दिया है, द्विक्परिवर्तक \(\left[\widehat{A}^2,\widehat{B}\right] =\left[\widehat{A},\widehat{B}\right]\widehat{A}+\widehat{A}\left[\widehat{A},\widehat{B}\right]\), \(\left[x,[\widehat{p^2_x},x]\right]\) का मान है

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This question was previously asked in

CSIR-UGC (NET) Chemical Science: Held on (18 Sept 2022)

Answer 1

सिद्धांत:-

रैखिक और स्थिति संवेग के लिए:

[x̂, p̂x] = iℏ ....(1)

[p̂x, x̂] = -iℏ ...(2)

[x̂n, p̂x] = nxn-1 [x, p̂x] ....(3)

[p̂xn, x̂] = npxn-1[p̂x, x̂] .....(4)

व्याख्या:-

\(\left[\widehat{A}^2,\widehat{B}\right] =\left[\widehat{A},\widehat{B}\right]\widehat{A}+\widehat{A}\left[\widehat{A},\widehat{B}\right]\)

= \(\left[x, \left[\widehat{p_x},\widehat{x}\right]\widehat{p_x}+\widehat{p_x}\left[\widehat{p_x},\widehat{x}\right ]\right]\)...........(5)

[p̂x, x̂] = -iℏ

\(\left[x, \left[\widehat{p_x},\widehat{x}\right]\widehat{p_x}+\widehat{p_x}\left[\widehat{p_x},\widehat{x}\right ]\right]\)

= \(\left[x, \left ( -i\hbar\right )\widehat{p_x}+\widehat{p_x}\left ( -i\hbar\right )\right]\)

= \(\left[x, -2i\hbar \widehat{p_x}\right]\)

= \( -2i\hbar \left[x, \widehat{p_x}\right]\)

= \( -2i\hbar \times i\hbar\)

= \( -2i^2\hbar^2 \)

= \( 2\hbar^2 \)

निष्कर्ष:-

इसलिए, \(\left[x,[\widehat{p^2_x},x]\right]\) का मान \( 2\hbar^2 \) है।