I What is the physical significance of <x'|x> in quantum mechanics?

[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' ?

 

[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.

 

[friend]

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..

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

 

[A. Neumaier]

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

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.

 

[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?

 

[A. Neumaier]

OK, so it is a probability amplitude.

This is consistent with my statement

One can force an interpretation upon it, but I don't know a natural setting where this would appear as a transition 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.

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.

 

[friend]

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.

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?

 

[Nugatory]

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?

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.

 

[A. Neumaier]

But the propagators that are actually used in quantum mechanics are all between momentum states

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!

 

[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>⋅⋅⋅<x_n|e^-iHε/ħ|x> dx₁dx²dx₃⋅⋅⋅dx_n

where the integration over x₁, x₂, x₃... x_n 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?

 

[A. Neumaier]

Are these virtual particles?

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]

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.

 

[vanhees71]

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.

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
$$\psi(t,x)=\int_{\mathbb{R}} \mathrm{d} x' U(t;x,x') \psi_0(x)=\langle x|\exp(-\mathrm{i} \hat{H}) \psi_0 \rangle.$$

 

[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 $|\ x \rangle$ ?

 

[A. Neumaier]

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 $|\ x \rangle$ ?

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

 

[friend]

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

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.

 

[A. Neumaier]

(It's been a while since I looked at this).

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]

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.

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]

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.

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.

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.

 

[friend]

Thank you. I still await some further insights.

 

[vanhees71]

I've given an answer. If you don't understand it, please take a good textbook and study it. We cannot rewrite textbooks in form of forum postings! A good one is Sakurai, Modern Quantum Mechanics.

 

[A. Neumaier]

I still await some further insights.

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]

So I remembered that this was addressed in a book I once read, https://www.amazon.com/dp/1441982663/?tag=pfamazon01-20. On page 194, the author summarizes how a "spinor" describes spin-1/2 particles, fermions. And the author says that the SU(2) symmetry of the Standard Model is the generator of spinor rotations. So it seems that fermions/bosons come from the symmetries of the Lagrangian of the Standard Model. And the Lagrangian can be rewritten as the Hamiltonian. So it seems, yes, the spin of a particle does come from H.

 

[A. Neumaier]

fermions/bosons come from the symmetries of the Lagrangian of the Standard Model.

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.

Also what you write has nothing to do with your original question, and hence doesn't belong into this thread. (And next time post with the rating B, not I, since what you are lacking are very basic things.)

 

[friend]

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.

As opposed to what, giving you my word alone for it.

Also what you write has nothing to do with your original question, and hence doesn't belong into this thread.

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.

And once again you've reduced the thread to condescension. I think this thread is over.

 

[friend]

What is |<x'|x>|² = <x'|x><x'|x>^* = <x'|x><x|x'> ? Is this a legitimate probability, or probability density?

 

[vanhees71]

No, it's obviously not. Again, generalized eigenstates of self-adjoint operators in the continuous part of their spectrum are never proper Hilbert-space vectors and thus do not represent (pure) quantum states. They are distributions and belong to the dual space of the domain of the self-adjoint operator. You must not multiply them. In your case of a position eigenvector you have
$$\langle x'|x \rangle=\delta(x'-x).$$
This clearly shows you that you must not take its square!

For an introduction to the modern treatment of these issues in terms of the "rigged-Hilbert space formalism" see Ballentine, Quantum Mechanics, Addison-Wesley. More mathematical details can be found in

A. Galindo and P. Pascual. Quantum Mechanics. Springer Verlag, Heidelberg, 1990. 2 Vols.

 

[friend]

As I mentioned in post #5 in this thread, <x'|e^-iHt/ħ|x> is a propagator which is a probability amplitude per this wikipedia.org article, where e^-iHt/ħ is the unitary operator U(t,t') of that article. And Nugatory in post #8 pointed out that <x'|x> is <x'|e^-iHt/ħ|x> as t→0.

My question is, at what value of t as it approached 0 does <x'|e^-iHt/ħ|x> stop being a probability amplitude such that it cannot be treated as any other probability amplitude so that we can take its modulus squared to get a probability?

