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PHY4221Fall05_Notes3

# PHY4221Fall05_Notes3 - PHY4221 Quantum Mechanics I Fall...

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PHY4221 Quantum Mechanics I Fall 2004 Examples in One Dimension: 1. Free Particle The Hamiltonian operator ˆ H of a free particle consists of the kinetic energy term only: ˆ H = ˆ p 2 2 m . (1) Then what are the eigenvalues and eigenvectors of ˆ H ? Since any eigenvector of ˆ p is also an eigenvector of ˆ H , we can easily see that the eigenvalues of ˆ H are simply given by E p = p 2 2 m (2) where p ( -∞ , ) and the corresponding eigenvectors are {| p i} . It should be noted that there exists a two-fold degeneracy for each eigenvalue E p ; in other words, there are two diﬀerent eigenvectors, namely {| ± p i} , associated with the same eigenvalue E p . Next, we consider the question: “How does the state vector of the free particle evolve in time?” To answer this question, we need to solve the time-dependent Schr¨ odinger equation: ˆ H | ψ ( t ) i = i ¯ h ∂t | ψ ( t ) i (3) whose solution is formally given by | ψ ( t ) i = exp - i t ¯ h ˆ H ± | ψ (0) i = Z -∞ dp | p ih p | exp - i t ¯ h ˆ H ± | ψ (0) i = Z -∞ dp | p i exp ( - i p 2 t 2 m ¯ h ) h p | ψ (0) i = ⇒ h p | ψ ( t ) i = exp ( - i p 2 t 2 m ¯ h ) h p | ψ (0) i . (4) 1

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If the initial state vector | ψ (0) i , or equivalently the initial wave function h p | ψ (0) i in the momentum space, of the free particle is given, then we can determine the subsequent time evolution of the system, namely determining the state vector | ψ ( t ) i . It is interesting to note that the wave function h p | ψ ( t ) i at any later time t > 0 diﬀers from the initial wave function simply by a phase factor. Furthermore, the evolution operator exp n - i t ¯ h ˆ H o of the system can be represented by exp - i t ¯ h ˆ H ± = Z -∞ dp | p i exp ( - i p 2 t 2 m ¯ h ) h p | . Similarly, if the initial wave function h x | ψ (0) i in the coordinate space is given instead, then the state vector | ψ ( t ) i can be formally expressed as | ψ ( t ) i = Z -∞ dx | x ih x | exp - i t ¯ h ˆ H ± | ψ (0) i = Z -∞ dx | x i exp ( i t ¯ h 2 m 2 ∂x 2 ) h x | ψ (0) i = Z -∞ dx | x ih x | exp - i t ¯ h ˆ H ±²Z -∞ dx 0 | x 0 ih x 0 | | ψ (0) i = Z -∞ dx | x i Z -∞ dx 0 K ( x,t ; x 0 , 0) h x 0 | ψ (0) i = ⇒ h x | ψ ( t ) i = Z -∞ dx 0 K ( ; x 0 , 0) h x 0 | ψ (0) i . (5) where the kernel K ( ; x 0 , 0) ≡ h x | exp n - i t ¯ h ˆ H o | x 0 i is given by K ( ; x 0 , 0) = h x | exp - i t ¯ h ˆ H -∞ dp | p ih p | | x 0 i = Z -∞ dp exp ( - i p 2 t 2 m ¯ h ) h x | p ih p | x 0 i = Z -∞ dp 2 π ¯ h exp ( - i p 2 t 2 m ¯ h ) exp ( i p ( x - x 0 ) ¯ h ) = r m 2 πi ¯ ht exp - m 2 i ¯ ht ( x - x 0 ) 2 ± . (6) The kernel K ( ; x 0 , 0) is also commonly known as the propagator of the free particle. Provided that the particle is initially localised at the point x 0 , i.e. | ψ (0) i is the eigenstate of the position operator ˆ x with eigenvalue x 0 , the wave function h x | ψ ( t ) i at any later 2
time t > 0 is simply given by h x | ψ ( t ) i = K ( x,t ; x 0 , 0) . (7) Exercise: How does the wave function h p | ψ ( t ) i look like if the particle is initially localised at the point x 0 ?

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PHY4221Fall05_Notes3 - PHY4221 Quantum Mechanics I Fall...

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