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kuahji

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- Thread starter kuahji
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kuahji

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- #2

Vanadium 50

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Because the derivative is momentum: a discontinuity means infinite momentum.

- #3

Meir Achuz

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- #4

Fredrik

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What Vanadium50 and Clem said are the physical reasons. I'll add that mathematically, the fact that the function satisfies the Schrödinger equation means that it's differentiable, and differentiable functions are continuous.

There's a related question that I still don't know the answer to: Is there a*mathematical* reason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?

(There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).

There's a related question that I still don't know the answer to: Is there a

(There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).

Last edited:

- #5

SpectraCat

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What Vanadium50 and Clem said are the physical reasons. I'll add that mathematically, the fact that the function satisfies the Schrödinger equation means that it's differentiable, and differentiable functions are continuous.

There's a related question that I still don't know the answer to: Is there amathematicalreason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow 0[/itex]?

(There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).

I guess you meant as x--> +/- infinity above? Otherwise I don't understand what you wrote at all.

If I am right and it was supposed to be x--> +/- infinity, then I would be interested to see an example of a square integrable smooth function that doesn't tend to zero in that limit.

- #6

Fredrik

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Consider the function:

[tex]

\psi(x) =

\begin{cases}

0 & x < 1 \\

1 & x \in [n, n + n^{-2}) \\

0 & x \in [n + n^{-2}, n+1)

\end{cases}

[/tex]

wherenranges over all positive integers. [itex]\psi(x)[/itex] does not converge to zero at [itex]+\infty[/itex]. However,

[tex]

\int_{-\infty}^{+\infty} |\psi(x)|^2 \, dx

= \sum_{n = 1}^{+\infty} \int_{n}^{n + n^{-2}} 1 \, dx

= \sum_{n = 1}^{+\infty} n^{-2} = \pi^2 / 6[/tex]

There are continuous, differentiable, and even smooth versions of such functions; you just replace the basic rectangle shape

[tex]with something smoother.

f(x) = \begin{cases} 0 & x < 0 \\

1 & 0 \leq x < 1 \\

0 & 1 \leq x \end{cases}[/tex]

- #7

George Jones

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There's a related question that I still don't know the answer to: Is there amathematicalreason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?.

Maybe:

1) [itex]\psi[/itex] is continuous and in the domain of the position operator, or;

2) (much stronger) [itex]\psi[/itex] is continuous and in the domain of all positive integral powers of the position operator?

- #8

Fredrik

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(Q and P are of course the position and momentum operators).

I think something very much like this (or exactly this) is standard in the rigged Hilbert space formalism.

- #9

strangerep

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(Q and P are of course the position and momentum operators).

I think something very much like this (or exactly this) is standard in the rigged Hilbert space formalism.

Indeed. We want an exponentiated momentum operator to get a representation of

finite translations, but we define an exponentiated operator in terms of a Taylor

series. Hence all powers of P must be accommodated sensibly. This notion (and a

similar one for Q) basically motivates how one defines the "small" nuclear space

in a Gelfand triple (aka rigged Hilbert space). It must be such that all powers

of the relevant operators are accommodated in the space. For the case being

discussed here, the elements of the space must not merely go to 0 at spatial

infinity, but must do so

- #10

jostpuur

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I like that general idea, but shouldn't we take it even further? E.g. say that the physical states are the ones such that [itex]f(Q,P)\psi[/itex] is square integrable for any polynomial f.

(Q and P are of course the position and momentum operators).

We know from earlier experience with the delta potential, that the derivative of the wave function can be discontinuous, and ill defined at some points. So your demands for physical states don't look good to me.

---

Another interesting fact is that the domain of the time evolution operator [itex]\exp(-\frac{it}{\hbar}H)[/itex] is usually larger than the domain of the Hamilton's operator [itex]H[/itex]. This is because always

[tex]

\|\exp\Big(-\frac{it}{\hbar}H\Big)\psi\|_2 = \|\psi\|_2

[/tex]

but sometimes

[tex]

\|H\psi\|_2 = \infty

[/tex]

even though [itex]\|\psi\|_2=1[/itex]. This is why I would not give the domain of [itex]H[/itex] too fundamental role. The actual time evolution is more fundamental, than the Hamilton's operator, and sometimes the time evolution can be well defined even when the Schrödinger's equation is not satisfied in vector space sense.

