Superconductivity at room temperature??

[ben m]

The fact remains that there cannot be an actual differentiable scalar field whose gradient can be integrated around a loop to obtain a non-zero result. An angle around a symmetry point is NOT a scalar field, and can thus not be treated as one mathematically.

Sorry, Johan, you're flailing.

The "scalar field" y, for which I wrote the equation S(theta) = 4*theta, is a scalar field in the same way that f(x) = 3x + 4 is a scalar field. It's a quantity that has different values at different coordinates. It's not by itself "an angle around a symmetry point".

You know what? It sounds like you've forgotten the difference between the phase f (or S, or y, or φ, or whatever) that's represented by the function---the quantity we're going to take spatial derivatives of---and the coordinates (R,theta), or (x,y) that we use to label locations in space. This is a very common thing for a mediocre high-school student to do, or a student racing through a final exam without thinking very hard. It's not a mistake you expect from a healthy and clear-thinking adult, which is why I ask if you are quite well.

Try this, Johan: go find a *human being* that you know, in real life, and ask them to sit down and explain this to you. On paper.

 

[Ziggurat]

So you claim that these mathematical components are actual solutions?

In the case of the hydrogen atom, YES! Those are the standard solutions. In fact, those complex solutions are generally preferred to the non-complex solutions, because the former are angular momentum eigenstates but the latter are not.

Let us make it simpler: Do you claim that there are actually two infintely long running waves on a violin string which superpose to form a standing wave. That is of course BS.

It's exactly equivalent to the superposition of two traveling waves. If the end points are fixed then only that particular linear combination satisfies the boundary conditions, making the traveling wave basis set a bad set to work with (ie, you have sin(x) solutions but not cos(x) solutions), but in the case of circular motion, the boundary conditions admit those traveling wave solutions, obviously.

This is REALLY basic physics here, johan. Hell, it's not even specifically quantum mechanics. You get the same thing with classical waves. Why are you getting it so badly wrong?

 

[sol invictus]

As usual you are trying to be deceptive. Where have you moved out of the Argand Plane into a plane in real space? I suppose that you think that this statement did:

2) Do you agree that if we write ψ=|ψ|exp(iS), we obtain |ψ|=r, S=φ, where r and φ are the standard polar coordinates in the x-y plane? Yes or no?

I do not agree that they are the standard polar coordinates in the real x-y plane within which the superconducting ring finds itself. How can they be when the expression for the wave-function contains an imaginary number i!

What in the world are you talking about, Johan? Have you changed your mind for that sentence and decided that the QM wavefunction ψ must be real?

QM wavefunctions are complex functions of the real spacetime variables x,y,z, and t. So we have ψ(x,y,z,t), and ψ has a real part and an imaginary part.

Would it make you happier if I re-wrote my function as

Re(ψ)=x, Im(ψ)=y? That's identical to what I wrote above, of course.

The wave function itself is complex

Yes, it's a complex function of x,y,z,t - just as in the example I gave you.

Now, will you answer those four questions, or will you continue to wriggle and squirm away from doing so?

 

[sol invictus]

You know what? It sounds like you've forgotten the difference between the phase f (or S, or y, or φ, or whatever) that's represented by the function---the quantity we're going to take spatial derivatives of---and the coordinates (R,theta), or (x,y) that we use to label locations in space.

Yes, you're right, that is (part) of his confusion. It's made even clearer in the post I just responded to.

 

[johanfprins]

What in the world are you talking about, Johan? Have you changed your mind for that sentence and decided that the QM wavefunction ψ must be real?

How long must I still try and teach you what wave amplitudes are and what the difference between a running wave and a stationary wave is? Really, I would like to know at which Universities you have studied so that I can advise my students to avoid those institutions.

Whether the wave has a real wave amplitude (like a light wave) or a complex amplitude (lke a matter wave) you can write for both at any point within the wave intensity that the wave amplitude PSI is:

PSI =(modPSI)*exp(itheta(x.t))=(modPSI)*exp[i(S-(omega)t)]

Firstly one should remember that the phase angle ALWAYS changes with time, and only changes with position in a certain manner which dependes on whether the wave is a running wave or a standing wave.

