Wave function deterministic yet not classical

[Blue Scallop]

We laymen and newbies are taught the Schrodinger equation was deterministic. So we tend to picture it’s like a classical thing.. some forever thinking it that way where the idea is etched deep in the mind. Yet when we are home with the idea it is deterministic (that is.. when not measured).. some scolded us and told us quantum state is not like classical state at all where there is one to one correspondence. In fact, you need to use operators to get the quantum state. For 80% of laymen. It’s too late as we thought the Schrodinger equation was deterministic and most can’t forget it and unable to get any analogy right.

To illustrate the point. Can you explain to laymen what occurs in the following.

Let’s say you have a c60 buckyball in superposition. Then you descramble its wave function (meaning make the deterministic evolution sporatic by let’s say perturbing it with a wave function disruptor).. what would happen to the buckyball.. would it become descramble too.

We laymen reason that since the wave function is deterministic. It’s like being in one to one classical correspondence with the object. So we can’t get why you need operators for example to get the momentum or spin of the particle. Using laymen explanations that only laymen can understand. Please explain what it means the schrodinger equation is deterministic yet you need operators to compute for the momentum. 95% of laymen don’t know this. Thank you.

 

[Nugatory]

Saying that the Schrodinger equation is deterministic just means that if we know what the wave function is at some given time, we can use the Schrodinger equation to calculate what the wave function will be in the future. This has nothing to do with any of the other stuff you're asking about.

 

[Blue Scallop]

Saying that the Schrodinger equation is deterministic just means that if we know what the wave function is at some given time, we can use the Schrodinger equation to calculate what the wave function will be in the future. This has nothing to do with any of the other stuff you're asking about.

Can you give an example of what it means if we know what the wave function is at some given time, we can use the Schrodinger equation to calculate what the wave function will be in the future. We laymen were not given any example so we thought they were talking about states like classical states.

 

[zonde]

I would say that wavefunction describes only one side of dynamics of physical configurations. It tracks phase relationship in complex system when it can be approximated as closed. And it describes how dynamics are restricted by these phase relationships. That's the deterministic part as I see.

 

[Nugatory]

Can you give an example of what it means if we know what the wave function is at some given time, we can use the Schrodinger equation to calculate what the wave function will be in the future. We laymen were not given any example so we thought they were talking about states like classical states.

If $\hat{H}$ is the Hamiltonian and $\psi(x,t)$ is the wave function, then when the Hamiltonian is not a function of time (which is the case for many important and interesting problems) solving Schrodinger's equation gives us:
$$\psi(x,t)=e^{\frac{-i\hat{H}t}{\hbar}}\psi(x,0)$$
That is, if we know $\psi(x,0)$, the wave function at a particular moment that we're calling time zero, then we can calculate $\psi(x,t)$, the wave function at the later time $t$.

(I doubt that there are any examples suitable for a B-level thread - this one is about the simplest there is).

 

[Nugatory]

Mentor's note: several off-topic and/or confused posts have been removed from this thread, along with some sensible replies to the off-topic posts. All posters are asked to try to focus on the original question in the thread.

 

[Blue Scallop]

If $\hat{H}$ is the Hamiltonian and $\psi(x,t)$ is the wave function, then when the Hamiltonian is not a function of time (which is the case for many important and interesting problems) solving Schrodinger's equation gives us:
$$\psi(x,t)=e^{\frac{-i\hat{H}t}{\hbar}}\psi(x,0)$$

May I know where is the phase variable in this equation? What would happen if you remove all the phases, what would happen to the wave function?
Do you end up with just a ray in Hilbert Space.. I just want to understand how a ray (or vector) in Hilbert space can be deterministic in time too. Thanks.

That is, if we know $\psi(x,0)$, the wave function at a particular moment that we're calling time zero, then we can calculate $\psi(x,t)$, the wave function at the later time $t$.

(I doubt that there are any examples suitable for a B-level thread - this one is about the simplest there is).

 

[zonde]

May I know where is the phase variable in this equation?

This is the phase term:
$$e^{\frac{-i\hat{H}t}{\hbar}}$$
Look at Euler's formula.

 

[Blue Scallop]

This is the phase term:
$$e^{\frac{-i\hat{H}t}{\hbar}}$$
Look at Euler's formula.

ok. but what situation when the phase term is zero? does the phase term only occur if the position basis is used? Without position, there is no phase, correct?

 

[zonde]

ok. but what situation when the phase term is zero?

You mean it does not change with time? I would say that you have different theory, not Quantum Mechanics.

does the phase term only occur if the position basis is used? Without position, there is no phase, correct?

No. Basis just specifies how you describe the vector but does not change the vector itself. It's like a change of coordinate system.

 

[Blue Scallop]

You mean it does not change with time? I would say that you have different theory, not Quantum Mechanics.

Ok. phase is natural in wave function because wave has phase.. but for vectors in Hilbert space.. what is the equivalent of phase? (besides the interference terms in density matrix)

No. Basis just specifies how you describe the vector but does not change the vector itself. It's like a change of coordinate system.

 

[zonde]

but for vectors in Hilbert space.. what is the equivalent of phase?

Hilbert space is complex space. I would say that you can imagine any basis vector as actually spanning two dimensions - real and imaginary. So even basis vector can be rotated between real and imaginary dimensions. I hope I didn't went wrong. If you want to be certain ask for someone more knowledgeable than me to confirm what I said (or correct me).

 

[Nugatory]

ok. but what situation when the phase term is zero?

There is no value of $t$ for which $e^\frac{-iEt}{\hbar}$ is zero, so that situation cannot arise.

for vectors in Hilbert space.. what is the equivalent of phase?

The phase is just a complex number that you've multiplied a vector in the Hilbert space by. If $A$ is a vector, and $B=\alpha{A}$ we'd say that $\alpha$ is a phase constant.

But don't be misled into thinking that in this example $A$ has a phase and $B$ doesn't. I could have just as easily written $A=\frac{1}{\alpha}B$ (safe because the phase cannot be zero, so no divide-by-zero embarrassment) and now it looks as if the phase is attached to $B$. The only sensible thing we can say is that $\alpha$ is the relative phase between $A$ and $B$.