Quantum Virial Theorem

## **Problem**

Use the formula

d

d
t

⟨

A
^

# ⟩

i
ℏ

⟨
[

H
^

,

A
^

]
⟩
⟨

∂

A
^

∂
t

⟩

{\displaystyle \frac{d}{dt}\langle \hat{A} \rangle =\frac{i}{\hbar}\langle [\hat{H},\hat{A}] \rangle +\langle \frac{\partial \hat{A}}{\partial t} \rangle }

 to show that

d

d
t

⟨

x
^

p
^

# ⟩
2
⟨

T
^

⟩
−
⟨

x
^

∂

V
^

∂
x

⟩
,

{\displaystyle {\frac {d}{dt}}\langle {\hat {x}}{\hat {p}}\rangle =2\langle {\hat {T}}\rangle -\langle {\hat {x}}{\frac {\partial {\hat {V}}}{\partial x}}\rangle ,}

where

T
^

{\displaystyle \hat{T}}

 is the kinetic energy and

V
^

{\displaystyle \hat{V}}

 is the potential energy. Then show that the expression reduces to

2
⟨
T
# ⟩
⟨

x
^

∂

V
^

∂
x

⟩

{\displaystyle 2\langle T \rangle = \langle \hat{x} \frac{\partial \hat{V}}{\partial x} \rangle }

 for a stationary state. This result is called the Virial theorem.

### **Hint**

(1) Remember

H
^

=

T
^

V
^

=

p
^

2

2
m

V
^

{\displaystyle \hat{H} = \hat{T}+\hat{V} = \frac{{\hat{p}}^{2}}{2m}+\hat{V} }

, where

V
^

=
V
(
x
)

{\displaystyle \hat{V} = V(x)}

.

## **Solution**

Start with the formula

d

d
t

⟨

A
^

# ⟩

i
ℏ

⟨
[

H
^

,

A
^

]
⟩
⟨

∂

A
^

∂
t

⟩

{\displaystyle \frac{d}{dt}\langle \hat{A} \rangle =\frac{i}{\hbar}\langle [\hat{H},\hat{A}] \rangle +\langle \frac{\partial \hat{A}}{\partial t} \rangle }

where

H
^

=

p
^

2

2
m

V
^

{\displaystyle \hat{H} = \frac{{\hat{p}}^{2}}{2m}+\hat{V} }

 and

V
^

=
V
(
x
)

{\displaystyle \hat{V} = V(x)}

..

Let

A
^

=

x
^

p
^

{\displaystyle {\hat {A}}={\hat {x}}{\hat {p}}}

 where

x
^

=
x

{\displaystyle \hat{x} = x }

 and

p
^

=
−
i
ℏ

∂

∂
x

{\displaystyle \hat{p} =-i\hbar\frac{\partial}{\partial x} }

, then we get

d

d
t

⟨

x
^

p
^

# ⟩

i
ℏ

⟨
[

H
^

,

x
^

p
^

]
⟩
⟨

∂

x
^

p
^

∂
t

⟩
.

{\displaystyle {\frac {d}{dt}}\langle {{\hat {x}}{\hat {p}}}\rangle ={\frac {i}{\hbar }}\langle [{\hat {H}},{\hat {x}}{\hat {p}}]\rangle +\langle {\frac {\partial {\hat {x}}{\hat {p}}}{\partial t}}\rangle .}

Since

x
^

p
^

{\displaystyle {\hat {x}}{\hat {p}}}

 does not depend on time

t

{\displaystyle t}

, the final term becomes

⟨

∂

x
^

p
^

∂
t

# ⟩
0

{\displaystyle \langle {\frac {\partial {\hat {x}}{\hat {p}}}{\partial t}}\rangle =0}

 and we get

d

d
t

⟨

x
^

p
^

# ⟩

i
ℏ

⟨
[

H
^

,

x
^

p
^

]
⟩
.

