मान लीजिए एक विमा में हैमिल्टनी H

H = \(\rm \frac{p^2_x}{2m}\) +V(x) है

H का x के साथ कम्यूटेटर [H,x] है

This question was previously asked in
CSIR-UGC (NET) Chemical Science: Held on (26 Nov 2020)
View all CSIR NET Papers >
  1. \(\rm \frac{-ih}{m}px\)
  2. \(\rm \frac{-ih}{2m}p^2_x\)
  3. \(\rm \frac{ih}{m}px\)
  4. \(\rm \frac{ih}{2m}px\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \frac{-ih}{m}px\)
Free
Seating Arrangement
3.8 K Users
10 Questions 20 Marks 15 Mins

Detailed Solution

Download Solution PDF

संकल्पना:

  • दो संकारकों \({{\rm{\hat A}}}\) और \({\hat B}\) के क्रमविनिमय को इस प्रकार दर्शाया जाता है,

\(\left[ {{\rm{\hat A,\hat B}}} \right]{\rm{ = }}\left[ {{\rm{\hat A\hat B - \hat B\hat A}}} \right]\)

  • किसी दिए गए क्रमविनिमय की गणना करने के लिए, एक स्वेच्छ फलन (\(\Psi \)) का उपयोग किया जाता है ताकि संकारक इस प्रकार कार्य कर सकें

\(\left[ {{\rm{\hat A,\hat B}}} \right]\Psi {\rm{ = }}\left[ {{\rm{\hat A\hat B - \hat B\hat A}}} \right]\Psi \)

  • दो संकारकों को क्रमविनिमय कहा जाता है जब उनके क्रमविनिमय शून्य के बराबर होते हैं, इसलिए

\(\hat A\hat B = \hat B\hat A\)

  • कोई भी क्रमविनिमय (\({\hat A}\)) स्वयं के साथ क्रमविनिमय होगा,

\(\left[ {{\rm{\hat A, \hat A}}} \right]{\rm{ = 0}}\)

व्याख्या:

  • हैमिल्टोनियन संकारक (H) इस प्रकार दिया गया है,

H = \(\rm \frac{p^2_x}{2m}\) +V(x).

  • अब, स्वेच्छ फलन \(\Psi \) का उपयोग H के साथ x के क्रमविनिमयक की गणना करने के लिए किया जाता है।
  • H के साथ x का क्रमविनिमय है:

\(\left[ {{\rm{\hat H,\hat x}}} \right]\Psi {\rm{ = }}\left[ {\hat H\hat x - \hat x\hat H} \right]\Psi \)

\( = \left[ {\left( { {{{P_x}^2} \over {2m}} + V(x)} \right)x - x\left( { {{{P_x}^2} \over {2m}} + V(x)} \right)} \right]\Psi \)

\( = \left( { {{{P_x}^2} \over {2m}} + V(x)} \right)x\Psi - x\left( { {{{P_x}^2} \over {2m}} + V(x)} \right)\Psi \)

\( = {{{P_x}^2} \over {2m}}\left( {x\Psi } \right) + V(x)x\Psi - x{{{P_x}^2} \over {2m}}\Psi - xV(x)\Psi \)

\( = {1 \over {2m}}{\left( { - i\hbar {\partial \over {\partial x}}} \right)^2}\left( {x\Psi } \right) - x{1 \over {2m}}{\left( { - i\hbar {\partial \over {\partial x}}} \right)^2}\Psi \), as\(\left[ {{P_x} = \left( { - i\hbar {{\partial \Psi } \over {\partial x}}} \right)} \right]\) \( = - {{{\hbar ^2}} \over {2m}}{\left( {{\partial \over {\partial x}}} \right)^2}\left( {x\Psi } \right) + {{x{\hbar ^2}} \over {2m}}{\left( {{\partial \over {\partial x}}} \right)^2}\Psi \)

\( = - {{{\hbar ^2}} \over {2m}}\left( {{\partial \over {\partial x}}} \right)\left( {\Psi + x{{\partial \Psi } \over {\partial x}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( = - {{{\hbar ^2}} \over {2m}}\left( {{{\partial \Psi } \over {\partial x}} + x{{{\partial ^2}\Psi } \over {\partial {x^2}}} + {{\partial \Psi } \over {\partial x}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( =- {{{\hbar ^2}} \over {2m}}\left( {2{{\partial \Psi } \over {\partial x}} + x{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( = -{{2{\hbar ^2}} \over {2m}}\left( {{{\partial \Psi } \over {\partial x}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right) + {{x{\hbar ^2}} \over {2m}}\left( {{{{\partial ^2}\Psi } \over {\partial {x^2}}}} \right)\)

\( = {{{i^2}{\hbar ^2}} \over m}\left( {{{\partial \Psi } \over {\partial x}}} \right)\) as, \({i^2} = - 1\)

\( = - {{i\hbar } \over m}\left( { - i\hbar {{\partial \Psi } \over {\partial x}}} \right)\)

\( = - {{i\hbar } \over m}{P_x}\), as \(\left[ {{P_x} = \left( { - i\hbar {{\partial \Psi } \over {\partial x}}} \right)} \right]\)

निष्कर्ष:

इसलिए, H के साथ x का क्रमविनिमय, \(\left[ {{\rm{\hat H,\hat x}}} \right]\Psi\), है

\( - {{i\hbar } \over m}{P_x}\)

Latest CSIR NET Updates

Last updated on Jul 8, 2025

-> The CSIR NET June 2025 Exam Schedule has been released on its official website.The exam will be held on 28th July 2025.

-> The CSIR UGC NET is conducted in five subjects -Chemical Sciences, Earth Sciences, Life Sciences, Mathematical Sciences, and Physical Sciences. 

-> Postgraduates in the relevant streams can apply for this exam.

-> Candidates must download and practice questions from the CSIR NET Previous year papers. Attempting the CSIR NET mock tests are also very helpful in preparation.

More Basic Principles of Quantum Mechanics Questions

Get Free Access Now
Hot Links: teen patti gold new version 2024 teen patti - 3patti cards game downloadable content teen patti master golden india teen patti octro 3 patti rummy teen patti joy 51 bonus