علاقة عدم التأكد المعممة The Generalized Uncertainty Principle !!!

[رجب مصطفى]

بسم الله الرحمن الرحيم

الحمد لله الذي صدق وعده، ونصر عبده، وأعز جنده، وهزم الأحزاب وحده، والصلاة السلام على من لا نبي بعده، رسوله الذي هدى به الأنام، وكشف به شبهات الأوهام، وعلى آله الطيبين الأطهار، وأصحابه المجاهدين الأبرار، الذين أغاظ الله بهم الكفار، وبسط بهم رحمته في جميع الأقطار


أما بعد:

فهذا إشتقاق لمبدأ عدم التأكد المعمم على أي زوج من المؤثرات !

*** ملحوظة هامة جداً كالعادة ... هذا الموضوع حصري لـ "منتدى الفيزياء التعليمي" فقط، غير ذلك سيكون واضعه سارقاً له !!!

وحتى لا أطُيل عليكم ... مع



علاقة عدم التأكد المعممة
The Generalized Uncertainty Principle



تعرض كل دارس لميكانيكا الكم لـ "مبدأ عدم التأكد Uncertainty Principle" الشهير لـ "هايزنبرج Heisenberg" والذي يربط مقدار الشك الحاصل في كمية التحرك momentum بذاك الخاص بالموضع position من خلال العلاقة:


http://latex.codecogs.com/gif.latex?%5CDelta&space;x&space;%5C;&space;%5CDelta&space; p&space;%5Cgeq&space;%5Cfrac%7B%5Chbar%7D%7B2%7 D


والآن ... سنعمل على تعميم هذه العلاقة لأي زوج من المؤثرات operators، وليكن المؤثران http://latex.codecogs.com/gif.latex?A,B !

وقبل أن نبدأ، نُذكِر بأن القيمة المتوقعة expectation value أو المتوسط mean لمؤثرٍ ما، وليكن المؤثر http://latex.codecogs.com/gif.latex?O ، يأخذ بإستخدام ’’أقواس ديراك Dirac‘‘ الصورة التالية:


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ومن دراسة الإحصاء، نجد أن الإنحراف المعياري standard deviation أو عدم التاكد أو الشك للمؤثر يُعطى من العلاقة:


مقدار الشك = الجذر التربيعي لمتوسط مربع الفرق بين القيمة ومتوسطها


وللمؤثران المعنيان، يكون مربع الشك هو:


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لاحظ أن الطرف الأيمن عبارة عن "القيمة المتوقعة" لما في داخله ! الأمر الذي يُمكننا من كتابته على الصورة:


http://latex.codecogs.com/gif.latex?%28%5CDelta&space;A%29%5E%7B2%7 D=%5Cleft&space;%5Clangle&space;%7B%5Ccolor%7Br ed%7D&space;%5Cleft&space;%28A-%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cran gle&space;%5Cright&space;%29%5E%7B2%7D%7D&space;%5Cri ght&space;%5Crangle=&space;%5Cleft&space;%5Clangle&space;%5 Cpsi&space;%7C%5Cleft&space;%28A-%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cran gle&space;%5Cright&space;%29%5E%7B2%7D%7C&space;%5Cps i&space;%5Cright&space;%5Crangle


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لاحظ مرة ثانية "التربيع" الموجود، لذا يمكن كتابتها مرة أخرى بالصورة:


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والآن ... نستطيع بكل سهولة أن نفرض الدالة الموجية الـ ket في الصورة:


http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7Bblue%7D&space;%5C left&space;%7C&space;%5Cchi&space;%5Cright&space;%5Crangle&space; =&space;%5Cleft&space;%5Cleft&space;%28A-%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cran gle&space;%5Cright&space;%29%7C&space;%5Cpsi&space;%5Cright &space;%5Crangle%7D


ومنها نجد أنّ مقابلها المزدوج الـ bra هو:


http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7Bred%7D&space;%5Cl eft&space;%5Clangle&space;%5Cchi&space;%7C&space;%5Cright&space;= &space;%5Cleft&space;%5Clangle&space;%5Cpsi&space;%7C&space;%5Cri ght&space;%5Cleft&space;%5Cleft&space;%28A-%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cran gle&space;%5Cright&space;%29&space;%7D


أي أن:


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بالمثل مع المؤثر الآخر ... لنجد أن:


http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7Bblue%7D&space;%5C left&space;%7C&space;%5Cphi&space;%5Cright&space;%5Crangle&space; =&space;%5Cleft&space;%5Cleft&space;%28B-%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;%5Cran gle&space;%5Cright&space;%29%7C&space;%5Cpsi&space;%5Cright &space;%5Crangle%7D


ومنها نجد أنّ مقابلها المزدوج الـ bra هو:


http://latex.codecogs.com/gif.latex?%7B%5Ccolor%7Bred%7D&space;%5Cl eft&space;%5Clangle&space;%5Cphi&space;%7C&space;%5Cright&space;= &space;%5Cleft&space;%5Clangle&space;%5Cpsi&space;%7C&space;%5Cri ght&space;%5Cleft&space;%5Cleft&space;%28B-%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;%5Cran gle&space;%5Cright&space;%29&space;%7D


أي أن:


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يُتبع ...