 

[A. Neumaier]

a propagator which is a probability amplitude per this wikipedia.org article

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]

...while the absolute square of a propagator value can be any number, including infinity, and hence cannot have a probability interpretation.

Are we missing a normalization factor in order to turn a propagator into a probability amplitude?

 

[vanhees71]

Again! The propagator NEVER is the probability distribution, it's a generalized function and must not be squared. The probability amplitude (wave function) is given by
$$\psi(t,x)=\int \mathrm{d} x G(t;x,x') \psi_0(x'),$$
where $\psi_0$ is a square integrable initial wave function.

 

[friend]

There may be some confusion on how various authors label the propagator. In the expression <x'|e^-iHt/ħ|x>, some authors say that U(t)=e^-iHt/ħ is the propagator that is used to advance the wave function in time. For example, Shankar, Ramamurti, Principles of Quantum Mechanics, page 55. And this U(t) by itself is not a wave function that in any way leads to a probability. But other authors say that <x'|e^-iHt/ħ|x> is the propagator, see Professor Robert Littlejohn notes here. I think the latter is just the matrix elements of the former. Wouldn't that mean that <x'|e^-iHt/ħ|x> has some physical meaning?

One might expect that <x'|x> ≈ lim(t→0) <x'|e^-iHt/ħ|x> would have to be complex (and not just δ(x'-x) ) so that <x|x'> could be its complex conjugate so that <x'|x><x|x'> would have some meaning in terms of probabilities.

 

[vanhees71]

Ok let's get this straight, because it's very important. You have to distinguish different forms to express quantum theory. The most general (and in my opinion best to start with) is the formalism using abstract rigged Hilbert spaces and operators. The only obstacle sometimes is that you have some arbitrariness to choose how the statistical operator (i.e., the state) and the self-adjoint operators representing observables depend on time. That's called the choice of the picture. In non-relativistic quantum theory one usually starts with the Schrödinger picture, where the full time evolution is on the statistical operator. Its time dependence is given by the time-evolution operator $\hat{U}(t,t')$,
$$\hat{\rho}(t)=\hat{U}(t,t_0) \hat{\rho}(t_0) \hat{U}^{\dagger}(t,t_0),$$
wherea the time-evolution operator is determined by the operator-valued initial-value problem
$$\mathrm{i} \partial_t \hat{U}(t,t_0)=\hat{H}(t) \hat{U}(t,t_0).$$
If $\hat{H}$ is not explicitly time dependent, then this is readily integrated to
$$\hat{U}(t,t_0)=\exp[-\mathrm{i} \hat{H}(t-t_0)], \quad \hat{U}(t_0+0^+,t_0)=1.$$
For wave mechanics you usually express everything in the position representation, i.e., you work with wavefunctions in the concrete function space $L^2$ of square-integrable functions.

For a pure state you have
$$\hat{\rho}(t)=|\psi,t \rangle \langle \psi,t|, \quad \text{with} \quad \langle \psi,t|\psi,t \rangle=1.$$
and you can choose
$$|\psi,t \rangle=\hat{U}(t,t_0) |\psi,t_0 \rangle$$
as representant of the state $\hat{\rho}(t)$. For the wave function this implies
$$\psi(t,x)=\langle x|\psi,t \rangle=\int \mathrm{d} x' \langle x|\hat{U}(t,t_0)|x' \rangle \langle x'|\psi,t_0 \rangle=\int \mathrm{d} x' U(x,t;x',t_0) \psi_0(x').$$
obviously the propagator $U(x,t;x',t_0)$ satisfies the initial-value problem of the Schrödinger equation,
$$\mathrm{i} \partial_t U(x,t;x',t_0)=\hat{H} U(x,t;x',t_0), \quad U(x,t_0+0^+;x',t_0)=\delta(x-x'),$$
where now $\hat{H}$ is the Hamiltonian in position representation.

