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The document discusses various properties of commutators in quantum mechanics, specifically focusing on the relationships between operators and their commutators. It proves several mathematical relations involving exponentials of operators and their commutators, including the Baker-Campbell-Hausdorff formula. Additionally, it expresses the commutator of the product of two operators with a third operator in terms of their individual commutators.

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22 views2 pages

2397523

The document discusses various properties of commutators in quantum mechanics, specifically focusing on the relationships between operators and their commutators. It proves several mathematical relations involving exponentials of operators and their commutators, including the Baker-Campbell-Hausdorff formula. Additionally, it expresses the commutator of the product of two operators with a third operator in terms of their individual commutators.

Uploaded by

dearhrhaerhaer
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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Using the product rule we have

" #
i dfˆ d d
[p̂, fˆ] = + fˆ − fˆ
~ dx dx dx
(6)
dfˆ
[p̂, fˆ] = −i~ .
dx
For the other commutator we may evaluate the x̂ operator in momentum space to give

[x̂, ĝ] = x̂ĝ − ĝx̂


i d i d
= ĝ − ĝ
~ dp ~ dp
(7)
 
i dĝ d d
= + ĝ −
~ dp dp dp
dg
[x̂, ĝ] = −i~ .
dp

4: Prove the following relation:


1 1
eL̂ âe−L̂ = â + [L̂, â] + [L̂, [L̂, â]] + [L̂, [L̂, [L̂, â]]] + ... (8)
2! 3!

h i h i
Define f (β) ≡ exp β L̂ â exp −β L̂ . Then

df
= L̂eβ L̂ âe−β L̂ − eβ L̂ âL̂e−β L̂ = [L̂, âeβ L̂ ]e−β L̂
dβ
d2 f
= L̂2 eβ L̂ âe−β L̂ − L̂eβ L̂ âL̂e−β L̂ − L̂eβ L̂ âL̂e−β L̂ + eβ L̂ âL̂2 e−β L̂ = [L̂, [L̂, â]]e−β L̂
dβ 2
d3 f
= L̂3 eβ L̂ âe−β L̂ − L̂2 eβ L̂ âL̂e−β L̂ − 2L̂2 eβ L̂ âL̂e−β L̂ + 2L̂eβ L̂ âL̂2 e−β L̂ + L̂eβ L̂ âL̂2 e−β L̂ − eβ L̂ âL̂3 e−β L̂
dβ 3
= [L̂, [L̂, [L̂, â]]]e−β L̂ .
(9)

Taylor expansion of our function is


∞
X f (k) (0) 1 00 1
f (β) = β k = f (0) + f 0 (0)β + f (0)β 2 + f (3) (0)β 3 + ... (10)
k! 2! 3!
k=0

If we set β = 1, we have
1 00 1
eL̂ âe−L̂ = f (0) + f 0 (0) + f (0) + f (3) (0) + ...
2! 3! (11)
1 1
eL̂ âe−L̂ = â + [L̂, â] + [L̂, [L̂, â]] + [L̂, [L̂, [L̂, â]]] + ...
2! 3!
which was our target.
5: Show that if [Â, B̂] = K where K ∈ C and if λ ∈ C,
1
eλ(Â+B̂ ) = eλ eλB̂ e− 2 λ
2K
. (12)

Let f be the right hand side. Taking the derivative of the right hand side with respect to λ we get

df 1 2 1 2 1 2
= Âeλ eλB̂ e− 2 λ K + eλ B̂eλB̂ e− 2 λ K − λKeλ eλB̂ e− 2 λ K
dλ n o 1 2
= Âeλ eλB̂ + eλ B̂eλB̂ − λKeλ eλB̂ e− 2 λ K (13)
n o 1 2
= Âeλ eλB̂ + eλ [B − λK] eλB̂ e− 2 λ K

We may use the result of last problem to state

e−λ B̂eλ ' B̂ − λ[Â, B̂] = B̂ − λK. (14)

Then

df n o 1 2
= Âeλ eλB̂ + eλ e−λ B̂eλ eλB̂ e− 2 λ K
dλ (15)
df n o
= Â + B̂ f.
dλ
The solution to this differential equation is an exponential with the factor (Â + B̂) as the argument:

f = eλ(Â+B̂ ) . (16)

Hence, the statement is true.

6: Let Â, B̂, and Ĉ be linear operators. Express the commutator of the product of ÂB̂ with Ĉ in terms of
the commutators [Â, Ĉ] and [B̂, Ĉ].

The commutator of ÂB̂ with Ĉ is

[ÂB̂, Ĉ] = ÂB̂ Ĉ − Ĉ ÂB̂. (17)

The suggested commutators are

[Â, Ĉ] = ÂĈ − Ĉ Â


(18)
[B̂, Ĉ] = B̂ Ĉ − Ĉ B̂.

If we multiply the first on the right by B̂ and the second one on the left by  then add, we have

[Â, Ĉ]B̂ + Â[B̂, Ĉ] = ÂĈ B̂ − Ĉ ÂB̂ + ÂB̂ Ĉ − ÂĈ B̂


[Â, Ĉ]B̂ + Â[B̂, Ĉ] = ÂB̂ Ĉ − Ĉ ÂB̂ (19)
[Â, Ĉ]B̂ + Â[B̂, Ĉ] = [ÂB̂, Ĉ].

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