass term, and it determines whether there is a repulsive or attractive force between similar particles we might label as |x1> and | x2>.A. Neumaier said:This notation only makes sense when the whole system consists of a single, spinless particle. H is the energy, nt the charge.
I think you'd first read a bit more and improve in this way your formal understanding, rather than using this forum to have pointed out your (many) shortcomings in your present understanding. It is a better use of your time, and of eveyone elses, too.friend said:(It's been a while since I looked at this).
Your comments only support the fact that I must be asking questions that do not have quick and easy answers. Otherwise, you'd simply answer them since you are so informed on the situation. I don't have a lot of time to go through many texts looking for connections that they don't focus on. If someone can help me with what exact words, theorems, equations, and concepts that I seem to be trying to find, then I can get into the books and do my own research. Or are you suggesting that this forum is the place where only experts in the field go to find answers? If I already knew it all, I wouldn't be here.A. Neumaier said:I think you'd first read a bit more and improve in this way your formal understanding, rather than using this forum to have pointed out your (many) shortcomings in your present understanding. It is a better use of your time, and of eveyone elses, too.
No, but you should not expect to get useful answers without accompanying self-study. Quantum mechanics cannot be learned through bed-time reading and superficial discussions on the web.friend said:Or are you suggesting that this forum is the place where only experts in the field go to find answers? If I already knew it all, I wouldn't be here.
You'll get it if you do some serious study based on what was discussed in this thread, not by asking further questions. Most insights do not come for free but only when you are sufficiently prepared.friend said:I still await some further insights.
No; they are an input to the standard model. You better do some thorough reading instead of posting poorly remembered and poorly assembled fragments of facts.friend said:fermions/bosons come from the symmetries of the Lagrangian of the Standard Model.
As opposed to what, giving you my word alone for it.A. Neumaier said:No; they are an input to the standard model. You better do some thorough reading instead of posting poorly remembered and poorly assembled fragments of facts.
I asked what is <x'|x> which seems to be for an unspecified generic particle. I wanted to know where the specificity comes from. And it seems that information (including charge and spin) comes from the Lagrangian.A. Neumaier said:Also what you write has nothing to do with your original question, and hence doesn't belong into this thread.
Don't trust everything written in wikipedia! The entries in wikipedia are written by fallible people, usually not even by experts. Therefore not everything wikipedia says is true. Here it is completely mistaken. See post #9.friend said:a propagator which is a probability amplitude per this wikipedia.org article
Are we missing a normalization factor in order to turn a propagator into a probability amplitude?A. Neumaier said:...while the absolute square of a propagator value can be any number, including infinity, and hence cannot have a probability interpretation.
Your obviously wrong expectations are a sure sign that you don't understand the algebra, let alone the meaning of what you talk about. And it seems to be impossible to explain it to you.friend said:One might expect
Wow! I'm impressed. Thank you, vanhees, for all the effort you put into your response. Now we have some math to refer to. [Point of order: you might wish to number your equations for easier reference]vanhees71 said:Ok let's get this straight, because it's very important...