For a free particle (and also for the harmonic oscillator) you can solve for the propgator explicitly. That's also a very good exercise to do for yourself. For the free particle we have
$$\hat{H}=\frac{\hat{p}^2}{2m}.$$
To evaluate the propagator it's thus most convenient to use the completeness of the generalized momentum eigenbasis:
$$U(t,x;t_0,x')=\int \mathrm{d} p \langle x|\exp \left[ -\mathrm{i} \frac{\hat{p}^2}{2m}(t-t_0) \right ]|p \rangle \langle p|x' \rangle=\int \mathrm{d} p \langle x|p \rangle \langle p|x' \rangle \exp\left[ -\mathrm{i} \frac{p^2}{2m} (t-t_0)\right ]=\int \frac{\mathrm{d} p}{2 \pi} \exp\left[ -\mathrm{i} \frac{p^2}{2m} (t-t_0)\right ] \exp[\mathrm{i} p(x-x')].$$
Now this is a Fourier transformation of a Gaussian, but to make sense of it you have to add a little imaginary part to the time to "regularize" it, i.e., you set $t \rightarrow t-\mathrm{i} 0^+$. Then you can solve for the Gaussian integral and get a unique result for $\hat{U}$:
$$U(t,x;t_0,x')=\sqrt{\frac{1}{2 \pi m \mathrm{i}(t-t_0)}} \exp \left [\frac{\mathrm{i} m}{2 (t-t_0)} (x-x_0)^2 \right ].$$
As you see, that's not square integrable as to be expected already from the initial condition.

 

[A. Neumaier]

One might expect

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]

Ok let's get this straight, because it's very important...

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]

I'm understanding from our conversation that $U(t;x,x')=\langle x|\exp(-\mathrm{i} \hat{H} t) x' \rangle,$ is not a wave function since it is strictly not square integrable. So no probabilities can be derived from it alone. I probably got confused because it seems to be a solution to the Schrodinger's equation as you indicate.

However, that may not be connected to my ultimate purpose here. I'm trying to understand the nature of the wave function itself in terms of (the here unpopular) virtual particles (whatever those might be). To that end I'm understanding that the Propagator is the Green's function associated with the Schrodinger's equation, and I found this quote from Prof. Matthew Strassler, Theoretical Physicists from Rutger's University, " [Students of math and physics will recognize real photons as solutions of a wave equation, and virtual photons as related to the Green function associated with this equation.]", at the bottom of the 5th paragraph, here.

For our purposes here, the propagator can be written,
<x'|e^-iHt/ħ|x> = ∫∫∫⋅⋅⋅∫<x'|e^-iHε/ħ|x₁><x₁|e^-iHε/ħ|x₂><x₂|e^-iHε/ħ|x₃>⋅⋅⋅<x_n|e^-iHε/ħ|x> dx₁dx²dx₃⋅⋅⋅dx_n

The question is: do these lim(ε→0) <x_i|e^-iHε/ħ|x_j> ≈(?) <x_i|x_j> represent the virtual particles referred to by Prof. Matt Strassler?

 

[vanhees71]

This latter idea is kind of a first step of the path-integral formulation. Do not study photons before you haven't fully understood non-relativistic quantum theory. Photons cannot described adequately in the wave-function formalism. It has no position observable to begin with. It's a quite complicated object, even already in the free case and can only described correctly in terms of quantum field theory.

In non-relativistic quantum theory, the meaning (and the only logically consistent meaning!) of the wave function is that
$$P(t,x)=|\psi(t,x)|^2$$
is the probability density function for the position of a single particle, provided that
$$\int_{\mathbb{R}} \mathrm{d} x P(t,x)=1.$$
So the wave function must be square integrable to have a well-defined definition of a physical state in sense of quantum theory.

Of course, not every solution of the Schrödinger equation is a wave function. One example is the above derived propagator for a free non-relativistic particle. It's a solution of the Schrödinger equation, but in the sense of a generalized function. It lives not in the Hilbert space of square integrable wave functions but in the dual of a smaller dense subspace, where the position and momentum operators are defined. Since the subspace is strictly smaller than the Hilbert space, it's dual is larger than the Hilbert space. The dual of the Hilbert space itself is isomorphic to the Hilbert space.

