Chapter 3: Q31P (page 126)

$Virial theorem . Use 3.71 to show that \frac{d}{dt} < xp > = 2 < T > - < x \frac{dV}{dx} > where T is the kinetic energy (H = T + V) . In a stationary state the left sideis zero (why ?) so 2 < T > = < x \frac{dV}{dx} > This is called the virial theorem . Use it to prove that < T > = < V > for stationary states of the harmonic oscillator, and check that this is consistent with the results you got in Problem 2.11 and 2.12 .$

Short Answer

Expert verified

It is proved that $\frac{d}{d t} <x p> = 2 <T> - <x \frac{d V}{d x}> .$ .

The reason is all expectation values for stationary states are time independent.

So, $\frac{d <x p>}{d t} = 0$

It is proved that $<T> = <V>$ .

Step by step solution

Equation 3.71 and reason for zero on the left side in a stationary

The equation 3.71 is given by,

$\frac{d}{d t} (Q) = \frac{i}{h} <[\hat{H}, \vec{Q}]> + (\frac{\partial \hat{Q}}{\partial t})$

Now replace Q = x p,

$\frac{d}{d t} (x p) = \frac{i}{h} ([H, x p]) + (\frac{\hat{\partial} (x p)}{\hat{c} t})$

There is no time dependence of x and p explicitly,

$\frac{d}{d t} <x p> = \frac{i}{h} <[H, x p]>$ .........(1)

Now, consider $[H, x p]$

$[H, x p] = [H, x] p + x [H, p]$

The standard results $[H, x] = \frac{- i h p}{m}$

$[H, p] = i h \frac{d V}{d x}$

Now use these values,

role="math" localid="1656331639566" $[I I, x p] = \frac{i h^{2}}{m} + x i h \frac{d V}{d x}$ .

Substitute the values of $[H, x p]$ into equation (1),

$\frac{d}{d t} <x p> = \frac{i}{h} [- \frac{i h}{m} <p^{2}> + i h <x \frac{d V}{d x}>] = \frac{(p^{2})}{m} - <x \frac{d V}{d x}> = 2 \frac{(p^{2})}{2 m} - <x \frac{d V}{d x}> = 2 <T> - <x \frac{d V}{d x}>$

All expectation values for stationary states are time independent.

Sorole="math" localid="1656332438187" $\frac{d <x p>}{d t} = 0 Thus, 2 <T> - <x \frac{dV}{dx}> = 0 2 <T> = <x \frac{dV}{dx}> (2)$

This is called the virial theorem.

Prove that <T>=<V>for stationary states of the harmonic oscillator

For a harmonic oscillator,

$V (x) = \frac{1}{2} {mω}^{2} x^{2} \frac{dV}{dx} = {mω}^{2} x = 2 \frac{V}{x} Thus, <x \frac{dV}{dx}> = 2 <V>$

Substitute these in equation (2),

$2 <T> = 2 <V> <T> = <V> It is known that <T> = \frac{1}{2} (n + \frac{1}{2}) hω While <x^{2}> = (n + \frac{1}{2}) \frac{h}{mω} Here,$

$<V> = \frac{1}{2} {mω}^{2} <x^{2}> = \frac{1}{2} (n + \frac{1}{2}) hω Thus, it is proved that <T> = <V> for all stationary states, and it is consistent with the problem 2.11 and 2.12 .$

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Short Answer

Step by step solution

Equation 3.71 and reason for zero on the left side in a stationary

Prove that <T>=<V>for stationary states of the harmonic oscillator

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