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Syed Shoaib posted an Question
January 24, 2022 • 01:10 am 30 points
  • IIT JAM
  • Physics (PH)

The expectation value of position and momentum of a particle having normalized wave function vr) = n exp (a) (r=0 (p.)0 (6) ()=0 p.)=hk (c) hk p.)=0 d)( hk p.)h

The expectation value of position and momentum of a particle having normalized wave function vr) = N exp (a) (r=0 (P.)0 (6) ()=0 P.)=hk (c) hk P.)=0 d)( hk P.)hk

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  • Himanshu best-answer

    answer is b because expection value of momentum of real wave function is zero always .but here wave is not real then expectation value of momentum has some answer and expectation value of x is zero because in integration form make a odd function so expectation value of x is zero and expectation value of momentum is answered.

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