 

[friend]

This latter idea is kind of a first step of the path-integral formulation. Do not study photons before you haven't fully understood non-relativistic quantum theory. Photons cannot described adequately in the wave-function formalism. It has no position observable to begin with. It's a quite complicated object, even already in the free case and can only described correctly in terms of quantum field theory.

Yea, I kind a picked up on that. I think Prof. Strassler must have meant real and virtual particles (not photons) since they have more to do with the Schrodinger equation and the Green's function of that.

 

[vanhees71]

Forget virtual particles. That's pop-sci speach. It describes mathematical terms in the perturbative expansion, which can be efficiently written in terms of Feynman diagrams, where the "virtual particles" are symbolized for the internal lines that stand for propagators, i.e., mathematical expressions, nothing else.

 

[friend]

Forget virtual particles. That's pop-sci speach. It describes mathematical terms in the perturbative expansion, which can be efficiently written in terms of Feynman diagrams, where the "virtual particles" are symbolized for the internal lines that stand for propagators, i.e., mathematical expressions, nothing else.

Yes, they are just mathematical devises, just as are manifolds, and differential lengths. We invent terms like force and mass to give some definition of what we are integrating in classical mechanics. So I'm trying to understand what those little pieces are that we are integrating in the Feynman path integral since that's what we do in classical integrals. And it seems unavoidable since we assume that the whole is constructed from the little pieces. So your advice seems to be to be saying that I should give up on understanding the basic structures of physical things. And the little pieces in the path integral seem to be called "virtual". The only problem left is to discover their properties and to understand the world in terms of them. So you are basically asking me to stop caring at all. Sorry, I can't do that.

 

[friend]

... For the wave function this implies
$$\psi(t,x)=\langle x|\psi,t \rangle=\int \mathrm{d} x' \langle x|\hat{U}(t,t_0)|x' \rangle \langle x'|\psi,t_0 \rangle=\int \mathrm{d} x' U(x,t;x',t_0) \psi_0(x').$$
obviously the propagator $U(x,t;x',t_0)$ satisfies the initial-value problem of the Schrödinger equation,
$$\mathrm{i} \partial_t U(x,t;x',t_0)=\hat{H} U(x,t;x',t_0), \quad U(x,t_0+0^+;x',t_0)=\delta(x-x'),$$
where now $\hat{H}$ is the Hamiltonian in position representation.

It seems what this says is that for the wave function to propagate through time at all means that one must take into account how every point in space interacts with the initial wave function. Or in other words, the wave function cannot ever exist in spacetime at all unless it moves through time, and this in turn means the wave function is always defined in terms of the propagator which takes into account every point in space.

 

[stevendaryl]

Yes, they are just mathematical devises, just as are manifolds, and differential lengths. We invent terms like force and mass to give some definition of what we are integrating in classical mechanics. So I'm trying to understand what those little pieces are that we are integrating in the Feynman path integral since that's what we do in classical integrals. And it seems unavoidable since we assume that the whole is constructed from the little pieces. So your advice seems to be to be saying that I should give up on understanding the basic structures of physical things. And the little pieces in the path integral seem to be called "virtual". The only problem left is to discover their properties and to understand the world in terms of them. So you are basically asking me to stop caring at all. Sorry, I can't do that.

I think you're mixing up two different topics. There is a path-integral formulation for nonrelativistic quantum mechanics, and that in no way involves "virtual particles". Virtual particles appear when one does perturbation theory in quantum field theory. The two subjects are related, in that both of them make use of something called the "propagator", but the expression that is being discussed in this thread, \langle x| \hat{U}(t, t_0)|x_0\rangle is from nonrelativistic quantum mechanics, and doesn't involve virtual particles at all.

As vanhees71 says, a propagator is not a transition amplitude, because it has different normalization rules. However, conceptually, it seems to me that it can be understood as a generalization of transition amplitudes, so I don't agree with him that it's so bad to call them that (as long as the normalization business is made clear).

