3. predavanje iz Kvantne mehanike
Table of Contents
1. Formalizem kvantne mehanike
Hermitsko adjungiran operator
Če velja za poljubna operatorja \( A, B \) in za vsaka vektorja \( \phi, \psi \)
\[ \left\langle \phi \middle| A \middle| \Psi \right\rangle = \left\langle B \phi \middle| \Psi \right\rangle \],
potem je \( B = A ^{\dagger} \) hermitsko adjungiran operator.
Lastnosti operatorja so:
za \( z \in \mathbb{C} \) je \( A = z B \) velja
\[ A^{\dagger} = z^{\ast} B ^{\dagger} \]
Dokaz: \( z \) daš na levo (cinično).
za \( A = \left| m \right\rangle \left\langle n \right|\) velja
\[ A ^{\dagger} = \left| n \right\rangle \left\langle m \right| \]
Dokaz:
\begin{align*} \left\langle \phi \middle| A \middle| \Psi \right\rangle &= \left\langle \phi \middle| m \right\rangle \left\langle n \middle| \Psi \right\rangle \\ &= \left( \left\langle \Psi \middle| n \right\rangle \left\langle m \middle| \phi \right\rangle\right) ^{\ast} \\ &= \left( \left\langle \Psi \middle| A \phi \right\rangle \right) ^{\ast} \\ &= \left\langle A^{\dagger} \phi \middle| \Psi \right\rangle \end{align*}velja
\[ \left( \mu A + B \right)^{\dagger} = \mu^{\ast} A^{\dagger} + \lambda^{\ast} B^{\dagger} \]
velja
\[ \left( AB \right) ^{\dagger} = B^{\dagger} A^{\dagger} \]
Dokaz:
\begin{align*} \left\langle \phi \middle| AB \middle| \Psi \right\rangle &= \left\langle A^{\dagger} \phi \middle| B \middle| \Psi \right\rangle \\ &= \left\langle B^{\dagger} A^{\dagger} \phi \middle| \Psi \right\rangle \end{align*}Kako najti hermitsko adjungiran operator \( A \)?
\( A \) zapišemo v bazi
\[ A = I A I = \sum\limits_{m, n}^{} \left| m \right\rangle A_{mn} \left\langle n \right| \]
Hermitsko adjungiramo bazo in dobimo
\[ A ^{\dagger} = \sum\limits_{m, n}^{} \left| n \right\rangle A_{mn}^{\ast} \left\langle n \right| \]
Za ta operator velja \( \left( A^{\dagger} \right)^{\dagger} = A \)
Za lepši zapis zamenjamo \( n \) z \( m \) in obratno. Tako se poenostavi(?) v
\[ A^{\dagger} = \sum\limits_{m, n}^{} \left| m \right\rangle A_{nm}^{\ast} \left\langle n \right| \]
kjer je \( A_{mn} \) postal \( A_{nm}^{\ast} \).
Sebi adjungirani operatorji so operatorji, če velja za operator \( A \)
\[ A = A^{\dagger} \quad \mathrm{in} \quad \mathcal{D} \left( A \right) = \mathcal{D} \left( A^{\dagger} \right) \]
Za \( \phi \) in \( \Psi \) \( \left\langle \phi \middle| A \middle| \Psi \right\rangle \) je enak robni pogoj, če je operator sebi adjungiran. Spektralni teorem
\[ A \left| n \right\rangle = a_n \left| n \right\rangle \]
kjer je \( \left\{ \left| n \right\rangle \right\} \) baza v \( L ^2 \). To je izredno pomembno, saj če iščemo bazo, so tukaj lastne funkcije kar baza.
Unitarni operator je operator, kjer velja
\[ U^{-1} = U^{\dagger} \]
Velja \( U^{-1} U = U U^{-1} = I \) in posledično tudi \( U^{\dagger} U = U U^{\dagger} = I \).