{\displaystyle {\frac {d}{dt}}\langle {{\hat {x}}{\hat {p}}}\rangle ={\frac {i}{\hbar }}\langle [{\hat {H}},{\hat {x}}{\hat {p}}]\rangle .}

Let us calculate the commutator

[

H
^

,

x
^

p
^

]

{\displaystyle [{\hat {H}},{\hat {x}}{\hat {p}}]}

first. By the commutator property

[
A
,
B
C
# ]
[
A
,
B
]
C
B
[
A
,
C
]

{\displaystyle [A, BC] = [A, B]C + B[A, C]}

, we find

[

H
^

,

x
^

p
^

# ]
[

H
^

,

x
^

]

p
^

x
^

[

H
^

,

p
^

]
.

{\displaystyle [{\hat {H}},{\hat {x}}{\hat {p}}]=[{\hat {H}},{\hat {x}}]{\hat {p}}+{\hat {x}}[{\hat {H}},{\hat {p}}].}

**Calculation for

[

H
^

,

x
^

]

{\displaystyle [\hat{H}, \hat{x}]}**

[

H
^

,

x
^

]

=
[

p
^

2

2
m

V
^

,

x
^

]

=
[

p
^

2

2
m

,

x
^

]
[

V
^

,

x
^

]

=

1

2
m

[

p
^

2

,

x
^

]
[

V
^

,

x
^

]

{\displaystyle \begin{align}
\left[\hat{H}, \hat{x}\right] &= [\frac{\hat{p}^2}{2m} + \hat{V} , \hat{x}] \\
&= [\frac{\hat{p}^2}{2m} , \hat{x} ] + [\hat{V} , \hat{x} ] \\
&= \frac{1}{2m}[\hat{p}^2 , \hat{x}] + [\hat{V} , \hat{x} ] \\
\end{align} }

The next step is to calculate

[

p
^

2

,

x
^

]

=

[

p
^

,

x
^

]

p
^

p
^

[

p
^

,

x
^

]

{\displaystyle \left[\hat{p}^2 , \hat{x} \right] = \left[\hat{p} , \hat{x} \right] \hat{p}+ \hat{p}\left[\hat{p} , \hat{x} \right] }

Since

[

x
^

,

p
^

]

=
i
ℏ

{\displaystyle \left[ \hat { x } ,\hat { p } \right] = i\hbar }

, we get

[

p
^

,

x
^

]

=
−
i
ℏ

{\displaystyle \left[\hat{p} , \hat{x} \right] = -i\hbar }

. Thus,

[

p
^

2

,

x
^

]

=
2
i
ℏ
p

{\displaystyle \left[\hat{p}^2 , \hat{x} \right] = 2i\hbar p }

. Since

V
^

{\displaystyle \hat{V}}

 is a function of

x

{\displaystyle x}

, we get

[

V
^

,

x
^

# ]
0

{\displaystyle [\hat{V},\hat{x}] = 0 }

.

Therefore,

[

p
^

2

2
m

V
^

,

x
^

]

=

1

2
m

(
−
2
i
ℏ

p
^

)

=
−

i
ℏ

m

p
^

{\displaystyle \begin{align}
\left[\frac{\hat{p}^2}{2m} + \hat{V} , \hat{x} \right] &= \frac{1}{2m}(-2i\hbar \hat{p})\\
&=-\frac{i\hbar}{ m}\hat{p}
\end{align} }

and

[

H
^

,

x
^

# ]
−

i
ℏ

m

p
^

{\displaystyle [\hat{H}, \hat{x}] = -\frac{i\hbar}{m} \hat{p} }

.