If you have a quantum system with a discrete number of states |psi_j\rangle, then you can compute a transition amplitude:

A_{ij}(t, t_0)

which is defined to be the probability amplitude that a system initially in state |\psi_j\rangle at time t_0 will be found in state |\psi_i\rangle at time t. The mathematical expression for this is: A_{ij}(t, t_0) = \langle \psi_i | \hat{U}(t, t_0) | \psi_j \rangle = \langle \psi_i | e^{-i \hat{H} (t-t_0)/\hbar} | \psi_j \rangle. The probability that a system initially prepared in state j will later be observed to be in state i is just P_{ij} = |A_{ij}|^2.

The laws of quantum mechanics allow us to relate A_{ij} at different times as follows:

A_{ij}(t, t_0) = \sum_k A_{ik}(t, t_1) A_{kj}(t_1, t_0)

where t_1 is any time between t_0 and t_1, and the index k runs over all possible intermediate states.

Obviously, you can continue to expand the amplitude to get something like:

A_{ij}(t, t_0) = \sum_{k_1, k_2, ..., k_N} A_{ik_N}(t, t_N) A_{k_N, k_{N-1}}(t_N, t_{N-1}) ... A_{k_1, j}(t_1, t_0)

where t_n = t + n \epsilon, where \epsilon = \frac{n (t-t_0)}{N+1}

If we define a "path" (not through physical space, but through state space) p to be a sequence of states \psi_{k_0}, \psi_{k_1}, ..., \psi_{k_{N+1}} of length N+2, then we can associate an amplitude with such a path:

\phi(p, t, t_0) = A_{k_{N+1}, k_N}(t, t_N) A_{k_N, k_{N-1}}(t_N, t_{N-1}) ... A_{k_1, k_0}(t_1, t_0)

Then our formula above can be summarized as:

A_{ij}(t, t_0) = \sum_p \phi(p, t, t_0)

where the sum is over all possible paths p that start at \psi_j and end at \psi_i. So you can interpret the equation for transition amplitudes as saying: "Take all possible paths from \psi_j to \psi_i, and add their amplitudes." I suppose you could say that these paths are "virtual", in that there is no sense in which the system actually takes any of those paths--it's just a calculational device.

The propagator is in some sense the continuum limit of such a transition amplitude, where the states \psi_j are states that are localized in physical space.

 

[friend]

I think you're mixing up two different topics. There is a path-integral formulation for nonrelativistic quantum mechanics, and that in no way involves "virtual particles". Virtual particles appear when one does perturbation theory in quantum field theory. The two subjects are related, in that both of them make use of something called the "propagator", but the expression that is being discussed in this thread, \langle x| \hat{U}(t, t_0)|x_0\rangle is from nonrelativistic quantum mechanics, and doesn't involve virtual particles at all.

It does seem that there are different senses in which the words virtual particles are used. There is the sense used in perturbation theory, as you say. But I think there is also the sense in which virtual particles automatically exist everywhere; they are also called quantum fluctuations. I think it is these that are used when talking about the Casimir effect and Hawking and Unruh radiation. My feeling is that the word virtual is invoked when talking about the little pieces that are being integrated in the Feynman Path Integral. I've even heard of virtual geometries when using the path integrals in quantum gravity when possible geometries of spacetime exist in superposition. You might call each of the infinite number of paths in the path integral virtual sense none of them can be said to really exist but they are nevertheless integrated in superposition. But then again each of those virtual paths consist of even smaller displacements (of what particles?). Since the wave function is a distributed thing, I'm trying to understand the wave function in terms of these tiny, virtual, things. What I suspect is that the wave function can be described in terms of these virtual particles of quantum mechanics. And the virtual particles of quantum mechanics can be explained in terms of the virtual particles of quantum field theory. I imagine it will be easy to confuse the two.

 

[stevendaryl]

It does seem that there are different senses in which the words virtual particles are used. There is the sense used in perturbation theory, as you say. But I think there is also the sense in which virtual particles automatically exist everywhere; they are also called quantum fluctuations. I think it is these that are used when talking about the Casimir effect and Hawking and Unruh radiation.

Those are all cases of the same thing: particles that appear in perturbation theory.

My feeling is that the word virtual is invoked when talking about the little pieces that are being integrated in the Feynman Path Integral.

There might be a sense in which they are related, but I don't think anyone uses the phrase "virtual particle" in talking about Feynman's path integral formulation of nonrelativistic quantum mechanics.

