Exploring the Derivative of y(x,t) in Quantum Mechanics

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• strawman
Now, ##u = x - vt##, so ##h' = \dots##.In summary, the correct application of the chain rule in this problem involves substituting u = x - vt and using h' instead of h in the derivative.
strawman
TL;DR Summary
Should be a simple chain rule problem I think.
y(x,t) = 1/2 h(x-vt) + 1/2 h(x+vt)
This is from the textbook "quantum mechancs" by Rae.
The derivative is given as dy/dt = -v1/2 h(x-vt) + v1/2 h(x+vt)
I'm not quite sure how this is? If I use the chain rule and set the function h(x-vt) = u
Then by dy/dt = dy/du x du/dt I will get (for the first part):

dy/du = 1/2
du/dt = -v
dy/dt = -1/2 v

I've obviously misunderstood this somewhere, being a bit rusty on my calculus. Where have I gone wrong?
Thanks!

strawman said:
Summary: Should be a simple chain rule problem I think.

y(x,t) = 1/2 h(x-vt) + 1/2 h(x+vt)
This is from the textbook "quantum mechancs" by Rae.
The derivative is given as dy/dt = -v1/2 h(x-vt) + v1/2 h(x+vt)
I'm not quite sure how this is? If I use the chain rule and set the function h(x-vt) = u
Then by dy/dt = dy/du x du/dt I will get (for the first part):

dy/du = 1/2
du/dt = -v
dy/dt = -1/2 v

I've obviously misunderstood this somewhere, being a bit rusty on my calculus. Where have I gone wrong?
Thanks!
First, in the correct answer, those should be derivatives of ##h##.

Second, your substitution ##u = h## changes nothing. You just replace ##h## by ##u##. The two functions are identical.

What you can do is let ##u = x - vt##. Then:

##y(x, t) = h(u(x, t)) + \dots ##

It's that function composition to which the chain rule applies:

##\frac{dy}{dt} = h'(u)\frac{du}{dt} + \dots##

Note that it's ##h'## there.

strawman

1. What is the derivative of y(x,t) in quantum mechanics?

The derivative of y(x,t) in quantum mechanics is a mathematical expression that represents the rate of change of a quantum mechanical system over time. It is used to describe the behavior of particles and their interactions in the quantum world.

2. Why is exploring the derivative of y(x,t) important in quantum mechanics?

Exploring the derivative of y(x,t) is important in quantum mechanics because it allows us to understand how quantum systems evolve and change over time. This is crucial for predicting and analyzing the behavior of particles and their interactions in the quantum world.

3. What are some applications of the derivative of y(x,t) in quantum mechanics?

The derivative of y(x,t) has many applications in quantum mechanics, including predicting the behavior of particles in quantum systems, analyzing quantum entanglement, and understanding quantum tunneling. It is also used in the development of quantum technologies such as quantum computing and quantum cryptography.

4. How is the derivative of y(x,t) calculated in quantum mechanics?

The derivative of y(x,t) is calculated using mathematical techniques such as the Schrödinger equation or the Heisenberg equation of motion. These equations take into account the properties and interactions of quantum particles to determine the rate of change of a quantum system over time.

5. Are there any limitations to exploring the derivative of y(x,t) in quantum mechanics?

Yes, there are limitations to exploring the derivative of y(x,t) in quantum mechanics. These include the uncertainty principle, which states that the precise values of certain properties of a particle cannot be known simultaneously, and the complexity of mathematical calculations involved in determining the derivative for complex quantum systems.

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