# How do I show a single state trying to cross a potential well has less frequency oscillation instead of more?

#### new-branch

This could be a branch of QM called "Physicality"

Holographic Mass, frequency oscillation, and when forces intervene are all major suspects in this investigation.

Holographic Mass: quantum wave mass in the form of information

<a|a> =1
post operations wavefunction
a*c0 <0|0>
c0 = 1, c1 = 0
<0|0>=1
the single state can be fed into potential well

That single state is the same as an observed particle, decoherence. And with it comes the inability to tunnel. Does it naturally have a frequency of ground state? You need energy to tunnel. Without frequency oscillation from a wave ..it isn't going to tunnel.

The kinetic energy is open to forces (because it is a single state - observed) and the mass is physical, so the potential well acts on it. Kinetic energy is constant if it was a wave. Unobserved quantum waves are immune to forces.

Is this frequency oscillation I'm asking about the same thing as "Quantum Harmonic Oscillation"? And that zero point vibration, uncertainty, isn't enough to allow a classical particle (normalized single state) to tunnel?

Symmetry is related to forces, so is there a connection to a difference between coherence and decoherence?

Could the strings in string theory be unobservable quantum waves? Holographic Mass?

#### topsquark

Math Team
The only states that cannot tunnel are those that involve an infinite potential barrier.

Could the strings in string theory be unobservable quantum waves?
Not in the way you are thinking. Strings cover all particle types. What you are calling "unobservable quantum waves" do not exist.

-Dan

#### new-branch

Non-zero probability is a wave only activity. A particle in duality is not going to tunnel because the quantum field only has the ability to make it ageless at that point.

In more detail, the calculations show that if atoms are treated as classical particles, that is, as simple points in space, many distortions of the structure tend to lower the energy of the system.

My goal is an equation that says observed particles do not tunnel. No, you only assume a physical particle has tunneled. Observing a particle after it has ended its journey is not causing it to decohere in flight. It wasn't physical in flight. Nature didn't consider it observation.

#### topsquark

Math Team
Non-zero probability is a wave only activity. A particle in duality is not going to tunnel because the quantum field only has the ability to make it ageless at that point.

My goal is an equation that says observed particles do not tunnel. No, you only assume a physical particle has tunneled. Observing a particle after it has ended its journey is not causing it to decohere in flight. It wasn't physical in flight. Nature didn't consider it observation.
Sorry, but this is pretty much all wrong. The only thing worth salvaging from this is that you cannot actually see a wave tunnel through a barrier during the tunneling. But your usage of the properties of real versus virtual particles is so messed up I don't think there's any way to explain this to you without tutoring you through basic QM.

-Dan

#### new-branch

Thanks for such valued input

#### topsquark

Math Team
All right. Let me give it a go. Here's the basic rundown.

Consider a particle moving in the +x direction encountering a finite potential barrier $$\displaystyle V = \begin{cases} 0 & \text{ for } x < 0 \\ V_0 & \text{ for } 0 < x < L \\ 0 & \text{ for } L < x \end{cases}$$

Let the incoming wave have energy E. If $$\displaystyle V_0 < E$$ then we have three domains where the wavefunction is given by
Region I: $$\displaystyle x < 0$$
$$\displaystyle \psi (x) = A e^{i \alpha x} + B e^{-i \alpha x }$$ where $$\displaystyle \alpha = \sqrt{ \dfrac{2m E}{ \hbar ^2} }$$
which implies an incident wave moving in the +x direction and a reflected wave moving in the -x direction.

Region II: $$\displaystyle 0 < x < L$$
$$\displaystyle \psi (x) = C e^{i \beta x} + D e^{-i \beta x}$$ where $$\displaystyle \beta = \sqrt{ \dfrac{2m (E - V_0 )}{ \hbar ^2} }$$
which implies a wave moving in the +x direction and a reflected wave moving in the -x direction. The far side of the barrier produces a reflected wave as well, meaning the wave "bounces around" over the barrier.

Region III: $$\displaystyle L < x$$
$$\displaystyle \psi (x) = F e^{i \alpha x}$$ where $$\displaystyle \alpha = \sqrt{ \dfrac{2m E}{ \hbar ^2} }$$
This is a transmitted wave only so we don't have the second term.

We can find the coefficients by matching boundary values for $$\displaystyle \psi (x)$$ and $$\displaystyle \psi ' (x)$$ at the boundaries. There is nothing special here and the calcuations are tedious, but doable.

Now, $$\displaystyle E < V_0$$. Classically there's nothing to say. The wave can't penetrate the barrier so there is only the incident and reflected wave in Region I. The wave does not exist in Regions II and III.

However, in the Quantum case, let's look a little closer. For $$\displaystyle V_0 < E$$ note that the wavefunction in Region II is given by
$$\displaystyle \psi (x) = C e^{i \beta x} + D e^{-i \beta x}$$ where $$\displaystyle \beta = \sqrt{ \dfrac{2m (E - V_0 )}{ \hbar ^2} }$$

If we take this solution and put in $$\displaystyle E < V_0$$ then $$\displaystyle \beta$$ is an imaginary number, leading to a wavefunction
$$\displaystyle \psi (x) = C e^{ \gamma x} + D e^{ - \gamma x}$$ where $$\displaystyle \gamma = \sqrt{ \dfrac{2m (V_0 - E )}{ \hbar ^2} }$$

Quantum mechanically we can have a wave inside the barrier but it is exponentially suppressed. (It's a possible solution to the Schrodinger wave equation so we have to take it seriously.) This also means that we can again have an appreciable (ie measureable) transmitted wave in Region III, provided the L is short enough. And again we can match boundary conditions and find the coefficients.

That's it. There's nothing else to say. The only catch is that we can't observe the wave while it's in the barrier. But theory turns out to match experiment so we are done.

I know you have your own ideas about observed and unobserved waves and you probably won't agree that the problem here is solved but there's nothing in the Science nor Nature that requires anything else to be involved.

-Dan

• new-branch

#### new-branch

Now, $$\displaystyle E < V_0$$. Classically there's nothing to say. The wave can't penetrate the barrier so there is only the incident and reflected wave in Region I. The wave does not exist in Regions II and III.
This sounds like you are agreeing with me, I must not be reading it right. I want the math of a decohered "wave" attempting to tunnel. You get that by letting the wave functions cancel out leaving you with a single state.

The holographic mass of a wave (not a particle in duality) ..is not physical. It doesn't have to answer to forces like a particle in duality.

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#### topsquark

Math Team
This sounds like you are agreeing with me, I must not be reading it right. I want the math of a decohered "wave" attempting to tunnel. You get that by letting the wave functions cancel out leaving you with a single state.
A coherent wave will simply be a sum (or integral probably) with the same properties so coherent or not coherent doesn't matter.

The holographic mass of a wave (not a particle in duality) ..is not physical. It doesn't have to answer to forces like a particle in duality.
Did you see any mention of forces or mass (holographic or otherwise) in the work that I posted? Why did you bring it up?

This is exactly why I feel you need the tutoring. The problem you are talking about is basic. They teach it in undergrad Intro QM. Stop trying to make it fancy to fit your ideas.

-Dan

#### new-branch

2m in your equations isn't mass?

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