It is also important to note that this expression is true at of EACH and EVERY point within the wave-intensity, not just at a selected point around a circular symmetry axis. It is also valid when there is no polar or spherical symmetry involved at all. This means that at EVERY point in the wave's intensity-field, no matter what shape or symmetry it has, this equation defines an Argand-plane within which (modPSI) is a radius-vector that is rotating around the origin of the Argand plane with an angular frequency (omega): Thus you CANNOT generally equate S with a polar angle around an axis that defines circular symmetry at all.

1. When PSI is a real amplitude (as for a light-wave) we can only observe the projection of the radius-vector (modPSI), at a point within the intensity-field of the wave, as the projection of (modPSI) along the real axis of the Argand plane. This allows one to solve the wave equation by starting of with a complex wave amplitude, as long as you remember that at the end of you calculation you must take the real part of the complex solution as the actual solution

Now as far as the position-dependent part S of the phase-angle is concerned: For a harmonic wave with a frequency (omega) one has only that S is continuous and therefore differentiable when you have a running wave. In this case S always changes with position perpendicularly to the moving wave fronts: For example, if the wave-fronts are flat and moving along a direction x, one has that S=kx where k is the wave-vector in reciprocal space. When the wave fronts are circular or spherical one has that S=kr. S only has a gradient perpendicular to the moving wave-fronts which always move in such a manner that they are parallel to each other: And only by integrating along such a perpendicular linear direction can you end up with 2n(PI) for the angular dependence. Thus S cannot change with position along a wave-front at all.

Furthermore, you cannot have wave fronts following each other along a circular path since they will not be parallel, and therefore you cannot write that S=ks, where s is measured perpendicularly to the wave fronts along the circumference. Thus to assume that you can integrate the gradient of S along a circular path and obtain 2n(Pi) is pure physics nonsense and it also violates the foundations of vector calculus.

When it is a stationary wave, the function S does not change between the nodes of this wave but flips discontinuously through PI at every node. Thus in this case you cannot differentiate this function at all.

2. When the wave amplitude is a complex function (like an electron wave) there is still an Argand plane at every point within the intensity of the wave within which (modPSI) forms a radius vector: BUT in this case we observe both projections along the real-axis and the imaginary axis simultaneously. Thus, although in this case you have the same phase angle changing in the same way with position, you, however, find that when you solve the wave equation as if it is a "free" wave which can move linearly through space, the intensity obtained within our three-dimensional space does NOT have crests and troughs which pushes the wave to move along the direction in which it is supposed to move. In fact, the solution that is obtained mandates that there is no change in intensity with position anywhere within our three-dimensional space: So the solution IS NOT that of a running wave; but an infinitely long stationary wave. The fact is that the Schroedinger wave-equation give solutions which can never be running waves at all.

I believe that the correct solution for an electron within its inertial refrence frame is a radially symmetric Gaussian wave around the electron's centre-of-mass. Although the (modPSI) radius vector changes with position, the phase angle at every point is exactly in tune: All the (modPSI) radius vectors move around their origins in phase: Like one-armed clocks with different lengths for their arms.

But we know that a moving electron does act like a running wave with crests and troughs. So where does the de Broglie wavelength then come? It comes from Einstein's special theory of relativity: It obviously MUST be determined by relativity since the momentum and thus de Broglie wavelength for a moving electron has different values within inertial refrence frames that moves relative to each other.

Now consider a stationary Gaussian wave within the electron's inertial refrence frame, and consider an observer doing measurements on this electron within another reference frame moving with a speed v relative to the stationary electron. Now when you Lorentz-transform the stationary Gaussian wave into the observer's moving reference frame, you find that the wave LENGTHENS (correct! there is no length contraction) and that the time along this length is not the same at every point: it varies linearly from past to future within the wave. Thus the radius vectors within the Argand planes at these different points are now not rotating in phase anymore: Crests and troughs develop that allows the electron to be diffracted.

Thus if you solve the Schroedinger wave equation assuming that you can have a macro-running wave with crests and troughs running around a ring, you are assuming a solution which is not possible at all. It is thus utter nonsense to maintain that you can derive the change in positional phase-angle by taking a circular integral within the gradient od S. The gradient of S can ONLY change perpendicularly to wave-fronts which are completely parallel to each other.

 

[johanfprins]

In the case of the hydrogen atom, YES! Those are the standard solutions. In fact, those complex solutions are generally preferred to the non-complex solutions, because the former are angular momentum eigenstates but the latter are not.

But they are not solutions which can actually physically manifest. Obviously, they can be useful to superpose mathematically in order to obtain a solution that is physically possible. But so is any function-space you can dream out. This does not imply that the functions spanning a function-space are physically real.

