What Is the Expected Value of Kinetic Energy for This Harmonic Oscillator?

In summary, the homework statement is to find the expected value of the kinetic energy of a harmonic oscillator. I tried to solve it using two diffrent methods and they don't give the same result, so something is wrong with one or both of them.
  • #1
castlemaster
38
0

Homework Statement



Calculate the expected value of the kinetic energy being

[tex]\varphi(x,0)=\frac{1}{\sqrt{3}}\Phi_0+\frac{1}{\sqrt{3}}\Phi_2-\frac{1}{\sqrt{3}}\Phi_3[/tex]

Homework Equations



[tex]K=\frac{P^2}{2m}[/tex]

The Attempt at a Solution



I tried to solve it using two diffrent methods and they don't give the same result, so something is wrong with one or both of them.
The first method is using matrix representations. I use the matrix of P and generate P^2.
Then I compute

[tex]\varphi(x,0)^*,P^2,\varphi(x,0)[/tex]

and end up with an expresion like

[tex]-3+2/3*\sqrt{2}cos (2wt)[/tex]

The second method is using the creation and annhilitation operators a+ and a.

[tex]P^2=a^{+}a^{+}+aa-a^{+}a-aa^{+}[/tex]

[tex]\phi(x)_na^{+}a^{+}\phi(x)_m=\sqrt{(n+1)(n+2)}\phi(x)_{n,m+2}[/tex]
[tex]\phi(x)_naa\phi(x)_m=\sqrt{(n-1)(n)}\phi(x)_{n,m-2}[/tex]
[tex]\phi(x)_na^{+}a\phi(x)_m=n\phi(x)_{n,m}[/tex]
[tex]\phi(x)_naa^{+}\phi(x)_m=(n+1)\phi(x)_{n,m}[/tex]

But here I only get values for

[tex]\phi(x)_0,\phi(x)_0;\phi(x)_2,\phi(x)_2;\phi(x)_3,\phi(x)_3[/tex]
those are giving a number and
[tex]\phi(x)_2,\phi(x)_0[/tex]
which is giving an expression like a*exp(2jwt)

The question is: where is the other exp(-2jwt) to form the cos(2wt) of the other solution and why I don't get the same result ... I guess I have to normalise somewhere.
 
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  • #2
How do you get a time dependence in your answer? You want to calculate the kinetic energy of a harmonic oscillator so using the creation and annihilation operator is a smart thing to do.
Ignoring all constants, you want to calculate [itex]\langle \varphi |p^2|\varphi \rangle[/itex]. You know that, assuming [itex]\phi_n[/itex] is normalized, that [itex]\langle \phi_n |\phi_m\rangle=\delta_{nm}[/itex]. Therefore every term will just be a number.

castlemaster said:
[tex]\phi(x)_2,\phi(x)_0[/tex]
which is giving an expression like a*exp(2jwt)

How did you get this?
 
  • #3
You are right, the energy should be constant in time.

Then

[tex]
\phi(x)_0a^{+}a^{+}\phi(x)_2=0
[/tex]

because the functions of the basis are orthonormal?
 
  • #4
Yes that is correct. On top of that [itex]a^{+}a^{+}\phi(x)_2[/itex] would yield [itex]const*\phi(x)_4[/itex], which isn't even an allowed state in this problem. Same for [itex]a \phi_0[/itex] etc.
 
  • #5
yes, i wanted to put

[tex]

\phi(x)_2a^{+}a^{+}\phi(x)_0=0

[/tex]

this gives an allowed state right?
 
  • #6
Yes, because [itex]a^{+}a^{+}\phi(x)_0=const*\phi(x)_2[/itex] and [itex]\phi(x)_2[/itex] is a part of [itex]\varphi(x,0)[/itex]. The entire thing should yield [itex]\sqrt{2}[/itex].
 

FAQ: What Is the Expected Value of Kinetic Energy for This Harmonic Oscillator?

What is a harmonic oscillator?

A harmonic oscillator is a physical system that follows a repetitive pattern of motion, typically back and forth, around a stable equilibrium point. It is often represented mathematically as a mass attached to a spring that is being stretched and compressed.

How does a harmonic oscillator behave?

A harmonic oscillator exhibits simple harmonic motion, which means that the acceleration is directly proportional to the displacement from the equilibrium point and is directed towards the equilibrium point. This results in a sinusoidal motion with a constant period and amplitude.

What is the kinetic energy of a harmonic oscillator?

The kinetic energy of a harmonic oscillator is the energy that it possesses due to its motion. It is directly related to the velocity of the oscillating mass and is given by the equation KE = 1/2mv^2, where m is the mass and v is the velocity.

How does the kinetic energy of a harmonic oscillator change over time?

The kinetic energy of a harmonic oscillator follows a sinusoidal pattern, just like its position and velocity. It is at its maximum when the displacement from the equilibrium point is zero (where the velocity is at its maximum), and it is at its minimum when the displacement is at its maximum (where the velocity is zero).

Can kinetic energy be converted into potential energy in a harmonic oscillator?

Yes, in a harmonic oscillator, kinetic energy can be converted into potential energy and vice versa as the mass oscillates back and forth. At the equilibrium point, all of the kinetic energy is converted into potential energy, and at the maximum displacement, all of the potential energy is converted back into kinetic energy.

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