What is the Minimum Energy of a One-Dimensional Linear Oscillator?

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Homework Help Overview

The discussion revolves around the minimum energy of a one-dimensional linear oscillator, focusing on the application of the uncertainty principle and the mathematical derivation involved in finding this minimum energy.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between momentum and position using the uncertainty relation, questioning how certain constants and terms arise in their equations. There is also discussion about finding the minimum of a function, with some participants suggesting the use of derivatives.

Discussion Status

The conversation is ongoing, with participants attempting to clarify their understanding of the problem and the mathematical relationships involved. Some guidance has been offered regarding the focus on the minimum energy expression, but there is no explicit consensus on the approach to take.

Contextual Notes

Participants are navigating the complexities of the problem, including the distinction between different constants (h and ħ) and the role of frequency and angular velocity in the energy expression. There is a sense of confusion regarding the derivation and simplification of the energy equation.

Abdul.119
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Homework Statement


The (classical) energy of one-dimensional linear oscillator is
2.gif

a) show, using the uncertainty relation, that the energy can be written as
gif.gif

b) Show that the minimum energy of the oscillator is
2.gif

Where
m%29.gif

Homework Equations


Δp Δx >= ħ/2
p ≈ ħ/2x

The Attempt at a Solution


I'm really stuck on this problem, I could say that the p_x = ħ^2 / (2x)^2 then substitute in the equation, but still where did the π and the 32 come from? I couldn't find any other relevant equations
Edit: Oh I just realized ħ = h/2π , then p^2 = h^2/(4πx)^2 , substituting in the equation would give the answer.
But now what about b)? I believe to find the minimum energy we set Δx >= ħ/2Δp , but it isn't clear to me how the frequency and angular velocity come in
 
Last edited:
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Are you familiar with the difference between h and ħ?
 
Orodruin said:
Are you familiar with the difference between h and ħ?
Oh yes, I just realized that a minute ago and edited my post, but now I'm still having trouble with part b)
 
Are you familiar with how to find the minimum of a function?
 
Orodruin said:
Are you familiar with how to find the minimum of a function?
Hmm I believe I have to take the derivative with respect to Δp, minimum energy dE/dΔp = 0
 
Δp is no longer part of your expression for the energy ... You have just expressed it as a function of x only.
 
Then the derivative of p_x = ħ^2 / (2x)^2 ?
 
No, it is the energy you wish to minimise.
 
use dE/dx = 0 ?
 
  • #10
Abdul.119 said:
use dE/dx = 0 ?
Why don't you try it and see what you get?
 
  • #11
Orodruin said:
Why don't you try it and see what you get?
Would get
E = -2h^2/32mx^3π^2 + 2Cx
Doesn't make sense because I still don't see where the omega and f come from
 
Last edited:
  • #12
You care far too much about the omega and f. You are just supposed to find the minimum energy, which is a certain given expression in terms of omega where omega is an expression in other variables which arethe actual variables you have.
 
  • #13
Orodruin said:
You care far too much about the omega and f. You are just supposed to find the minimum energy, which is a certain given expression in terms of omega where omega is an expression in other variables which arethe actual variables you have.
So you mean E = -2h^2/32mx^3π^2 + 2Cx can be simplified into
2.gif
? how?
 
  • #14
No, that is your energy derivative. You need to put it to zero to find the minimum.
 

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