Average kinetic energy quantum harmonic oscillator
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OK, let's unpack this a bit. The result in Griffiths is, if we write it out more carefully, $$\langle n | \hat{V} | n \rangle = \frac{\hbar \omega}{2}(n + 1/2)$$ where $\hat{V}$ is the potential energy operator $\frac{1}{2} m \omega^2 x^2$ and $|n\rangle$ is the eigenstate of the Hamiltonian, $\hat{H} |n\rangle = E_n |n \rangle$, with $E_n = \hbar \omega(n + 1/2)$.
"Is the potential energy of the quantum harmonic oscillator always one half the oscillator's total energy?"
No. Remember, a state only has a definite value of an operator if it is an eigenstate of that operator - the state $|n\rangle$ does not have a well-defined potential energy, since $\hat{V}$ and $\hat{H}$ do not commute. If we attempt to measure the potential energy $V$ of a quantum harmonic oscillator, we would get a value $V$ with a probability proportional to $|\langle V | n \rangle|^2$, and collapse the oscillator into the eigenstate of potential energy (also the eigenstate of position for a harmonic oscillator).
OK, let's unpack this a bit. The result in Griffiths is, if we write it out more carefully, $$\langle n | \hat{V} | n \rangle = \frac{\hbar \omega}{2}(n + 1/2)$$ where $\hat{V}$ is the potential energy operator $\frac{1}{2} m \omega^2 x^2$ and $|n\rangle$ is the eigenstate of the Hamiltonian, $\hat{H} |n\rangle = E_n |n \rangle$, with $E_n = \hbar \omega(n + 1/2)$.
"Is the potential energy of the quantum harmonic oscillator always one half the oscillator's total energy?"
No. Remember, a state only has a definite value of an operator if it is an eigenstate of that operator - the state $|n\rangle$ does not have a well-defined potential energy, since $\hat{V}$ and $\hat{H}$ do not commute. If we attempt to measure the potential energy $V$ of a quantum harmonic oscillator, we would get a value $V$ with a probability proportional to $|\langle V | n \rangle|^2$, and collapse the oscillator into the eigenstate of potential energy (also the eigenstate of position for a harmonic oscillator).
"Is the potential energy of the quantum harmonic oscillator always one half the oscillator's total energy?"
No. Remember, a state only has a definite value of an operator if it is an eigenstate of that operator - the state $|n\rangle$ does not have a well-defined potential energy, since $\hat{V}$ and $\hat{H}$ do not commute. If we attempt to measure the potential energy $V$ of a quantum harmonic oscillator, we would get a value $V$ with a probability proportional to $|\langle V | n \rangle|^2$, and collapse the oscillator into the eigenstate of potential energy (also the eigenstate of position for a harmonic oscillator).
OK, let's unpack this a bit. The result in Griffiths is, if we write it out more carefully, $$\langle n | \hat{V} | n \rangle = \frac{\hbar \omega}{2}(n + 1/2)$$ where $\hat{V}$ is the potential energy operator $\frac{1}{2} m \omega^2 x^2$ and $|n\rangle$ is the eigenstate of the Hamiltonian, $\hat{H} |n\rangle = E_n |n \rangle$, with $E_n = \hbar \omega(n + 1/2)$.
"Is the potential energy of the quantum harmonic oscillator always one half the oscillator's total energy?"
No. Remember, a state only has a definite value of an operator if it is an eigenstate of that operator - the state $|n\rangle$ does not have a well-defined potential energy, since $\hat{V}$ and $\hat{H}$ do not commute. If we attempt to measure the potential energy $V$ of a quantum harmonic oscillator, we would get a value $V$ with a probability proportional to $|\langle V | n \rangle|^2$, and collapse the oscillator into the eigenstate of potential energy (also the eigenstate of position for a harmonic oscillator).
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