Imamo vektorja \( \left| \phi \right\rangle \) in \( \left| \psi \right\rangle \). Zapišemo nov vektor
\[ \left| \tilde{\phi} \right\rangle = U \left| \phi \right\rangle \]
Zaradi definicije unitarnega operatorja velja
\[ \left| \phi \right\rangle = U^{\dagger} \left| \tilde{\phi} \right\rangle \]
Potem zapišemo
\[ \left\langle \phi \middle| \Psi \right\rangle = \left\langle U ^{\dagger} \tilde{\phi} \middle| U^{\dagger} \tilde{\Psi} \right\rangle = \left\langle \tilde{\phi} \middle| U U^{\dagger} \middle| \tilde{\Psi} \right\rangle = \left\langle \tilde{\phi} \middle| \tilde{\Psi} \right\rangle \]
bla
\begin{align*} \left\langle \phi \middle| \Psi \right\rangle &= \left\langle U^{\dagger} \tilde{\phi} \middle| A \middle| U^{\dagger} \tilde{\Psi} \right\rangle \\ &= \left\langle \tilde{\phi} \middle| U A U ^{\dagger} \middle| \tilde{\Psi} \right\rangle \\ &= \left\langle \tilde{\phi} \middle| \tilde{A} \middle| \tilde{\Psi} \right\rangle \end{align*}kjer je \( \tilde{A} = UA U^{\dagger} \).
Naj velja \( A \left| a \right\rangle = a \left| a \right\rangle\). Potem velja, da so \( a \) tudi lastne vrednosti operatorja \( \tilde{A} \)
\begin{align*} UAU^{\dagger}U \left| a \right\rangle &= a U \left| a \right\rangle \\ \tilde{A} \left| \tilde{a} \right\rangle &= a \left| \tilde{a} \right\rangle \end{align*}Če je \( K = K^{\dagger} \) (hermitski), potem je operator \( U \) definiran kot
\[ U = e ^{\mathrm{i} K} \]
unitaren operator.
Brez dokaza.
Vsak enoparametrični unitarni operator ima naslednjo obliko
\[ U(s) = e^{\mathrm{i} s K} \]
kjer je \( K = K^{\dagger} \). Operatorju \( K \) rečemo generator.
Časovni razvoj stanja z unitarnim operatorjem
Operator lahko razvijemo v vrsto. Za \( x \in \mathbb{C} \) lahko zapišemo
\[ f(x) = \sum\limits_n^{} c_n x^n \]
Definiramo operator \( \hat{A} \) in velja
\[ f \left( \hat{A} \right) = \sum\limits_n^{} c_n \hat{A}^n \]
Kot primer vzamemo funkcijo \( f(x) = e^x \) in operator \( \hat{A} = \frac{\partial }{\partial x} \). Potem je
\[ e^{\frac{\partial }{\partial x} } = 1 + \frac{\partial }{\partial x} + \frac{1}{2!} \frac{\partial ^2 }{\partial x ^2} + \ldots + \frac{1}{n!} \frac{\partial ^n }{\partial x ^n} \]
\begin{align*} f \left( \hat{A} \right) \left| a_n \right\rangle &= \sum\limits_m^{} c_m \hat{A}^m \left| a_n \right\rangle \\ \hat{A} \left| a_n \right\rangle &= a_n \left| a_n \right\rangle \\ &= \sum\limits_m^{} c_m \left( a_n \right)^m \left| a_n \right\rangle = f \left( a_n \right) \left| a_n \right\rangle \end{align*}Actual stuff nimamo \( \Psi(t) \), vendar imamo začetni pogoj \( \Psi(0) \).
