Ehrenfest 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 prove the Ehrenfest theorem

{

m

d

d
t

⟨

x
^

⟩

=
⟨

p
^

⟩

d

d
t

⟨

p
^

⟩

=
−
⟨

∂

V
^

∂
x

⟩

{\displaystyle \begin{cases}
m\frac{d}{dt}\langle \hat{x}\rangle &= \langle \hat{p} \rangle \\[5 pt]
\frac{d}{dt}\langle \hat{p}\rangle &= -\langle \frac{\partial \hat{V}}{\partial x} \rangle \\

\end{cases}}

where

V
^

=
V
(
x
)

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

 is the potential energy. [**Hint:** Remember

H
^

=

p
^

2

2
m

V
^

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

.]

## **Solution**

Since

x
^

{\displaystyle \hat x}

 does not depend on time

t

{\displaystyle t}

, the final term becomes

⟨

∂

x
^

∂
t

# ⟩
0

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

 and we get

d

d
t

⟨

x
^

# ⟩

i
ℏ

⟨
[

H
^

,

x
^

]
⟩
.

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

Now calculate

[

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
^

=
V
(
x
)

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

 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} }

. Consequently,

d

d
t

⟨

x
^

⟩

=

i
ℏ

⋅
⟨
−

i
ℏ

m

p
^

⟩

=
⟨

p
^

m

⟩

{\displaystyle \begin{align}
\frac{d}{dt}\langle {\hat{x}} \rangle &=\frac{i}{\hbar} \cdot\langle -\frac{i\hbar}{m} \hat{p}\rangle \\[5 pt]
&= \langle \frac{\hat{p}}{m} \rangle
\end{align} }

Therefore,

m

d

d
t

⟨

x
^

# ⟩
⟨

p
^

⟩
.

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

Since

p
^

{\displaystyle \hat p}

 does not depend on time

t

{\displaystyle t}

, the final term becomes

⟨

∂

p
^

∂
t

# ⟩
0

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

 and we get

d

d
t

⟨

p
^

# ⟩

i
ℏ

⟨
[

H
^

,

p
^

]
⟩
.

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

Now calculate

[

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} }

. Consequently,

d

d
t

⟨

p
^

⟩

=

i
ℏ

⋅
⟨
i
ℏ

∂

V
^

∂
x

⟩

=
−
⟨

∂

V
^

∂
x

⟩

{\displaystyle \begin{align}
\frac{d}{dt}\langle {\hat{p}} \rangle &=\frac{i}{\hbar} \cdot \langle i\hbar \frac{\partial \hat{V}}{\partial x} \rangle \\[5 pt]
&= -\langle \frac{\partial \hat{V}}{\partial x} \rangle
\end{align} }