It's exactly equivalent to the superposition of two traveling waves.

I have not questioned this at all. The fact however remains that they are not solutions one can physically observe and therefore, to paraphrase Roger Penrose, they live in Plato's mathematical universe, not within our universe.

If the end points are fixed then only that particular linear combination satisfies the boundary conditions, making the traveling wave basis set a bad set to work with (ie, you have sin(x) solutions but not cos(x) solutions),

Correct, the real solutions are only obtained when one correctly identifies the relevant boundary conditions and then correctly apply them.

but in the case of circular motion, the boundary conditions admit those traveling wave solutions, obviously.

NO it is not the case! The boundary conditions do not allow such a circular running wave at all. They are not physically possible, since it is assumed that wave fronts can move around a circular path while staying parallel to each other. This is nonsense!

This is REALLY basic physics here, johan. Hell, it's not even specifically quantum mechanics. You get the same thing with classical waves. Why are you getting it so badly wrong?

I agree: Just a pity that you are incapable of understanding basic physics. Please demonstrate to me a classical wave with wave fronts following each other around a circle while still staying parallel to each other. Hell, man, this is just simple common sense that it is not possible!

 

[Ziggurat]

But they are not solutions which can actually physically manifest.

As I said before, you are objecting to standard quantum mechanics.

Obviously, they can be useful to superpose mathematically in order to obtain a solution that is physically possible. But so is any function-space you can dream out. This does not imply that the functions spanning a function-space are physically real.

In the case of the eigenfunctions of z-axis angular momentum, they are very real. They are physically measurable. One can prepare atoms with electrons in exactly such a state and observe them.

NO it is not the case! The boundary conditions do not allow such a circular running wave at all. They are not physically possible, since it is assumed that wave fronts can move around a circular path while staying parallel to each other. This is nonsense!

First, that's not the boundary condition, it's the STANDARD solution which you yourself claimed one obtains from separation of variables with or without satisfying the boundary condition. The boundary condition merely quantizes the angular momentum, it does not require that it's zero.

You are objecting to standard quantum mechanics. Go ahead and do that if you want, but there's not much point in discussing superconductivity if you don't accept standard quantum mechanics to begin with.

I agree: Just a pity that you are incapable of understanding basic physics.

Please demonstrate to me a classical wave with wave fronts following each other around a circle while still staying parallel to each other. Hell, man, this is just simple common sense that it is not possible!

Then perhaps your common sense needs a tuneup. What do you THINK the vibrational modes of a solid ring look like? You think they are only standing waves? No. That's only one basis set. You can get traveling waves moving around the ring in a circle. And they won't dephase with themselves either. THAT would be nonsensical.

 

[W.D.Clinger]

In the case of the hydrogen atom, YES! Those are the standard solutions. In fact, those complex solutions are generally preferred to the non-complex solutions, because the former are angular momentum eigenstates but the latter are not.

But they are not solutions which can actually physically manifest. Obviously, they can be useful to superpose mathematically in order to obtain a solution that is physically possible.

Wait a second. You're telling us that eigenstates of the angular momentum operator "are not solutions which can actually physically manifest", but non-eigenstates are "physically possible". Did you really mean to say that?

This is REALLY basic physics here, johan. Hell, it's not even specifically quantum mechanics. You get the same thing with classical waves. Why are you getting it so badly wrong?

I agree: Just a pity that you are incapable of understanding basic physics. Please demonstrate to me a classical wave with wave fronts following each other around a circle while still staying parallel to each other. Hell, man, this is just simple common sense that it is not possible!

Since you didn't like my r sin θ example from four months ago, I'll try a more complicated example.

Let
g(r) = r when r <= 1
g(r) = 2 - r when 1 <= r <= 2
g(r) = 0 when r >= 2

Let
f(r,θ,t) = g(r) sin (2θ+t)

The example is f(r,θ,t).

 

[sol invictus]

How long must I still try and teach you what wave amplitudes are and what the difference between a running wave and a stationary wave is? Really, I would like to know at which Universities you have studied so that I can advise my students to avoid those institutions.

Whether the wave has a real wave amplitude (like a light wave) or a complex amplitude (lke a matter wave) you can write for both at any point within the wave intensity that the wave amplitude PSI is:

PSI =(modPSI)*exp(itheta(x.t))=(modPSI)*exp[i(S-(omega)t)]

Firstly one should remember that the phase angle ALWAYS changes with time, and only changes with position in a certain manner which dependes on whether the wave is a running wave or a standing wave.