Preko
\[ H \left| \phi_n \right\rangle = E_n \left| \phi_n \right\rangle \]
potem
\[ \left| \Psi(t) \right\rangle = \sum\limits_n^{} \left\langle \phi_n \middle| \Psi(0) \right\rangle \exp \left\{ - \mathrm{i \frac{E_n t}{\hbar}} \right\} \left| \phi_n \right\rangle \]
kjer je npr. \( \phi_n (x) = C_n \sin k_n x \)
Kakor smo zapisali višje v tej točki, lahko nadomestimo \( E_n \) z operatorjem
\[ \left| \Psi(t) \right\rangle = \sum\limits_n^{} \left\langle \phi_n \middle| \Psi(0) \right\rangle \exp \left\{ - \mathrm{i \frac{H t}{\hbar}} \right\} \left| \phi_n \right\rangle = \exp \left\{ - \mathrm{i} \frac{H t}{\hbar} \right\} \sum\limits_n^{} \left\langle \phi_n \middle| \Psi(0) \right\rangle \left| \phi_n \right\rangle \]
Ker velja komutacija pri skalarnem produktu, lahko vsoto pretvorimo v
\[ \left| \Psi(t) \right\rangle = \exp \left\{ - \mathrm{i} \frac{Ht}{\hbar} \right\} \sum\limits_n^{} \left| \phi_n \right\rangle \left\langle \phi_n \right| \left| \Psi(0) \right\rangle \]
Nadalje \( \left| \phi_n \right\rangle \left\langle \phi_n \right| = I \) in dobimo zapis
\[ \left| \Psi(t) \right\rangle = \exp \left\{ - \mathrm{i} \frac{Ht}{\hbar} \right\} \left| \Psi(0) \right\rangle \]
Dobimo časovni razvoj
\[ U(t) = \exp \left\{ - \mathrm{i} \frac{Ht}{\hbar} \right\} \]
Reprezentacija \( p \) in \( x \) (ang. x- or p-representation)
Spomnimo se Fourierovega integrala
\begin{equation} \label{eq:1} f(x) = \int\limits_{-\infty}^{\infty} \tilde{f} \left( k \right) e^{\mathrm{i} k x} \, \mathrm{d} k \end{equation}Za dovolj pohlevne funkcije predpostavimo obstoj inverza
\[ \tilde{f}(k) = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} f(x') e^{- \mathrm{i} k x'} \, \mathrm{d} x' \]
V enačbo \ref{eq:1} vstavimo definicijo inverz in zamenjamo vrstni red integriranja
\begin{equation} \label{eq:2} f(x) = \int\limits_{-\infty}^{\infty} \left( \frac{1}{2\pi} \int\limits_{- \infty}^{\infty} e^{- \mathrm{i} k \left( x' - x \right)} \, \mathrm{d} k \right) f(x') \, \mathrm{d} x' \end{equation}Po stopinjah Diraca, uvedemo objekt
\[ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{- \mathrm{i}k \left( x' - x \right) } \, \mathrm{d} k = \delta (x ' - x) \]
kot Diracovo delta funkcijo.
Enačbo \ref{eq:2} z novo definicijo zapišemo
\[ f(x) = \int\limits_{-\infty}^{\infty} \delta \left( x' - x \right) f(x') \, \mathrm{d} x' \]
Vzemimo kot primer prost delec s potencialom \( V(x) = 0 \). Zanimata nas energija \( H = \frac{p ^2}{2m} \) in lastna funkcija energije
\[ p \left| \phi_{p_0} \right\rangle = p_0 \left| \phi_{p_0} \right\rangle, \quad p_0 \in \mathbb{R} \]
kjer je \( p \) operator in \( p_0 \) lastna vrednost. Upoštevamo definicijo operatorja gibalne količine
\[ - \mathrm{i} \hbar \frac{\partial }{\partial x} \phi_{p_0} (x) = p_0 \phi_{p_0} (x) \]
katere rešitev je
\[ \phi_{p_0} = C e^{\mathrm{i} \frac{p_0}{\hbar} x} \]
in velja \( \left| \phi_{p_0} \right| ^2 = \left| C \right| ^2\). Funkcija torej ni normirana in se, I guess, ne da integrirati. Težavo lahko rešimo s Fourierovo transformacijo. Definiramo \( C = \frac{1}{\sqrt{2 \pi \hbar}} \). Faktor \( 2\pi \) smo uporabili zaradi definicije delta funkcije, medtem ko je valovni vektor definiran \( k = \frac{p}{\hbar} \). Hkrati pa za delta funkcijo velja \( \delta(ax) = \frac{1}{\left| a \right|} \delta(x) \). Rešitev se preobrazi v