[رجب مصطفى]

***


بضرب المعادلتين (1) و (2)، نجد أن:


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ولكن من "متباينة كوشي – شفارتز Cauchy–Schwarz inequality"، نجد أن:


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Cauchy–Schwarz inequality - From Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Cauchy%E2%80%93Schwarz_inequality)


وأيضاً من جبر الأعداد المركبة complex numbers، نجد أن:


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ومن نفس الجبر، نجد أنه لأي عدد مركب z يكون:


http://latex.codecogs.com/gif.latex?%5Cleft&space;%7C&space;z&space;%5Cright&space;%7 C%5E%7B2%7D=z&space;z%5E%7B%5Cast%7D=&space;Re&space; %28z%29%5E%7B2%7D+Im%28z%29%5E%7B2% 7D&space;%5Cgeq&space;Im%28z%29%5E%7B2%7D


حيث:


http://latex.codecogs.com/gif.latex?Im%28z%29=&space;%5Cfrac%7Bz-z%5E%7B%5Cast%7D%7D%7B2i%7D


إذن:


http://latex.codecogs.com/gif.latex?%5Cleft&space;%7C&space;z&space;%5Cright&space;%7 C%5E%7B2%7D%5Cgeq%5Cleft&space;%28&space;%5Cfra c%7Bz-z%5E%7B%5Cast%7D%7D%7B2i%7D&space;%5Crigh t&space;%29%5E%7B2%7D


وعليه يُصبح:


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أو:


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وبأخذ الجذر التربيعي، نحصل على العلاقة:


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يُتبع ...

[رجب مصطفى]

***


والآن سنعمل على إيجاد قيمة كل حد من الطرف الأيمن على النحو التالي:


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أي:


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أو:


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وهنا يجب التذكِرة بأن القيمة المتوقعة هي قيمة ""ثابته (يعني "عدد")""، مما يجعلها تخرج خلال عمليات التفاضل !

وعليه، نجد أن:


http://latex.codecogs.com/gif.latex?%5Cleft&space;%5Clangle&space;%5Cpsi&space; %7CAB%7C&space;%5Cpsi&space;%5Cright&space;%5Crangle= %5Cleft&space;%5Clangle&space;AB&space;%5Cright&space;%5Cra ngle&space;%5Cquad,


http://latex.codecogs.com/gif.latex?%5Cleft&space;%5Clangle&space;%5Cpsi&space; %7CA%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;%5 Crangle%7C&space;%5Cpsi&space;%5Cright&space;%5Crangl e=&space;%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;%5C rangle&space;%5Cleft&space;%5Clangle&space;%5Cpsi&space;%7C A%7C&space;%5Cpsi&space;%5Cright&space;%5Crangle&space;=&space;%5 Cleft&space;%5Clangle&space;B&space;%5Cright&space;%5Crangl e&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cr angle&space;%5Cquad,


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http://latex.codecogs.com/gif.latex?%5Cleft&space;%5Clangle&space;%5Cpsi&space; %7C&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5 Crangle&space;%5Cleft&space;%5Clangle&space;B&space;%5Crigh t&space;%5Crangle%7C&space;%5Cpsi&space;%5Cright&space;%5Cr angle=&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright &space;%5Crangle&space;%5Cleft&space;%5Clangle&space;B&space;%5Cr ight&space;%5Crangle&space;%5Cleft&space;%5Clangle&space;%5 Cpsi&space;%7C&space;%5Cpsi&space;%5Cright&space;%5Crangle&space; =&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cr angle&space;%5Cleft&space;%5Clangle&space;B&space;%5Cright&space; %5Crangle


حيث:


http://latex.codecogs.com/gif.latex?%5Cleft&space;%5Clangle&space;%5Cpsi% 7C%5Cpsi&space;%5Cright&space;%5Crangle&space;=1&space;%5Cq uad,&space;%5Cquad&space;%5Cleft&space;%5Clangle&space;A&space;%5 Cright&space;%5Crangle&space;%5Cleft&space;%5Clangle&space; B&space;%5Cright&space;%5Crangle&space;=&space;%5Cleft&space;%5Cl angle&space;B&space;%5Cright&space;%5Crangle&space;%5Cleft&space; %5Clangle&space;A&space;%5Cright&space;%5Crangle


إذن:


http://latex.codecogs.com/gif.latex?%5Cleft&space;%5Clangle&space;%5Cchi% 7C%5Cphi&space;%5Cright&space;%5Crangle&space;=%5Clef t&space;%5Clangle&space;AB&space;%5Cright&space;%5Crang le&space;-&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cra ngle&space;%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;% 5Crangle&space;%5Cquad&space;%5Cquad&space;%5Cto&space;%286 %29


وبالمثل مع الحد الآخر، لنجد أن:


http://latex.codecogs.com/gif.latex?%5Cleft&space;%5Clangle&space;%5Cphi% 7C%5Cchi&space;%5Cright&space;%5Crangle&space;=%5Clef t&space;%5Clangle&space;BA&space;%5Cright&space;%5Crang le&space;-&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cra ngle&space;%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;% 5Crangle&space;%5Cquad&space;%5Cquad&space;%5Cto&space;%287 %29


وهذا تمرين بسيط !

والآن نعوض في المعادلة رقم (5):


http://latex.codecogs.com/gif.latex?%5CDelta&space;A%5C;&space;%5CDelta&space;B &space;%5Cgeq&space;%5Cfrac%7B%5Cleft&space;%5Clangle &space;AB&space;%5Cright&space;%5Crangle&space;-&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright&space;%5Cra ngle&space;%5Cleft&space;%5Clangle&space;B&space;%5Cright&space;% 5Crangle-%5Cleft&space;%5Clangle&space;BA&space;%5Cright&space;%5Cra ngle&space;+&space;%5Cleft&space;%5Clangle&space;A&space;%5Cright &space;%5Crangle&space;%5Cleft&space;%5Clangle&space;B&space;%5Cr ight&space;%5Crangle%7D%7B2i%7D


إذن:


http://latex.codecogs.com/gif.latex?%5CDelta&space;A%5C;&space;%5CDelta&space;B &space;%5Cgeq&space;%5Cfrac%7B%5Cleft&space;%5Clangle &space;AB&space;%5Cright&space;%5Crangle&space;-%5Cleft&space;%5Clangle&space;BA&space;%5Cright&space;%5Cra ngle&space;%7D%7B2i%7D=&space;&space;%5Cfrac%7B%5Clef t&space;%5Clangle&space;AB&space;-BA&space;%5Cright&space;%5Crangle&space;%7D%7B2i%7 D


أو:


http://latex.codecogs.com/gif.latex?%5CLARGE&space;%5CDelta&space;A%5C;&space;% 5CDelta&space;B&space;&space;%5Cgeq&space;%5Cfrac%7B%5Cleft &space;%5Clangle&space;%5Cleft&space;[A,B&space;%5Cright&space;]%5Crangle&space;%7D%7B2i%7D


وهي العلاقة المطلوبة، حيث:


http://latex.codecogs.com/gif.latex?%5Cleft&space;[A,B&space;%5Cright&space;]=AB-BA


هي معادلة التبادلية commutator للمؤثرين !

هذا ... والله أعلى وأعلم


وآخر دعوانا أن الحمد لله رب العالمين
وصلِّ اللهم وبارك على نبي الرحمة سيد ولد أدم أبى القاسم "محمد بن عبد الله" وعلى آله الطيبين الطاهرين وعلى أصحابه الغر الميامين أجمعين وسلم تسليماً كثيراً
لا تنسونا من صالح دعائكم

[محمد عريف]

بسم الله ماشاء الله يارجب

بجد تسلم اياديك ... مجهود رائع وكبييييييييييييير

اتذكر أننا أخذنا فقط المعادلة الأخيرة فقط مباشرة

ولم أكن أعلم أن إثباتها بهذه الصورة

بارك الله فيك .. وزادك علماً ... ونفع بك الأمة

وكل عام وانت بخير

مع وافر احترامي وتقديري

[رجب مصطفى]

بسم الله ماشاء الله يارجب

بجد تسلم اياديك ... مجهود رائع وكبييييييييييييير

اتذكر أننا أخذنا فقط المعادلة الأخيرة فقط مباشرة

ولم أكن أعلم أن إثباتها بهذه الصورة

بارك الله فيك .. وزادك علماً ... ونفع بك الأمة

وكل عام وانت بخير

مع وافر احترامي وتقديري


أهلاً ومرحباً ... أخي وحبيب قلبي محمد باشا عريف

في أمان الله

[ammar alrawi]

الله يبارك فيكم ويجزيكم كل خير