A classical harmonic oscillator of mass m and spring constant k is known to have a total energy

A classical harmonic oscillator of mass m and spring...

Key Concepts

Harmonic Oscillator

The harmonic oscillator is a fundamental model in classical mechanics that describes a system where the restoring force is directly proportional to displacement. Its dynamics are governed by Newton’s equations of motion with a quadratic potential energy, making it an essential concept for understanding oscillatory behavior in physical systems.

Energy Conservation

In a conservative system like the harmonic oscillator, the total mechanical energy (kinetic plus potential) remains constant over time. This principle allows one to relate the position and velocity of the oscillator since the energy at any point is partitioned into kinetic energy and potential energy, with the latter being a quadratic function of displacement.

Velocity as a Function of Position

For a harmonic oscillator, the velocity at a given position can be derived using energy conservation. The speed, being the square root of a function of the total energy minus the potential energy at that position, indicates that the oscillator slows down near the turning points (where potential energy is maximum) and speeds up as it passes through the equilibrium position.

Time to Spatial Distribution Transformation

When the starting phase or time of the oscillator is unknown, the probability of finding the oscillator at a given position is proportional to the time it spends near that position. This concept leads to the transformation from a uniform distribution in time to a non-uniform spatial probability density, achieved by considering the inverse of the velocity at each position.

Normalization of Probability Density

The probability density function must be normalized so that the total probability of finding the oscillator within its allowed range is one. This involves integrating the derived spatial probability density over the entire range of motion (from one turning point to the other) and ensuring that the integral sums to unity, which is an essential requirement for any probability distribution.

Transcript

00:01 Now in this question i'm looking at a simple harmonic oscillator, that is a particle in a simple harmonic potential.

00:10 We are given that this is the solution, to the wave function, to the problem of the symbol harmonic oscillator.

00:18 You want to find the energy of this state given that the particle has this wave function.

00:25 So to find the energy, we need to make use of the schrodinger's equation.

00:29 Negative hb square over to m times the second derivative of the wave function plus potential u times si this equals to e times si so we want to find what is our e over here now we know the potential for a harmonic oscillator u this is equals to half m omega square x square so later we can substitute this in right to our expression but first off we need to find what is the second derivative of our way function so we got to do some chain rule because there's two terms with containing x so let me do the first derivative first differentiate the first term and then we differentiate the second term.

02:19 Alright, so we simplify this second term a bit.

02:33 I'm gonna just simplify it over here.

02:36 So we bring the x terms together, bring out the negative sign in front.

02:58 So this is the expression.

03:02 Now we've got to differentiate it again.

03:33 Alright, so this is the first term and the second term.

03:41 So i'm not going to bore you with a lot of the chain rule, but i'm going to just skip all the way to the final simplified expression.

03:58 But all of this is just chain rule again and again.

04:32 So this is the second deferitive.

04:35 And i'm going to simplify this by replacing what i can find as si in these expressions to kind of take out some of the unsightly terms.

04:54 So b, x, and the exponential term will come together to form the si term.

05:08 Over here we have b, one of the x and the exponential term.

05:17 So i left with m omega over hbar square times x squared times side.

05:35 So now we compare it with putting it into the shuringus equation.

05:50 We have negative h bar square over 2 m multiplied.

05:55 Applied to all these terms.

05:57 Alright, so this is the first term after substituting it to the shrolincus equation multiplying by negative h bar square over 2m.

06:52 Then we add the potential which is half m omega square x square x x x.

07:02 Equate this to e times si.

07:07 So first thing that we can already notice is that these two terms are the same.

07:14 Since one is negative one is positive they will cancel each other out and we are left with 3 over 2 hb omega times si equals to e times si right with 1 to 1 to 1 correspondence e must be therefore it goes to 3 over 2 h bar omega now to find what is the position that you will least likely to find the particle that is we want 1 the probability density or finding the particle over there to be 0 or minimum smallest amounts and 0 is a possible value for this because we can get psi to be 0 when x is 0 so when x is 0 this will be the point where our wave function is 0 which means that the probability density is 0 and therefore it is least likely to find the particle over there.

08:32 On the other hand, what is the points where we will most likely find the particle? well, there will be when our wave function is maximized.

08:45 So to find when our wave function is max, we will have to differentiate wave function with respect to x and find a point when this is equal.

08:59 Goes to 0.

09:00 This will be when our wave function is at the maximum point...

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