\begin{equation} \label{eq:3} \phi_{p_0} (x) = \frac{1}{\sqrt{2\pi\hbar}} e^{\mathrm{i} \frac{p_0 x}{\hbar}} \end{equation}Velja
\begin{equation} \label{eq:4} \int\limits_{}^{}\phi_{p_0} ^{\ast} (x) \phi_p (x) \, \mathrm{d} x = \delta \left( p_0 - p \right) \end{equation}Delujemo z operatorjem \( \hat{p} \) na \( \Psi(x) \), ki ga zapišemo s Fourierovim integralom:
\[ \hat{p} \Psi(x) = - \mathrm{i} \hbar \frac{\partial }{\partial x} \Psi(x) = \int\limits_{}^{} \tilde{\Psi} (p) \left( - \mathrm{i} \hbar \frac{\partial }{\partial x} \phi_p (x) \right) \, \mathrm{d} p \]
Note: Fourierovo transformacijo bomo pisali oblike
\[ \Psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int\limits_{-\infty}^{\infty} \tilde{\Psi} (p) e^{\mathrm{i} \frac{p x}{\hbar}} \, \mathrm{d} p, \quad \text{in}\quad \tilde{\Psi}(p) = \frac{1}{\sqrt{2\pi\hbar}} \int\limits_{-\infty}^{\infty} \Psi(x) e^{- \mathrm{i} \frac{px}{\hbar}} \, \mathrm{d} x \]
Poudarek na ulomke pred integrali. Zaradi definicije \ref{eq:3} zapišemo
\begin{align*} \Psi(x) &= 1 \int\limits_{-\infty}^{\infty} \tilde{\Psi}(p) \phi_p (x) \, \mathrm{d} p \\ \tilde{\Psi} (p) &= 1 \int\limits_{-\infty}^{\infty} \Psi(x) \phi_p^{\ast} (x) \, \mathrm{d} p \end{align*}Koeficient v razvoju dobimo z integriranjem produkta funkcije in operatorja.
Nadaljujemo naprej s \( \hat{p} \Psi(x) \). Odvod \( - \mathrm{i} \hbar \frac{\partial }{\partial x} \phi_p (x) \) nam bo zaradi definicije podal integral
\[ \hat{p} \Psi(x) = \int\limits_{}^{} p \p \tilde{\Psi} (p) \phi_p (x) \, \mathrm{d} p \]
Iz tega sledi
\[ \hat{p}^n \Psi(x) = \int\limits_{}^{}p^n \tilde{\Psi}(p) \phi_p (x) \, \mathrm{d} p \]
Poglejmo si še primer, ko množimo valovno funkcijo \( \Psi(x) \) z \( x \). Dobimo
\begin{align*} x \Psi(x) &= \int\limits_{}^{} \vec{\Psi}(p) x \phi_p (x) \, \mathrm{d} p \\ &= \int\limits_{}^{} \tilde{\Psi}(p) \left( - \mathrm{i} \hbar \frac{\partial }{\partial p} \phi_p (x) \right) \, \mathrm{d} p \\ &= \int\limits_{}^{} \left( \mathrm{i} \hbar \frac{\partial }{\partial p} \tilde{\Psi}(p) \right) \phi_p (x) \, \mathrm{d} p \end{align*}Tako dobimo relacijo
\[ x \Psi(x) \iff \mathrm{i} \hbar \frac{\partial }{\partial p} \tilde{\Psi}(p) \]
Prav tako velja
\[ x^n \Psi(x) \iff \left( \mathrm{i} \hbar \frac{\partial }{\partial p} \right)^n \tilde{\Psi}(p) \]
Ta formalizem nam omogoči obravnavo in normiranost funkcij, ki niso iz \( L ^2 \).
Verjetnostna amplituda \( \left\langle x \middle| \Psi \right\rangle \) in \( \left\langle p \middle| \Psi \right\rangle \)
Naj bo funkcijo \( \Psi(x, t) \in L ^2 \)
Enačbo
\[ \left| \Psi \right\rangle = \int\limits_{}^{} \tilde{\Psi} (p) \left| p \right\rangle \, \mathrm{d} p \]
kjer je \( \left| p \right\rangle \) oz. \( \frac{1}{\sqrt{2\pi\hbar}} e^{\mathrm{i} \frac{px}{\hbar}} \), pomnožimo s \( \left\langle p_1 \right| \), zapišemo
\[ \left\langle p_1 \middle| \Psi \right\rangle = \int\limits_{}^{} \tilde{\Psi}(p) \left\langle p_1 \middle| p_0 \right\rangle \, \mathrm{d} p = \tilde{\Psi} (p_1) \]
kjer smo uporabili \ref{eq:4} in velja \( \left\langle p_0 \middle| p \right\rangle = \delta \left( p_0 - p \right)\)
Sledeče spada pod točko 13, ampak oh, well.
