I What is <x'|x> ?

1. Feb 4, 2016

friend

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' ?

2. Feb 4, 2016

A. Neumaier

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.

Note that not every inner product has such an interpretation. only those arising from scattering theory (or its simplified toy versions) can sensibly be interpreted as transition amplitudes.

3. Feb 4, 2016

friend

How does it compare to <x'|e-iHt|x> ? I think this does have a physical interpretation, right?

4. Feb 4, 2016

A. Neumaier

One can force an interpretation upon it, but I don't know a natural setting where this would appear as a transition amplitude.
The good transition amplitudes are $\langle p|S|p'\rangle$, where $p$ and $p'$ are momentum or spin states and $S$ is the S-matrix (a complicated limit with a physical meaning), and in some approximations, you can replace $S$ by an interaction potential.

5. Feb 4, 2016

friend

The Wikipedia article says that <x'|e-iHt/ħ|x> is the propagator that "gives the probability amplitude for a particle to travel from one place to another in a given time...", where e-iH(t'-t)/ħ is a "unitary time-evolution operator for the system taking states at time t to states at time t′ ", . OK, so it is a probability amplitude. Why is that not so obvious?

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 does not have any permanent energy?

Last edited: Feb 4, 2016
6. Feb 4, 2016

A. Neumaier

This is consistent with my statement
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.

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. One doesn't observe particles jumping from one place to another.

So you can choose from the literature what you like to use. The truth is in the formulas, not in the way people think or talk about them. The latter is often highly subjective.

Last edited: Feb 4, 2016
7. Feb 4, 2016

friend

Thanks, that makes a lot of sense.

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?

8. Feb 4, 2016

Staff: Mentor

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.

9. Feb 4, 2016

A. Neumaier

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.

It is highly misleading to try to interpret every inner product as an amplitude
! The physical meaning of a formula is determined by the way it is used in an argument leading to physical results, and not by making up stories about the symbols and associated virtual objects!!! So if you want to understand the physical meaning of a formula you need to study the context of it until you see its connection with something of true physical relevance!

10. Feb 4, 2016

friend

Just a minute, I often see the path integral derived using this inner product by inserting the identity many time...

<x'|e-iHt/ħ|x> = ∫∫∫⋅⋅⋅∫<x'|e-iHε/ħ|x1><x1|e-iHε/ħ|x2><x2|e-iHε/ħ|x3>⋅⋅⋅<xn|e-iHε/ħ|x> dx1dx2dx3⋅⋅⋅dxn

where the integration over x1, x2, x3... xn is from - to + infinity, and ε approaches 0. So basically we have <x1|e-iHε/ħ|x2> ≈ <x1|x2>. What then are these <x1|e-iHε/ħ|x2> ≈ <x1|x2>? They don't seem to be measurable, but they do seem to be micro-wave-functions for presumably micro excursions of a particle that contribute to the total wave-function. Are these virtual particles?

11. Feb 4, 2016

A. Neumaier

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.

Last edited: Feb 6, 2016
12. Feb 4, 2016

friend

An alternative interpretation of the path integral might be that real particles traveling through space do so by transferring energy from one stationary "virtual particle", <x1|e-iHε/ħ|x2>, to the next. Otherwise, I think we are ignoring quantum fluctuations' effect on traveling particles. I know some people have real difficulty thinking in terms of virtual particles. But I think that's because there's not enough effort put into identifying where in the math they are. I've made an attempt here that seems reasonable to me. In order to say that these are not virtual particles, one would have to show the math for true virtual particles. That doesn't seem to be readily available. But I think an attempt should be made because it would offer us a way to mathematically visualize exactly what's going on in the quantum world. I think it would explain where the wave-function comes from and, thus, exactly how entanglement works. And now I'm seriously beginning to think (I need a little more time on this) that such a virtual description of the math will allow us to unite spacetime, matter, and energy.

Last edited: Feb 4, 2016
13. Feb 6, 2016

vanhees71

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).

Further, of course
$U(t;x,x')=\langle x|\exp(-\mathrm{i} \hat{H} t) x' \rangle,$
thus is a distribution. It only has a meaning when applied to a true state. As already stated above in this thread, it's the propagator in the position representation, i.e., if you have a system that is at $t=0$ prepared in a true pure quantum state, represented by a normalized Hilbert-space vector $|\psi_0 \rangle$, then in the position representation you have the wave function
$\psi_0(x)=\langle x|\psi \rangle,$
which is a square-integrable function with norm 1. Then at any later time $t$ the state of the system is represented by the wave function
$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} x' U(t;x,x') \psi_0(x)=\langle x|\exp(-\mathrm{i} \hat{H}) \psi_0 \rangle.$$

14. Feb 8, 2016

friend

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 $|\ x \rangle$ ?

15. Feb 9, 2016

A. Neumaier

This notation only makes sense when the whole system consists of a single, spinless particle. H is the energy, nt the charge.

16. Feb 9, 2016

friend

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>.

The reason I ask is because the |x1> notation is not specific to what kind of particle, fermion or boson, the particle might be. So I'm wondering where that specification comes in, whether we have to assume it to even write the Lagrangian or does it come in with the introduction of coupling constants.

Last edited: Feb 9, 2016
17. Feb 10, 2016

A. Neumaier

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.

18. Feb 10, 2016

friend

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.

19. Feb 10, 2016

A. Neumaier

No, but you should not expect to get useful answers without accompanying self-study. Quantum mechanics cannot be learnt through bed-time reading and superficial discussions on the web.

Each answer given to you should trigger enough motivation to read something more systematic. Otherwise you'll end up with half-baked pseudo truths that don't make sense to anyone.

20. Feb 10, 2016

friend

Thank you. I still await some further insights.