 

[friend]

... There might be a sense in which they are related, but I don't think anyone uses the phrase "virtual particle" in talking about Feynman's path integral formulation of nonrelativistic quantum mechanics.

Well for example, for any wave function to propagate we must have,

<x'|e^-iHt/ħ|x> = ∫∫∫⋅⋅⋅∫<x'|e^-iHε/ħ|x₁><x₁|e^-iHε/ħ|x₂><x₂|e^-iHε/ħ|x₃>⋅⋅⋅<x_n|e^-iHε/ħ|x> dx₁dx²dx₃⋅⋅⋅dx_n

The question is for each <x_i|e^-iHε/ħ|x_j> is there a complex conjugate <x_i|e^-iHε/ħ|x_j>^* ? This would be a kind of virtual antiparticle, perhaps.

 

[friend]

No, it's obviously not. Again, generalized eigenstates of self-adjoint operators in the continuous part of their spectrum are never proper Hilbert-space vectors and thus do not represent (pure) quantum states. They are distributions and belong to the dual space of the domain of the self-adjoint operator. You must not multiply them. In your case of a position eigenvector you have
$$\langle x'|x \rangle=\delta(x'-x).$$
This clearly shows you that you must not take its square!

Shankar, Principles of Quantum Mechanics, page 145, says,
"The other set of improper kets we will use in the same spirit are the position eigenkets |x>, which also form a convenient basis. Again, when we speak of a particle being in a state |x₀> we shall mean that its wave function is so sharply peaked at x=x₀ that it may be treated as a delta function to a good accuracy."
So I think that this means that <x'|x> = δ(x-x') is a very close approximation for the purposes of any calculation to a physical entity, a transition amplitude.

 

[vanhees71]

Well, there are many sloppy quantum mechanics textbooks out there. That's a pity, but you can't help it. Just one last time: It doesn't make sense to claim that non-normalizable distributions represent states in any way. In this example it's obvious. You cannot even square the $\delta$ distribution, let alone integrate over this undefined quantity.

 

[stevendaryl]

So I think that this means that <x'|x> = δ(x-x') is a very close approximation for the purposes of any calculation to a physical entity, a transition amplitude.

Well, you can sort-of approximate a delta function by a very localized, square-integrable function. But the normalization is not right for a wave function. \delta(x-x&#039;) is normalized so that \int \delta(x-x&#039;) dx = 1, while a wave function is normalized so that \int |\psi(x)|^2 dx = 1.

 

[friend]

Well, you can sort-of approximate a delta function by a very localized, square-integrable function. But the normalization is not right for a wave function. \delta(x-x&#039;) is normalized so that \int \delta(x-x&#039;) dx = 1, while a wave function is normalized so that \int |\psi(x)|^2 dx = 1.

I find it strange to insist that there is no physical content to |x>. For we certainly consider it legitimate to write <x|ψ> = ψ(x). If <x| is meaningful in <x|ψ>, then shouldn't we expect that its dual, |x> is also physically meaningful.

 

[vanhees71]

This is the very point of the entire discussion! A distribution has physical meaning only when folded with a test function. Note that $|\psi \rangle$ must be in the domain of the position operator which is a dense subspace of the Hilbert space, so that $\langle x|\psi \rangle$ is well-defined!

 

[friend]

This is the very point of the entire discussion! A distribution has physical meaning only when folded with a test function. Note that $|\psi \rangle$ must be in the domain of the position operator which is a dense subspace of the Hilbert space, so that $\langle x|\psi \rangle$ is well-defined!

Yea, that all sounds familiar. Thanks for the reminder.

So my understanding from all this is that |x₀> is not a wave-function in the sense that you cannot extract a probability from it because it is not square integrable. That would only mean that we cannot ever predict the probabilities that |x₀> will occur, right? I think that is to be expected. It represents the extreme of the Uncertainty Principle ΔxΔp≥ħ/2. Since |x₀> represents a wave-function located precisely at x₀, then the Δx of it would be zero, which is impossible. Same can be said of |p> for a wave function of exact momentum, p.