It is also important to note that this expression is true at of EACH and EVERY point within the wave-intensity, not just at a selected point around a circular symmetry axis. It is also valid when there is no polar or spherical symmetry involved at all. This means that at EVERY point in the wave's intensity-field, no matter what shape or symmetry it has, this equation defines an Argand-plane within which (modPSI) is a radius-vector that is rotating around the origin of the Argand plane with an angular frequency (omega): Thus you CANNOT generally equate S with a polar angle around an axis that defines circular symmetry at all.

1. When PSI is a real amplitude (as for a light-wave) we can only observe the projection of the radius-vector (modPSI), at a point within the intensity-field of the wave, as the projection of (modPSI) along the real axis of the Argand plane. This allows one to solve the wave equation by starting of with a complex wave amplitude, as long as you remember that at the end of you calculation you must take the real part of the complex solution as the actual solution

Now as far as the position-dependent part S of the phase-angle is concerned: For a harmonic wave with a frequency (omega) one has only that S is continuous and therefore differentiable when you have a running wave. In this case S always changes with position perpendicularly to the moving wave fronts: For example, if the wave-fronts are flat and moving along a direction x, one has that S=kx where k is the wave-vector in reciprocal space. When the wave fronts are circular or spherical one has that S=kr. S only has a gradient perpendicular to the moving wave-fronts which always move in such a manner that they are parallel to each other: And only by integrating along such a perpendicular linear direction can you end up with 2n(PI) for the angular dependence. Thus S cannot change with position along a wave-front at all.

Furthermore, you cannot have wave fronts following each other along a circular path since they will not be parallel, and therefore you cannot write that S=ks, where s is measured perpendicularly to the wave fronts along the circumference. Thus to assume that you can integrate the gradient of S along a circular path and obtain 2n(Pi) is pure physics nonsense and it also violates the foundations of vector calculus.

When it is a stationary wave, the function S does not change between the nodes of this wave but flips discontinuously through PI at every node. Thus in this case you cannot differentiate this function at all.

2. When the wave amplitude is a complex function (like an electron wave) there is still an Argand plane at every point within the intensity of the wave within which (modPSI) forms a radius vector: BUT in this case we observe both projections along the real-axis and the imaginary axis simultaneously. Thus, although in this case you have the same phase angle changing in the same way with position, you, however, find that when you solve the wave equation as if it is a "free" wave which can move linearly through space, the intensity obtained within our three-dimensional space does NOT have crests and troughs which pushes the wave to move along the direction in which it is supposed to move. In fact, the solution that is obtained mandates that there is no change in intensity with position anywhere within our three-dimensional space: So the solution IS NOT that of a running wave; but an infinitely long stationary wave. The fact is that the Schroedinger wave-equation give solutions which can never be running waves at all.

I believe that the correct solution for an electron within its inertial refrence frame is a radially symmetric Gaussian wave around the electron's centre-of-mass. Although the (modPSI) radius vector changes with position, the phase angle at every point is exactly in tune: All the (modPSI) radius vectors move around their origins in phase: Like one-armed clocks with different lengths for their arms.

But we know that a moving electron does act like a running wave with crests and troughs. So where does the de Broglie wavelength then come? It comes from Einstein's special theory of relativity: It obviously MUST be determined by relativity since the momentum and thus de Broglie wavelength for a moving electron has different values within inertial refrence frames that moves relative to each other.

Now consider a stationary Gaussian wave within the electron's inertial refrence frame, and consider an observer doing measurements on this electron within another reference frame moving with a speed v relative to the stationary electron. Now when you Lorentz-transform the stationary Gaussian wave into the observer's moving reference frame, you find that the wave LENGTHENS (correct! there is no length contraction) and that the time along this length is not the same at every point: it varies linearly from past to future within the wave. Thus the radius vectors within the Argand planes at these different points are now not rotating in phase anymore: Crests and troughs develop that allows the electron to be diffracted.

Thus if you solve the Schroedinger wave equation assuming that you can have a macro-running wave with crests and troughs running around a ring, you are assuming a solution which is not possible at all. It is thus utter nonsense to maintain that you can derive the change in positional phase-angle by taking a circular integral within the gradient od S. The gradient of S can ONLY change perpendicularly to wave-fronts which are completely parallel to each other.

