Quantum Griffiths Example

[kuahji]

Find the expectation value of the potential energy in the nth state of the harmonic oscillator.

This is his example 2.5 in the book, he uses a[itex]\pm[/itex]=1/Sqrt[2hmw]([itex]\mp[/itex]ip+mwx) to get x=Sqrt[h/2mw](a[itex]_{+}[/itex]+a[itex]_{-}[/itex])

My question how does he do this? I can't seem to make the algebraic manipulations to get it into that form to follow out his example.

 

[G01]

You have two equations, one for [itex]a_+[/itex] and one for [itex]a_-[/itex].

All you have to do is solve them simultaneously for x and p. i.e. Solve one for x, solve the other for p, plug the first into the second, etc. etc.

It may help to simplify things by defining some constants A and B such that:

[tex]a_{\pm}=Ap\mp Bx[/tex]

 

[kuahji]

That was easy enough, thank you for the tip!

 

[G01]

No problem!

 

[paranormal]

The expectation value of the potential energy in the nth state of the harmonic oscillator can be calculated using the Schrödinger equation and the Hamiltonian operator. However, in this specific example, Griffiths uses the creation and annihilation operators, a+ and a-, to simplify the calculations.

To understand how he arrives at the expression x=Sqrt[h/2mw](a_{+}+a_{-}), we need to look at the definitions of these operators. The creation operator, a+, is defined as a linear combination of the position and momentum operators:

a_{+}=\sqrt{\frac{m\omega}{2\hbar}}(x+\frac{i}{m\omega}p)

Similarly, the annihilation operator, a-, is defined as:

a_{-}=\sqrt{\frac{m\omega}{2\hbar}}(x-\frac{i}{m\omega}p)

Substituting these definitions into the expression for x, we get:

x=\frac{1}{2}\sqrt{\frac{2\hbar}{m\omega}}(a_{+}+a_{-})

Simplifying further, we get:

x=\sqrt{\frac{\hbar}{2m\omega}}(a_{+}+a_{-})

Therefore, the expectation value of the potential energy in the nth state can be calculated as:

<V>=\frac{1}{2}m\omega^{2}<x^{2}>=\frac{\hbar\omega}{4}(<a_{+}^{2}>+<a_{-}^{2}>+2<a_{+}a_{-}>)

Using the commutation relation for these operators, we can simplify this expression to:

<V>=\frac{\hbar\omega}{4}(2n+1)

This is the final expression for the expectation value of potential energy in the nth state of the harmonic oscillator, which can be obtained using the creation and annihilation operators.