(a) Prove the following commutator identity:

[AB.C]=A[B.C]+[A.C]B

b) Show that

[xn,p]=ihnxn-1

(c) Show more generally that

[f(x),p]=ihdfdx

for any functionf(x).

Short Answer

Expert verified

a) [AB.C]=A[B.C]+[A.C]B

b)[xn,p]=ihnxn-1

c) [f(x),p]=ihdfdx

Step by step solution

01

Concept used

Commutator of two quantities and is defined:

A,B=AB-BA

AB,C=ABC-CAB ……. (1)

02

 Prove Commutator quantity

We can add ACB-ACBto equation (1):

localid="1658124315043" AB,C=ABC-CAB+ACB-ACB=ABC-CB+AC-CAB=AB,C+A,CB=AB,C+A,CB

03

Use mathematical induction

We use mathematical induction:

Mathematical induction:

n=1x,p=ih

We assume that following identity is valid for some n:

xn,p=ihnxn-1

Induction step: n+1

localid="1658124506145" xn+1,p=x·xnp=xn,p+x,pxn=x·ihnxn-1+ihxn=ihnxn+ihxn=ihn+1xn

04

To prove the given relation, after commutator have a test function

c)

After commutator have a test function:

fx,pψx=fx,-ihddxψx=-ihfx,ddxψx=-ihfdψdx-ddxfxψx=-ihfdψdx-fdψdx-dfdxψ=ihdfdxψ.

Since this relation must be valid for any test function ψx, it follows:

fx,p=ihdfdx

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