A huge wall of text, and you still have not answered even one of the four simple, yes/no questions I asked you. Just answer them, Johan, stop posting huge amounts of nonsense that I do not have the time to respond to point by point. And by the way, if you insist on including time dependence, go ahead an multiply the ψ below by a time-dependent phase. It makes no difference to this discussion.

Here they are again:

1) Is ψ = x+iy an acceptable complex scalar field by your criteria (x and y are Caretesian coordinates in the plane of the loop)? Yes or no?

2) Do you agree that if we write ψ=|ψ|exp(iS), we obtain |ψ|=r, S=φ, where r and φ are the standard polar coordinates in the x-y plane? Yes or no?

3) Do you agree that the gradient of S is 1/r, pointing in the φ direction? Yes or no?

4) Do you agree that the integral around a circle of radius r centered at x=y=0 of the gradient of S is 2 pi r/r = 2 pi? Yes or no?

 

[ben m]

NO it is not the case! The boundary conditions do not allow such a circular running wave at all. They are not physically possible, since it is assumed that wave fronts can move around a circular path while staying parallel to each other. This is nonsense!

I refute it thus:

(I am swirling my cup of tea)

 

[johanfprins]

A huge wall of text, and you still have not answered even one of the four simple, yes/no questions I asked you.

Because your yes-know questions are based ignorance.

Just answer them, Johan, stop posting huge amounts of nonsense that I do not have the time to respond to point by point.

It is NOT "huge amounts of nonsense": If you think this, you are even more ignorant than I thought in my wildest dreams! But seeing that you think that your questions have relevance, let me patronize you: If it could be of any help! Although I doubt that you have the intellectual ability!

1) Is ψ = x+iy an acceptable complex scalar field by your criteria (x and y are Caretesian coordinates in the plane of the loop)? Yes or no?

They are not in this case Cartesian coordinates within real space: They are two components of a complex function on an Argand plane.

2) Do you agree that if we write ψ=|ψ|exp(iS), we obtain |ψ|=r, S=φ, where r and φ are the standard polar coordinates in the x-y plane? Yes or no?

As I have already said time and again, I do not agree with this: If you are able to understand waves and read my "huge amounts of nonsense" and have an IQ above 100, you should be able to understand in detail why it is not correct.

3) Do you agree that the gradient of S is 1/r, pointing in the φ direction? Yes or no?

No I do not. There is no gradient of S in the Argand plane: There is an Argand plane at each point within the intensity of a wave, and within each separate Argand plane, S is a constant, and a constant does not have a gradient. You can only have a gradient of S in real space and in the case of radial symmetry S changes along the radial direction when you have a circular running wave. S can never have a circular gradient since one cannot have consecutive parallel wave fronts following a circular path (see my "huge amounts of nonsense" above).

4) Do you agree that the integral around a circle of radius r centered at x=y=0 of the gradient of S is 2 pi r/r = 2 pi? Yes or no?

No I do not, since S cannot have a circular gradient; and this so since a set of parallel wave fronts cannot move around a circle (see my "huge amounts of nonsense" above).

If you cannot understand that S can only have a linear gradient perpendicular to moving wave fronts (along the direction in which they are moving), and that these wave fronts must always be parallel to each other, and you have ever been to a class in physics, you should consider suing your teachers.

 

[johanfprins]

I refute it thus:

(I am swirling my cup of tea)

Really, and when you do this you see crests and troughs along the circumference (like wheel spokes turning around the centre of the cup) or do you see crest and troughs along the radial direction (as rings around the centre-point of your cup)? If you are able to do the first and create spokes, there awaits a lucrative career for you as an illusionist.

 

[sol invictus]

They are not in this case Cartesian coordinates within real space: They are two components of a complex function on an Argand plane.

I guess we need to back up even more, then.

(-3) Do you agree that the QM wavefunction ψ is a complex function of the real variables x,y,z,t, where (x,y,z) are Cartesian coordinates on space, and t is time - so ψ=ψ(x,y,z,t)?

(-2) Do you agree that the QM wavefunction ψ has a real part and an imaginary part, ψ=Re(ψ)+iIm(ψ)?

(-1) Do you agree that given (-2) and (-1), Re(ψ) and Im(ψ) are themselves functions of x,y,z, and t?

