Question:
What is the wavefunction of the first excited state of the quantum harmonic oscillator, |1⟩?
Solution:
The answer can easily be read from the Quantum Harmonic Oscillator page from Wikipedia. But since this is a blog of doing exercise, what is the point of copying the answer? Let us derive it from the ground state using the creation operator.
First, we have the ground state wavefunction ψ0(x)=(mωπℏ)14e−mωx22ℏ.
Now we apply the creation operator ˆa†=√mω2ℏ(ˆx−iˆpmω)=√mω2ℏ(x−ℏmω∂∂x).
Let's first compute the derivative:
∂∂xψ0(x)=(mωπℏ)14e−mωx22ℏ−mωxℏ
So we have the momentum term:
−ℏmω∂∂xψ0(x)=(mωπℏ)14e−mωx22ℏx=xψ0(x)
The position term is of course also xψ0(x). Combining using the creation operator we have
ˆa†ψ(x)=√mω2ℏ(2xψ0(x))=1√2ψ0(x)H1(mωxℏ).
H1 is the physicist Hermite polynomial of order 1 which is simply H1(x)=2x.
What is the wavefunction of the first excited state of the quantum harmonic oscillator, |1⟩?
Solution:
The answer can easily be read from the Quantum Harmonic Oscillator page from Wikipedia. But since this is a blog of doing exercise, what is the point of copying the answer? Let us derive it from the ground state using the creation operator.
First, we have the ground state wavefunction ψ0(x)=(mωπℏ)14e−mωx22ℏ.
Now we apply the creation operator ˆa†=√mω2ℏ(ˆx−iˆpmω)=√mω2ℏ(x−ℏmω∂∂x).
Let's first compute the derivative:
∂∂xψ0(x)=(mωπℏ)14e−mωx22ℏ−mωxℏ
So we have the momentum term:
−ℏmω∂∂xψ0(x)=(mωπℏ)14e−mωx22ℏx=xψ0(x)
The position term is of course also xψ0(x). Combining using the creation operator we have
ˆa†ψ(x)=√mω2ℏ(2xψ0(x))=1√2ψ0(x)H1(mωxℏ).
H1 is the physicist Hermite polynomial of order 1 which is simply H1(x)=2x.
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