- #11

jostpuur

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If Psi is described as a function, then the probability for a particle to be in a point will always be zero. Discontinuity doesn't lead to a probability contradiction.

- #12

jostpuur

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Suppose that the wave function is a following square with discontinuities

[tex]

\psi(x) =\frac{1}{\sqrt{2L}}\chi_{[-L,L]}(x)

[/tex]

Here chi is the characteristic function which equals 1 on the interval [-L,L] and equals 0 outside it.

The Fourier transform is

[tex]

\hat{\psi}(p) = \sqrt{\frac{2}{L}} \frac{\sin(pL)}{p}

[/tex]

This might look like that the momentum is approximately zero, but the expectation value of the momentum is actually

[tex]

\frac{1}{2\pi} \int\limits_{-\infty}^{\infty} p|\hat{\psi}(p)|^2 dp = \frac{1}{L\pi} \int\limits_{-\infty}^{\infty}

\frac{\sin^2(pL)}{p} dp

[/tex]

Oh! Its [itex]+\infty - \infty[/itex]

But how serious is this? I'm not sure. If it is decided that this is the wave function at some instant [itex]t=0[/itex], then the time evolution would be well defined. IMO it could be an acceptable situation where the expectation value of a momentum doesn't exist. The probability density would still be well defined, so isn't everything ok then?

- #13

guerom00

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There's a related question that I still don't know the answer to: Is there amathematicalreason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?

Continuum wavefunctions do not go to 0 at infinity

Who said our beloved plane waves are not physical?

- #14

jostpuur

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We know from earlier experience with the delta potential, that the derivative of the wave function can be discontinuous, and ill defined at some points. So your demands for physical states don't look good to me.I like that general idea, but shouldn't we take it even further? E.g. say that the physical states are the ones such that [itex]f(Q,P)\psi[/itex] is square integrable for any polynomial f.

(Q and P are of course the position and momentum operators).

I was left slightly disturbed by my own comment, because I'm not sure how to give a rigor definition for the delta potential from the operator point of view. It could be that the delta potential problem must be defined in some other way than as an eigenvalue problem for some operator [itex]D(H)\to L_2(\mathbb{R})[/itex], where [itex]D(H)\subset L_2(\mathbb{R})[/itex].

But notice that even if a potential [itex]V[/itex] is a well defined measurable function so that the multiplication operator [itex]M_V[/itex] exists, then if the potential contains discontinuities, the second derivatives of the eigenfunctions [itex]\partial_x^2\psi[/itex] will contain discontinuities too! And points of non-existence.

So this idea for "physical states" is too strict.

- #15

strangerep

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Another interesting fact is that the domain of the time evolution operator [itex]\exp(-\frac{it}{\hbar}H)[/itex] is usually larger than the domain of the Hamilton's operator [itex]H[/itex]. This is because always

[tex]

\|\exp\Big(-\frac{it}{\hbar}H\Big)\psi\|_2 = \|\psi\|_2

[/tex]

but sometimes

[tex]

\|H\psi\|_2 = \infty

[/tex]

even though [itex]\|\psi\|_2=1[/itex]. This is why I would not give the domain of [itex]H[/itex] too fundamental role. The actual time evolution is more fundamental, than the Hamilton's operator, and sometimes the time evolution can be well defined even when the Schrödinger's equation is not satisfied in vector space sense.

But how do you define the exponential of an operator, if not via a Taylor series:

[tex]

\exp(A) ~:=~ 1 + A + \frac{1}{2}A^2 + ..... ~~~~~(?)

[/tex]

If so, then all terms in the series must make sense when acting on the state.