**Calculation

[

H
^

,

p
^

]

{\displaystyle [\hat{H} , \hat{p}]}**

[

H
^

,

p
^

]

=
[

p
^

2

2
m

V
^

,

p
^

]

=
[

p
^

2

2
m

,

p
^

]
[

V
^

,

p
^

]

=

1

2
m

[

p
^

2

,

p
^

]
[

V
^

,

p
^

]

{\displaystyle \begin{align}
\left[\hat{H} , \hat{p}\right] &= [\frac{\hat{p}^2}{2m} + \hat{V} , \hat{p} ] \\
&= [\frac{\hat{p}^2}{2m} , \hat{p} ] + [\hat{V} , \hat{p}] \\
&= \frac{1}{2m}[\hat{p}^2 , \hat{p}] + [\hat{V} , \hat{p}]
\end{align} }

Since

[

p
^

2

,

p
^

# ]
0

{\displaystyle [\hat{p}^2, \hat{p}] = 0 }

 and

[

V
^

,

p
^

# ]
i
ℏ

∂

V
^

∂
x

{\displaystyle [\hat{V},\hat{p}] = i\hbar \frac{\partial \hat{V}}{\partial x} }

, we get

[

H
^

,

p
^

# ]
i
ℏ

∂

V
^

∂
x

{\displaystyle [\hat{H},\hat{p}] = i\hbar \frac{\partial \hat{V}}{\partial x} }

.

Subsequently,

[

H
^

,

x
^

p
^

# ]
−

i
ℏ

m

⟨

p
^

2

⟩
i
ℏ
⟨

x
^

d

V
^

d
x

⟩

{\displaystyle [{\hat {H}},{\hat {x}}{\hat {p}}]=-{\frac {i\hbar }{m}}\langle {\hat {p}}^{2}\rangle +i\hbar \langle {\hat {x}}{\frac {d{\hat {V}}}{dx}}\rangle }

and

d

d
t

⟨

x
^

p
^

⟩

=

i
ℏ

(

−

i
ℏ

m

⟨

p
^

2

⟩
i
ℏ
⟨

x
^

d

V
^

d
x

⟩

)

=
2
⟨

p
^

2

2
m

⟩
−
⟨

x
^

d

V
^

d
x

⟩

=
2
⟨
T
⟩
−
⟨

x
^

d

V
^

d
x

⟩

{\displaystyle {\begin{aligned}{\frac {d}{dt}}\langle {{\hat {x}}{\hat {p}}}\rangle &={\frac {i}{\hbar }}\left(-{\frac {i\hbar }{m}}\langle {\hat {p}}^{2}\rangle +i\hbar \langle {\hat {x}}{\frac {d{\hat {V}}}{dx}}\rangle \right)\\&=2\langle {\frac {{\hat {p}}^{2}}{2m}}\rangle -\langle {\hat {x}}{\frac {d{\hat {V}}}{dx}}\rangle \\&=2\langle T\rangle -\langle {\hat {x}}{\frac {d{\hat {V}}}{dx}}\rangle \end{aligned}}}

Therefore,

d

d
t

⟨

x
^

p
^

# ⟩
2
⟨

T
^

⟩
−
⟨

x
^

∂

V
^

∂
x

⟩
.

{\displaystyle {\frac {d}{dt}}\langle {\hat {x}}{\hat {p}}\rangle =2\langle {\hat {T}}\rangle -\langle {\hat {x}}{\frac {\partial {\hat {V}}}{\partial x}}\rangle .}

In a stationary state, all expectation values are time-independent , so

d

d
t

⟨

x
^

p
^

# ⟩
0

{\displaystyle {\frac {d}{dt}}\langle {\hat {x}}{\hat {p}}\rangle =0}

. Therefore,

# 0
2
⟨

T
^

⟩
−
⟨

x
^

∂

V
^

∂
x

⟩

{\displaystyle 0=2\langle {\hat {T}}\rangle -\langle {\hat {x}}{\frac {\partial {\hat {V}}}{\partial x}}\rangle }

and

2
⟨

T
^

# ⟩
⟨

x
^

∂

V
^

∂
x

⟩
.

{\displaystyle 2\langle {\hat {T}}\rangle =\langle {\hat {x}}{\frac {\partial {\hat {V}}}{\partial x}}\rangle .}