(0) Do you agree that Re(ψ)=x and Im(ψ)=y solves the Schrodinger equation for a free particle?

 

[ben m]

Really, and when you do this you see crests and troughs along the circumference (like wheel spokes turning around the centre of the cup) or do you see crest and troughs along the radial direction (as rings around the centre-point of your cup)? If you are able to do the first and create spokes, there awaits a lucrative career for you as an illusionist.

A) pick up cup
B) move hand rapidly in a circle in the plane of the floor at about 1Hz.

The depth of the tea around the circumference varies, approximately as sin(theta + w.t), with w in the ballpark of 5/sec.

 

[johanfprins]

As I said before, you are objecting to standard quantum mechanics. In the case of the eigenfunctions of z-axis angular momentum, they are very real. They are physically measurable. One can prepare atoms with electrons in exactly such a state and observe them.

I suppose you are talking about Rydberg atoms. They are not solutions of Schroedinger's equation since Schroedinger's equation cannot model running waves. They are localised electron waves with energy larger than an electron's mass energy, and are therefore able to circle the nucleus just like a planet circles the sun; but they are not able form stable orbits. One has to pump them with energy to keep them going.

First, that's not the boundary condition, it's the STANDARD solution which you yourself claimed one obtains from separation of variables with or without satisfying the boundary condition.

Not for a running wave. There is no standard solution of the Schroedinger equation that gives you a running wave circling the nuicleus.

The boundary condition merely quantizes the angular momentum, it does not require that it's zero.

Are you now talking of stationary Schroedinger waves or Bohr's obsolete and physically impossible model?

You are objecting to standard quantum mechanics. Go ahead and do that if you want, but there's not much point in discussing superconductivity if you don't accept standard quantum mechanics to begin with.

There is not much use to argue with a person who believes that Schroedinger's equation can give a solution that is a coherent wave moving along a ring; or moving along any direction for that matter.

Then perhaps your common sense needs a tuneup. What do you THINK the vibrational modes of a solid ring look like? You think they are only standing waves? No. That's only one basis set. You can get traveling waves moving around the ring in a circle.

These waves are not coherent since the ring now acts as a waveguide. In the case of the superconducting phase you claim that the wave is coherent: i.e. that it has parallel wave fronts moving around the ring: This is not possible for any harmonic wave and it cannot be derived from Schroedinger's equation since Schroedinger's equation cannot be solved to obtain any coherently moving wave fronts in ANY direction: Not even a linear direction.

 

[ben m]

This is not possible for any harmonic wave and it cannot be derived from Schroedinger's equation since Schroedinger's equation cannot be solved to obtain any coherently moving wave fronts in ANY direction: Not even a linear direction.

That's your problem right there.

 

[johanfprins]

A) pick up cup
B) move hand rapidly in a circle in the plane of the floor at about 1Hz.

The depth of the tea around the circumference varies, approximately as sin(theta + w.t), with w in the ballpark of 5/sec.

My you are good to circle a cup at a rate of 1 Hz. I have tried it but could not achieve it. But if you can do it would if you would be kind enough to write down a differential wave equation for this movement and prove that it is a harmonic wave that is generated in this manner! I am wiating with bated breath. Please enlighten me.

 

[johanfprins]

That's your problem right there.

Thern please enlighten me further. Please solve Schroedinger's equation and demonstrate a solution for which you have crests and troughs moving along any direction.

 

[ben m]

My you are good to circle a cup at a rate of 1 Hz. I have tried it but could not achieve it. But if you can do it would if you would be kind enough to write down a differential wave equation for this movement and prove that it is a harmonic wave that is generated in this manner! I am wiating with bated breath. Please enlighten me.

What happened in your teacup? Did the water sit still and flat?

(Remember, you are moving the cup, not spinning it around its axis)

 

[sol invictus]

Thern please enlighten me further. Please solve Schroedinger's equation and demonstrate a solution for which you have crests and troughs moving along any direction.

You can't possibly be serious.... oh wait, it's Johan.

Any solution with <p>=/=0 has that property, which of course is all but a set of measure zero of the solutions. There are infinitely many examples.

Here's a website that shows some solutions to the simple harmonic oscillator

http://eve.physics.ox.ac.uk/Personal/artur/Keble/Quanta/Applets/quantum/qhomain.html

Of course in that case <p> is a function of time, since there is a potential. But it clearly shows peaks and troughs moving, unless you choose special initial conditions.