Hence, the domain of the exponentiated operator is restricted to the much

smaller intersection of the domains of all powers of A.

- #16

jostpuur

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For a free non-relativistic particle this means that the definition comes most nicely in the Fourier basis, like this:

[tex]

\hat{\psi} \mapsto \exp\Big(\frac{-it}{\hbar}H\Big)\hat{\psi},\quad\quad

\Big(\exp\Big(-\frac{it}{\hbar}H\Big) \hat{\psi}\Big)(p) = \exp\Big(-\frac{it}{\hbar} \frac{p^2}{2m}\Big) \hat{\psi}(p)

[/tex]

I'm not sure what's the most general definition for [itex]\exp(A)[/itex]. It seems that there exists different definitions with different domains of validity, such that these definitions still agree under some simple conditions.

- #17

Fredrik

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Does anyone know if the nuclear space H' (i.e., the space of all [itex]\psi\in L^2(\mathbb R)[/itex] such that [itex]f(Q,P)\psi\in L^2(\mathbb R)[/itex] for all polynomials f) is a Hilbert space too? Is H' a "big" or a "small" supspace of H? I mean, is it dense in H or something like that?

Joostpuur, I don't see how the things you did change anything. Since p^2/2m is an eigenvalue of H, you can replace p^2/2m with H on the right, and now we have recovered the problem you tried to avoid.

- #18

jostpuur

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The use of a mapping

[tex]

L_2(\mathbb{R})\to L_2(\mathbb{R}),\quad \hat{\psi}\mapsto e^{- itH}\hat{\psi},\quad \Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)

[/tex]

avoids all problems, related to situations where

[tex]

\int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty

[/tex]

with some [itex]n[/itex].

----

Why do you keep this interest to the condition where [itex]f(Q,P)\psi\in L_2[/itex] with all polynomials [itex]f[/itex]? If a potential [itex]V[/itex] contains discontinuities, then the eigenstates, which will be physical states, will satisfy following conditions:

[tex]

\psi \in C,\quad \psi \in C^1,\quad \psi\notin C^2

[/tex]

This demand [itex]f(Q,P)\psi\in L_2[/itex] may produce some interesting function space, but it doesn't make sense to pose this demand for "physical states".

[tex]

L_2(\mathbb{R})\to L_2(\mathbb{R}),\quad \hat{\psi}\mapsto e^{- itH}\hat{\psi},\quad \Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)

[/tex]

avoids all problems, related to situations where

[tex]

\int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty

[/tex]

with some [itex]n[/itex].

----

Why do you keep this interest to the condition where [itex]f(Q,P)\psi\in L_2[/itex] with all polynomials [itex]f[/itex]? If a potential [itex]V[/itex] contains discontinuities, then the eigenstates, which will be physical states, will satisfy following conditions:

[tex]

\psi \in C,\quad \psi \in C^1,\quad \psi\notin C^2

[/tex]

This demand [itex]f(Q,P)\psi\in L_2[/itex] may produce some interesting function space, but it doesn't make sense to pose this demand for "physical states".

Last edited:

- #19

Fredrik

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Why would you consider solutions to Schrödinger equations with discontinuous potentials to be physical states? I'd say those are just as unrealistic as plane waves. Where do you find a discontinous potential in the real world? How would you design an experiment that prepares a system in such a state?

I don't see the point of defining the exponential exp(-iHt) by

[tex]\Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)[/tex]

because if [tex]\hat\psi[/tex] is the momentum space wavefunction,

[tex]e^{-it\frac{p^2}{2m}}\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-it\frac{p^2}{2m}}\Big)^n\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-itH\Big)^n\hat{\psi}[/tex]

So we end up with exp(-iHt) being equal to a power series in H whether we want it or not.

I got that polynomial condition from an article about the rigged Hilbert space formalism that I read some time ago. I think it was the article that Strangerep posted in the thread I started (a year ago?) about unbounded operators.