Povedali smo, kaj je
\[ \hat{p} \phi_{p_0} (x) = p_0 \phi_{p_0} (x) \]
Zanima pa nas, kaj je v tem primeru operator \( \hat{x} \).
\[ \hat{x} \Psi_0 (x) = x_0 \Psi_0 (x) = x \Psi(x) \]
Funkcijo \( \Psi_0 \) razvijemo po Fourieru in jo lahko zapišemo kot
\[ \hat{x} \Psi_0 (x) = x \int\limits_{}^{} \tilde{\Psi}_0 (p) \phi_p (x) \, \mathrm{d} p \]
Hkrati pa vemo od prej, da mora biti množenje z \( x \) enak odvodu po \( p \), torej
\[ x \int\limits_{}^{} \tilde{\Psi}_0 (p) \phi_p (x) \, \mathrm{d} p = \int\limits_{}^{} \left( \mathrm{i} \hbar \frac{\partial }{\partial p} \tilde{\Psi}_0 (p) \right) \phi_p (x) \, \mathrm{d} p = x_0 \int\limits_{}^{} \tilde{\Psi}_0 (p) \phi_p (x) \, \mathrm{d} p \]
Sedaj smo vzpostavili identiteto
\[ \mathrm{i} \hbar \frac{\partial }{\partial p} \tilde{\Psi}_0 (p) = x_0 \tilde{\Psi}_0 (p) \]
ki nam poda
\[ \frac{1}{\sqrt{2 \pi \hbar}} e^{-\mathrm{i} \frac{p x_0}{ \hbar} } = \phi^{\ast}_p (x_0) \]
Tako sledi, da lahko zapišemo
\[ \Psi_0 (x) = \int\limits_{}^{} \phi^{\ast}_p (x_0) \phi_p (x) \, \mathrm{d} p = \delta (x_0 - x) = \left| x_0 \right\rangle \]
To je lastni
Poljuben vektor razvijemo po lastnih stanjih gibalne količine
\[ \left| \Psi \right\rangle = \int\limits_{}^{} \tilde{\Psi}(p) \left| p \right\rangle \, \mathrm{d} p \]
Pomnožimo ga z lastno funkcijo \( x \) \( \left\langle x_0 \right| \)
\[ \left\langle x_0 \middle| \Psi \right\rangle = \int\limits_{}^{} \tilde{\Psi}(p) \left\langle x_0 \middle| p \right\rangle \, \mathrm{d} p \]
Iz prej izpeljanih stvari vidimo, da je
\begin{align*} \left\langle x_0 \middle| p \right\rangle &= \int\limits_{}^{} \delta (x_0 - x) \frac{1}{\sqrt{2 \pi \hbar}} e^{\mathrm{i} \frac{px}{\hbar}} \, \mathrm{d} x \\ &= \frac{1}{\sqrt{2\pi \hbar}} e^{\mathrm{i} \frac{px_0}{\hbar}} \end{align*}Identiteto upoštevamo v integralu, dobimo
\[ \left\langle x_0 \middle| \Psi \right\rangle = \int\limits_{}^{} \tilde{\Psi} (p) \frac{1}{\sqrt{2 \pi \hbar}} e^{\mathrm{i} \frac{p x_0}{\hbar}} \, \mathrm{d} p = \Psi(x_0) \]
Torej to velja
\begin{align*} \left\langle x \middle| \Psi \right\rangle &= \Psi(x) \\ \left\langle p \middle| \Psi \right\rangle &= \tilde{\Psi}(p) \end{align*}Z upoštevanjem komutacije skalarnega produkta lahko preko zgornje zveze torej zapišemo
\begin{align*} \left| \Psi \right\rangle &= \int\limits_{}^{} \Psi(x) \left| x \right\rangle \, \mathrm{d} x \\ &= \int\limits_{}^{} \left\langle x \middle| \Psi \right\rangle \left| x \right\rangle \, \mathrm{d} x \\ &= \left( \int\limits_{}^{} \left| x \right\rangle \left\langle x \right|\right) \, \mathrm{d} x \left| \Psi \right\rangle \\ &= \left( \int\limits_{}^{} \left| p \right\rangle \left\langle p \right|\right) \, \mathrm{d} p \left| \Psi \right\rangle \end{align*}saj je skalarni produkt v oklepajih enak identiteti.