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

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Homework Help Overview

The discussion revolves around calculating the expected value of kinetic energy for a harmonic oscillator, specifically using a given wave function expressed as a linear combination of basis states. The problem involves quantum mechanics concepts, particularly the use of creation and annihilation operators and matrix representations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss two different methods for calculating the expected value of kinetic energy, one using matrix representations and the other using creation and annihilation operators. Questions arise regarding discrepancies in results and the time dependence of the kinetic energy expression.

Discussion Status

Participants are actively exploring the problem, questioning assumptions about normalization and orthonormality of states. Some guidance has been provided regarding the implications of using the creation and annihilation operators, but no consensus has been reached on the correct approach or resolution of the discrepancies noted.

Contextual Notes

There are discussions about the normalization of states and the implications of using specific operators, as well as the requirement for the states to be allowed within the context of the problem. The original poster expresses uncertainty about the results obtained from different methods.

castlemaster
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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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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?
 
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?
 
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.
 
yes, i wanted to put

[tex] <br /> \phi(x)_2a^{+}a^{+}\phi(x)_0=0<br /> [/tex]

this gives an allowed state right?
 
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].
 

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