I don't see the point of defining the exponential exp(-iHt) by

[tex]\Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)[/tex]

because if [tex]\hat\psi[/tex] is the momentum space wavefunction,

[tex]e^{-it\frac{p^2}{2m}}\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-it\frac{p^2}{2m}}\Big)^n\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-itH\Big)^n\hat{\psi}[/tex]

So we end up with exp(-iHt) being equal to a power series in H whether we want it or not.

I got that polynomial condition from an article about the rigged Hilbert space formalism that I read some time ago. I think it was the article that Strangerep posted in the thread I started (a year ago?) about unbounded operators.

Last edited:

- #20

jostpuur

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I don't see the point of defining the exponential exp(-iHt) by

[tex]\Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)[/tex]

because if [tex]\hat\psi[/tex] is the momentum space wavefunction,

[tex]e^{-it\frac{p^2}{2m}}\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-it\frac{p^2}{2m}}\Big)^n\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-itH\Big)^n\hat{\psi}[/tex]

So we end up with exp(-iHt) being equal to a power series in H whether we want it or not.

The series

[tex]

e^{it\frac{p^2}{2m}} = \sum_{n=0}^{\infty} \frac{(it)^n}{n!}\Big(\frac{p^2}{2m}\Big)^n

[/tex]

will converge towards a complex number such that [itex]|z|=1[/itex]. So

[tex]

\int\limits_{-\infty}^{\infty} \Big| e^{it\frac{p^2}{2m}} \hat{\psi}(p)\Big|^2 dp = \int\limits_{-\infty}^{\infty} |\hat{\psi}(p)|^2 dp

[/tex]

will be guaranteed. Even if

[tex]

\int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty

[/tex]

with some n, the equation

[tex]

\int\limits_{-\infty}^{\infty} \Big|\Big(\sum_{n=0}^{\infty} \frac{(it)^n}{n!} \Big(\frac{p^2}{2m}\Big)^n\Big)

\hat{\psi}(p)\Big|^2 dp = \int\limits_{-\infty}^{\infty} |\hat{\psi}(p)|^2 dp

[/tex]

is still going to remain true.

Why would you consider solutions to Schrödinger equations with discontinuous potentials to be physical states? I'd say those are just as unrealistic as plane waves. Where do you find a discontinous potential in the real world?

I was thinking about the finitely deep square potential

[tex]

V(x) = -|V_0| \chi_{[-L,L]}(x)

[/tex]

but if you raise doubts about its physicalness, then I'm not sure how to proceed. I think you might as well ask where do you find smooth potentials in the real world? All mathematical potentials are approximations anyway.

I got that polynomial condition from an article about the rigged Hilbert space formalism that I read some time ago. I think it was the article that Strangerep posted in the thread I started (a year ago?) about unbounded operators.

I remember this thread. I think we clashed already there...

I didn't like the physicists' way of insisting that the unboundedness leads to continuous spectra.

- #21

PhaseShifter

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...and in the second, [tex]\psi(x)=0[/tex] for all x>1.

Am I missing something obvious, or is this a pedantic distinction between "function equals zero" and "function approaches zero"?

- #22

strangerep

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The use of a mapping

[tex]

L_2(\mathbb{R})\to L_2(\mathbb{R}),\quad \hat{\psi}\mapsto e^{-

itH}\hat{\psi},\quad \Big(e^{-itH}\hat{\psi}\Big)(p) =

e^{-it\frac{p^2}{2m}}\hat{\psi}(p)

[/tex]

avoids all problems, related to situations where

[tex]

\int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty

[/tex]

with some [itex]n[/itex].

"Avoids all problems..."?? But you don't have a representation

of the Galilei Lie group as a group of operators on the Hilbert

space. Physically relevant Lie groups are continuous and arbitrarily

differentiable in their parameters so (among other things),

[tex]

\partial^{(n)}_t \, e^{-itH}

[/tex]

must make sense. In your case, this gives terms involving

[tex]

p^k \, \hat{\psi}(p)

[/tex]

which don't necessarily remain in the Hilbert space.

