Heisenberg Uncertainty Principle

## **Problem**

   This problem is a step-by-step derivation of the famous uncertainty principle in quantum mechanics.

**Part 1:** Let

# f
(

x
^

−
⟨

x
^

⟩
)
Ψ

{\displaystyle f=(\hat{x}-\langle\hat{x}\rangle)\Psi }

and

# g
(

p
^

−
⟨

p
^

⟩
)
Ψ

{\displaystyle g=(\hat{p}-\langle\hat{p}\rangle)\Psi }

, show that

σ

x

2

σ

p

2

≥

[

1

2
i

(

⟨
f

g
⟩
−
⟨
g

f
⟩

)

]

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {\left[{\frac {1}{2i}}\left(\langle f|g\rangle -\langle g|f\rangle \right)\right]}^{2}}

.
**Hint:**

z

∗

z
≥

[

1

2
i

(
z
−

z

∗

)

]

2

{\displaystyle {z}^{\*}z \ge {\left[\frac{1}{2i}(z-{z}^{\*})\right]}^{2} }

**Part 2:** Show (by calculation) that

⟨
f

g
# ⟩
⟨

x
^

p
^

⟩
−
⟨

p
^

⟩
⟨

x
^

⟩

{\displaystyle \langle f|g\rangle =\langle {\hat {x}}{\hat {p}}\rangle -\langle {\hat {p}}\rangle \langle {\hat {x}}\rangle }

and

⟨
g

f
# ⟩
⟨

p
^

x
^

⟩
−
⟨

x
^

⟩
⟨

p
^

⟩

{\displaystyle \langle g|f\rangle=\langle\hat{p}\hat{x}\rangle-\langle\hat{x}\rangle\langle\hat{p}\rangle }

.
Then show that

σ

x

2

σ

p

2

≥

(

1

2
i

⟨

[

x
^

,

p
^

]

⟩

)

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {\left({\frac {1}{2i}}\left\langle [{\hat {x}},{\hat {p}}]\right\rangle \right)}^{2}}

.
**Part 3:** It is proven in this problem that

[

x
^

,

p
^

# ]
i
ℏ

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

, show that

σ

x

σ

p

≥

ℏ
2

{\displaystyle {\sigma}\_{x}{\sigma}\_{p} \ge \frac{\hbar}{2} }

.

## **Solution**

**Part 1**

The standard deviation of position is

σ

x

2

=

⟨

Ψ

(

x
^

−
⟨

x
^

⟩

)

2

Ψ

⟩

=
⟨
f

f
⟩

{\displaystyle {\sigma }\_{x}^{2}=\left\langle \Psi \left|\left({\hat {x}}-\langle {\hat {x}}\rangle \right)^{2}\right|\Psi \right\rangle =\langle f|f\rangle }

and the standard deviation of momentum is

σ

p

2

=

⟨

Ψ

(

p
^

−
⟨

p
^

⟩

)

2

Ψ

⟩

=
⟨
g

g
⟩

{\displaystyle {\sigma }\_{p}^{2}=\left\langle \Psi \left|\left({\hat {p}}-\langle {\hat {p}}\rangle \right)^{2}\right|\Psi \right\rangle =\langle g|g\rangle }

.
Now this means,

σ

x

2

σ

p

2

=
⟨
f

f
⟩
⟨
g

g
⟩

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}=\langle f|f\rangle \langle g|g\rangle }

.
The Cauchy-Schwarz inequality states

⟨
f

f
⟩
⟨
g

g
⟩
≥

⟨
f

g
⟩

2

, where

⟨
f

g
⟩
⟨
g

f
# ⟩

⟨
f

g
⟩

2

.

Therefore,

σ

x

2

σ

p

2

≥

⟨
f

g
⟩

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {|\langle f|g\rangle |}^{2}}

.

