Ch4 - SHO

Ch4 - SHO - Chapter 4 Simple harmonic oscillators Energy...

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Simple harmonic oscillators Chapter 4
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Energy eigenfunctions of a harmonic oscillator Schrödinger equation for energy states: 22 2 1 n nn n d mx E m dx ϕ ω ϕϕ −+ = = 1/2 /2 () ( ) 2! x n xe H x n α π ⎛⎞ = ⎜⎟ ⎝⎠ m = = 1 2 n En =+ = n = 0, 1, 2,…. Mathematical reasons for discreteness ?
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Creation and annihilation operators 2 22 1 p Hm x m ω =+ The Hamiltonian of a 1D simple harmonic oscillator is given by, Define two non-hermitian operators: 2 mi p ax m ⎛⎞ ⎜⎟ ⎝⎠ = 2 p m + =− = annihilation operator creation operator Then the Hamiltonian can be rewritten as 1 2 Ha a + = + = where is called the number operator . aa +
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Commutation relations 2 mi p ax m ω ⎛⎞ =+ ⎜⎟ ⎝⎠ = 2 p m + =− = ,1 aa + = ,, , 1 22 2 2 m ip m ip m ip m ip xx x x mm m m ωω ⎡⎤ ⎛⎞⎡⎤ +− = + = ⎢⎥ ⎥⎢⎥ ⎝⎠⎣⎦ ⎣⎦ == = = (2) ( ) 2 2 , a a aa a a aa a a a a aa a a ++ + + + + + + + =−=− + ( )( ) 2 aa a a a a aa a a a + +++++ + + −= (3) ,( 1 ) aaa a aa aa a aaa aa a a + + + =−= + (4) 1 ) a aa a aaa a a + + + + + + = MOST important !!
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Operator method of SHO 1 2 Ha a ω + = + = Let be an energy eigenket of H , so it must be an eigenvector of n aan n λ + = n aa + By , , aaa a + ⎡⎤ = ⎣⎦ (1 ) an aaa n ++ =− ( ) ( ) ) n a aan an + , a a + = − Similarly, by ( ) ( ) ) n a a + =+ is an (unnormalized) eigenvector of is an (unnormalized) eigenvector of + Let be the lowest eigenvector 0 0 00 + = then the spectrum of must be discrete as 0 1 2 n ⎛⎞ ⎜⎟ ⎝⎠ = integer But what is ? 0
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0 00 aa λ + = All eigenvalues of must be non-negative 0 1 λ≥ ( ) ( ) 0 0( 1 ) 0 aaa a + =− the lowest eigenvalue is contradiction! An eigenvalue smaller than + Case 1 0 01 ≤< Case 2 ( ) ( ) 0 1 ) 0 a + Negative eigenvalue unless a = 0 0 + = = 0 0 λ= 1 2 n En ω ⎛⎞ =+ ⎜⎟ ⎝⎠ = n = 0, 1, 2, 3,…. . aan nn + = Therefore where is the n -th excitation state of the harmonic oscillator with the eigen-energy n 0 0 But what is ? 0
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The ground state |0 > in position space 00 a = 2 mi p x m ω ⎛⎞ + = ⎜⎟ ⎝⎠ = 0 () ( ) 0 2 xi x mx ψ ⎡⎤ +− = ⎢⎥ ⎣⎦ = = 2 /2 0 xA e = = 0 '0 ( ' ) x x = How to obtain an excited state vector |n> in position space?
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Creation and annihilation operators () ( ) (1 ) aaan n an + =− 1 n c n ⇒= lowering one excitation * 1 n na n c + 2 || n na an c + = By , we have by choosing real numbers. naan n = n cn = ( ) ) n a a λ ++ + =+ Similarly, 1 n dn + + 2 n naa n d + = By , we have 1 n + 1 n = + raising one excitation 1 nn = 11 n n + =
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,1 mn man n δ = 01 0 .. 00 2 0. 3 . a ⎛⎞ ⎜⎟ = ⎝⎠ 000 . . 10 0 02 . 0 .3 . . . a + = 1 ma n n + + =+ 0 1 Work out the matrix representation of the position operator in number state basis.
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2 mi p ax m ω ⎛⎞ =+ ⎜⎟ ⎝⎠ = 2 p m + =− = () 2 xa a m + = ( ) 2 m p ia a + =
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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Ch4 - SHO - Chapter 4 Simple harmonic oscillators Energy...

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