Working in units such that m = 1 and ℏ = 1 , a one-dimensional quantum-mechanical particle moves under the influence of the potential V ( x ) = 2 1 x 2 + 2 1 x so that the time-independent Schrodinger equation for the wave-function is
− 2 1 d 2 x d 2 ψ + 2 1 x 2 ψ ( x ) + 2 1 x ψ ( x ) = E ψ ( x )
This system has countably many energy levels E n for n ∈ N ∪ { 0 } . It can be shown that n = 0 ∑ ∞ E n 2 1 = π a + b G where G is the Catalan constant , and a , b are positive integers. Find the value of a + b .
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We shall obtain the energy levels in two ways: firstly using differential equation methods, and then using differential operators.
First note that ψ ( x ) = φ ( x + ε ) satisfies the differential equation − 2 1 ψ ′ ′ ( x ) + 2 1 ( x + ε ) 2 ψ ( x ) = E ψ ( x ) precisely when φ satisfies the differential equation − 2 1 φ ′ ′ ( x ) + 2 1 x 2 φ ( x ) = E φ ( x ) Thus E is an energy level for the Schrodinger equation − 2 1 ψ ′ ′ ( x ) + 2 1 x 2 ψ ( x ) + ε x ψ ( x ) = E ψ ( x ) precisely when E = E − 2 1 ε 2 where E is an energy level for the Schrodinger equation − 2 1 ψ ′ ′ ( x ) + 2 1 x 2 ψ ( x ) = E ψ ( x ) ( ⋆ ) Let us find the energy levels of ( ⋆ ) . If we write ψ ( x ) = e − 2 1 x 2 f ( x ) , the differential equation becomes f ′ ′ ( x ) − 2 x f ′ ( x ) + ( 2 E − 1 ) f ( x ) = 0 Looking for a series expansion solution of this differential equation of the form f ( x ) = n = 0 ∑ ∞ a n x n we obtain the following recurrence relation for the coefficients: ( n + 2 ) ( n + 1 ) a n + 2 + ( 2 E − 2 n − 1 ) a n = 0 If 2 E = 2 N + 1 for some N ≥ 0 then we can obtain a polynomial solution for f ( x ) as follows: if N is even/odd then a n will have to be zero for all odd/even n , but we see that a N + 2 k = 0 for all k ≥ 1 , and hence f ( x ) is a polynomial of degree N of parity ( − 1 ) N . In fact, f ( x ) is the N th Hermite polynomial.
If 2 E is not an odd integer, then we will deduce that a n = 0 for all n , and that a n a n + 2 ∼ n 2 n → ∞ and this implies that f ( x ) ∼ e x 2 as x → ∞ , which means that we do not have a bound state. Thus 2 E must be an odd integer, and so the energy levels of ( ⋆ ) are E n = n + 2 1 for all n ∈ N ∪ { 0 } . Thus the energy levels of the original differential equation are E n = n + 2 1 − 2 1 ε 2 for n ∈ N ∪ { 0 } . In the particular case we have here, ε = 2 1 , and hence E n = n + 4 1 .
Now for the differential operator method. Let us first consider the standard harmonic oscillator Hamiltonian H 0 = − 2 1 ∂ 2 x ