Ana Sayfa Kuantum Mekaniği No Cloning Theorem: Kuantum Durumlar Neden Kopyalanamaz?

No Cloning Theorem: Kuantum Durumlar Neden Kopyalanamaz?

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Neden karlı bir pazar sabahı oturup bu yazıyı yazdığımı merak ediyorsanız bunun iki sebebi var. Birincisi COVID vakalarındaki artıştan dolayı Avusturya’nın aşılı aşısız farketmez herkesi kapsayan 3. kapanma döneminde dışarıda yapacak pek bir işimin olmaması ikincisi ise Psibilim’de yaptığım son Twitch yayınında bu konunun gündeme gelmiş olması ve canlı yayında bu işin matematiksel detayları hakkında basit bir yazı yazacağıma dair söz vermiş olmam.

Görsel: minutephysics
              

Sizler de kuantum şifreleme sistemleri, kuantum üstünlüğü, kuantum güvenliği gibi konularla ilgili bir şeyler okuduysanız veya izlediyseniz, “no cloning theorem” kavramıyla mutlaka karşılaşmışsınızdır. Kuantum kriptografiyle hiç alakam yok derseniz, o zaman da popüler bilim adı altında üretilen saçma sapan içeriklerde “kuantum klonlama” gibi spekülatif ve clickbait bir şekilde karşınıza çıkmış olması muhtemeldir. Bu kavramı Türkçe’ye en doğru şekilde çevirmek için daha önce bu konuda akademik çalışmalar yapmış hocalarımızın önerilerini kullanabiliriz. “Kopyalanamazlık Teoremi” veya “Kopya Yok Teoremi” olarak çevirmek yeterli olacaktır.

Kopya yok teoreminin bizlere söylediği şey basitçe şöyle : “Bir kuantum durumunun birebir ve bağımsız bir kopyasını yapabilecek evrensel bir kuantum kopyalama işlemi tanımlanamaz.” Bu nedenle eğer izole edilmiş tek bir kuantum durumunu bir yerden bir yere gönderiyorsak bu durumun yolda kopyalanamayacağını biliyoruz.

Örnek: tek fotonlar ile gerçekleştirilen BB84 Kuantum Kriptografi protokolü

Peki bir kuantum durum tam olarak nedir? Kuantum durum, üzerinde ölçüm yapılabilen bir sistemin olası tüm sonuçlarının olasılık dağılımını gösteren matematiksel bir ifadedir. Örnek vermek gerekirse bir kübit (qubit ) 0 ve 1 olma durumlarının lineer kombinasyonu olarak tanımlanır. Bir kübit üzerinde ölçüm yapıldığında 0 veya 1 sonuçlarından birisini elde ederiz. Aşağıdaki görsel saf bir kübit durumunun matematiksel temsilini görüyorsunuz. Alfa ve beta harfleri olasılık genliklerinden(Probability Amplitudes) başka bir şey değildir.

 
Bir Kübit Durumu

Kopya yok teoreminin matematiksel detaylarına biraz daha girmek istiyorum. Elimizdeki |Fi> (aşağıdaki görseldeki çizgili o harfi) durumunu |Boş> (blank) durumuna kopyalayacak bir evrensel operatörümüzün olduğunu düşünelim.

 
Denklem 1

Üstteki formülde gördüğümüz üzere U operatörü Fi durumuna kopyalıyor ve eşitliğin sağ tarafında elimizde iki tane Fi durumu oluyor. Bu operatör evrensel bir operatör olduğu için, kopyalama işleminin sadece Fi durumu için değil aynı zamanda Psi gibi farklı bir durum için de aynı işlemi gerçekleştirmesi gerekli.

 
Denklem 2

Üstteki formülde U operatörü aracılığıyla Psi durumunun da Boş durumuna kopyalandığını görüyorsunuz. Şimdi gelin Fi ve Psi durumlarının Hilbert uzayındaki durumlarına biraz daha yakından bakalım. Peki bu iki durumun birbiriyle iç çarpımını alırsak ne olur? Burdaki çember içerisindeki x işareti Kronecker çarpımı anlamına gelir.

 
Denklem 3

Üstteki denklemde U dagger olarak tanımlanan operatör aslında U operatörünüm Hermityen transpozundan başka bir şey değil. U operatörünün aynı zamanda üniter bir operatör olmak zorunda olduğunu unutmayın, o nedenle U dagger U = birim matris.

U dagger U işleminin sonucu bize birim operatörü verir. Bu diagonal elemanları 1 olan bir matristen ibarettir ve herhangi bir sayıyı 1 ile çarpmaktan farkı yoktur. Şimdi elimizde kalan denkleme tekrar bakalım. U dagger U’dan kurtulduğumuza göre elimizde kalanlar aşağıdaki formüle gördüğünüz gibi sadece Fi, Psi ve Blank durumlarından ibaret. Bu parantezi dağıtırken kronecker çarpımının dağılım kuralına bakmamız gerekli. (AxB)(CxD) işleminin sonucu (AC)x(BD) olarak bulunur. Aşağıdaki denklemi bu kurala göre çözdüğümüz zaman <Psi|Fi> elde ederiz.

 
Denklem 4

Peki U operatörlerini birbirleriyle çarpıp birim operatöre dönüştürmeden parantezlerin içerisine dağıtıp kopyalama işlemini yapsaydık ne olurdu? O zaman aşağıda gördüğümüz denklemi elde ederdik.

 
Denklem 5

Şimdi elimizdeki iki sonucu karşılaştırırsak bir gariplik göreceğiz. Aynı denklemden iki farklı sonuç elde ediyoruz. Denklem 5’in sonucu, Denklem 4’ün sonucunun karesine eşit, bu demek ki sonuç denklem Fi ve Psi’in iç çarpımı ya bir olmalı ya da sıfır. Sonuç eğer 1 ise bu Fi ve Psi birbirinin aynısı demektir, sonuç eğer 0 ise bu da bu iki durum birbirine diktir(orthogonal) demektir. Tüm bu sonuçlara göre basitçe şöyle söyleyebiliriz ki genel bir kuantum durumu kopyalabilecek tek bir üniter U operatörü tanımlanamaz.

 

Yusuf Karli

Universität Innsbruck

Not: Bir pazar sabahı popüler bilim yazısı olması amacıyla yazılmıştır. Bir hata gördüyseniz lütfen belirtmekten çekinmeyin.

İleri Okumalar:

  • Wootters, W.K. and Zurek, W.H.: A Single Quantum Cannot be Cloned. Nature 299 (1982), pp. 802–803
  • Dieks, D.: Communication by EPR devices. Physics Letters A, vol. 92(6) (1982), pp. 271–272
  • Buzek, V. and Hillery, M.: “Quantum cloning”. Physics World 14 (11) (2001), pp. 25–29
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