Since

z

∗

z
≥

[

1

2
i

(
z
−

z

∗

)

]

2

{\displaystyle {z}^{\*}z \ge {\left[\frac{1}{2i}(z-{z}^{\*})\right]}^{2} }

,

replacing

z

{\displaystyle z}

 with

⟨
f

g
⟩

{\displaystyle \langle f|g\rangle }

 and

z

∗

{\displaystyle {z}^{\*} }

 with

⟨
g

f
⟩

{\displaystyle \langle g|f \rangle }

 yields

σ

x

2

σ

p

2

≥

[

1

2
i

(

⟨
f

g
⟩
−
⟨
g

f
⟩

)

]

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {\left[{\frac {1}{2i}}\left(\langle f|g\rangle -\langle g|f\rangle \right)\right]}^{2}}

.
**Part 2**

⟨
f

g
⟩

=
⟨
(

x
^

−
⟨

x
^

⟩
)
Ψ

(

p
^

−
⟨

p
^

⟩
)
Ψ
⟩

=
⟨
Ψ

(

x
^

−
⟨

x
^

⟩
)
(

p
^

−
⟨

p
^

⟩
)
Ψ
⟩

=
⟨
Ψ

(

x
^

p
^

−

x
^

⟨

p
^

⟩
)
−

p
^

⟨

x
^

⟩
⟨

x
^

⟩
⟨

p
^

⟩
)
Ψ
⟩

=
⟨
Ψ

x
^

p
^

Ψ
⟩
−
⟨
p
⟩
⟨
Ψ

x
^

Ψ
⟩
−
⟨
x
⟩
⟨
Ψ

p
^

Ψ
⟩
⟨
x
⟩
⟨
p
⟩
⟨
Ψ

Ψ
⟩

=
⟨

x
^

p
^

⟩
−
⟨

p
^

⟩
⟨

x
^

⟩
−
⟨

x
^

⟩
⟨

p
^

⟩
⟨

x
^

⟩
⟨

p
^

⟩

=
⟨

x
^

p
^

⟩
−
⟨

p
^

⟩
⟨

x
^

⟩
.

Taking the complex conjugate yields

⟨
g

f
# ⟩
⟨

p
^

x
^

⟩
−
⟨

x
^

⟩
⟨

p
^

⟩

{\displaystyle \langle g|f\rangle=\langle\hat{p}\hat{x}\rangle-\langle\hat{x}\rangle\langle\hat{p}\rangle }

.
This means

⟨
f

g
⟩
−
⟨
g

f
# ⟩
⟨

x
^

p
^

⟩
−
⟨

p
^

⟩
⟨

x
^

⟩
−

(

⟨

p
^

x
^

⟩
−
⟨

x
^

⟩
⟨

p
^

⟩

)

{\displaystyle \langle f|g\rangle -\langle g|f\rangle =\langle {\hat {x}}{\hat {p}}\rangle -\langle {\hat {p}}\rangle \langle {\hat {x}}\rangle -\left(\langle {\hat {p}}{\hat {x}}\rangle -\langle {\hat {x}}\rangle \langle {\hat {p}}\rangle \right)}

⟨
f

g
⟩
−
⟨
g

f
# ⟩
⟨

x
^

p
^

⟩
−
⟨

p
^

x
^

⟩

{\displaystyle \langle f|g\rangle -\langle g|f\rangle =\langle {\hat {x}}{\hat {p}}\rangle -\langle {\hat {p}}{\hat {x}}\rangle }

.

Subsequently,

σ

x

2

σ

p

2

≥

(

1

2
i

⟨

[

x
^

,

p
^

]

⟩

)

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {\left({\frac {1}{2i}}\left\langle [{\hat {x}},{\hat {p}}]\right\rangle \right)}^{2}}

.
**Part 3**

Since

[

x
^

,

p
^

# ]
i
ℏ

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

,

σ

x

2

σ

p

2

≥

(

1

2
i

(
i
ℏ
)

)

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {\left({\frac {1}{2i}}(i\hbar )\right)}^{2}}

σ

x

2

σ

p

2

≥

(

ℏ
2

)

2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}^{2}\geq {\left({\frac {\hbar }{2}}\right)}^{2}}

σ

x

2

σ

p

≥

ℏ
2

{\displaystyle {\sigma }\_{x}^{2}{\sigma }\_{p}\geq {\frac {\hbar }{2}}}

.