So you've pretty much abandoned the principle of constructing a

unitary representation of a physically relevant Lie group.

Last edited:

- #23

strangerep

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Does anyone know if the nuclear space H' (i.e., the space of all

[itex]\psi\in L^2(\mathbb R)[/itex] such that

[itex]f(Q,P)\psi\in L^2(\mathbb R)[/itex] for all polynomials f) is a

Hilbert space too? Is H' a "big" or a "small" supspace

of H? I mean, is it dense in H or something like that?

Start with a Hilbert space [itex]{\mathcal H}[/itex] of functions

[itex]\psi(p)[/itex] which are square-integrable:

[tex]

\int dp \, |\psi(p)|^2 ~<~ \infty

[/tex]

Then consider the subspace of those functions such that

[itex]p\,\psi(p)[/itex] is also square-integrable, i.e.,

[tex]

\int dp \, |p \, \psi(p)|^2 ~<~ \infty

[/tex]

Call the space of such functions [itex]{\mathcal H}_1[/itex].

Clearly, [itex]{\mathcal H}_1 \subset {\mathcal H}[/itex].

One can also show that [itex]{\mathcal H}_1[/itex] is dense in [itex]{\mathcal H}[/itex].

Similarly, for higher powers of p, one gets smaller and smaller

spaces (I'll call them [itex]{\mathcal H}_n[/itex]), each dense

in the earlier ones. Taking an inductive limit of arbitrarily high powers,

one gets a nuclear space [itex]{\mathcal H}_\infty[/itex] which is still dense

in [itex]{\mathcal H}[/itex]. In fact, [itex]{\mathcal H}_\infty[/itex] is the

"small" space in a Gel'fand triple corresponding to this example.

Yes, one first restricts to the largest space on which arbitrary powers of all theBut now Wigner's symmetry representation theorem confuses me instead. We

define a symmetry as a probability-preserving bijection on the set of unit

rays of some Hilbert space H, and we impose the requirement that there's a

symmetry for each member of the Galilei group (when we want to end up with

non-relativistic QM) or Poincaré group (when we want to end up with special

relativistic QM). The theorem says (roughly) that there exists a unitary

representation of the group of symmetries into GL(H') where H' is a Hilbert

space. I always thought that H'=H, but now it seems to me that H' must be a

subset of H, at least if the H we started with is the whole space L^2(R) [...]

interesting operators (Lie algebra elements) are well-defined. (This is the "small"

space in a Gelfand triple.) Then we can take its dual and antidual spaces (the large

space(s) in which Dirac's bras and kets live as distributions) and (if the small

space is a nuclear space) it can be shown rigorously that operations on the small

space can be extended to operations on the large space (of distributions) in a

natural way by virtue of the dual-pairing between the spaces.

That's (one of) the main points of distribution theory -- that one can extend

well-defined stuff on the small space naturally to similar stuff on the large

space (provided one remembers we're dealing with distributions in the dual

spaces).

- #24

unusualname

- 664

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What Vanadium50 and Clem said are the physical reasons. I'll add that mathematically, the fact that the function satisfies the Schrödinger equation means that it's differentiable, and differentiable functions are continuous.

There's a related question that I still don't know the answer to: Is there amathematicalreason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?

(There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).

The rigorous proof of this is quite delicate, eg see Chapter 2 of Berezin & Schubin - 'The Schrödinger Equation' (Kluwer AP 1991).

Essentially, for an

The proof involves careful analytic bounds and is rather formal, the details don't give any extra physical insight into the Schrödinger Eqn (as far as I can see).

(I recall tom.stoer asked about this a few months back and I don't think there was a simple (rigorous) solution)

- #25

DrDu

Science Advisor

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In the easiest case H=-1/2m (p-A)^2.

This can be made precise in the sense of a self adjoint extension of the Hamiltonian. If one wants to insist in representing the delta function as the limit of continuous potentials, one has to be carefull with terms which would lead to the ill defined square of a delta function.

The situation is simpler in relativistic QM where H is linear in p.

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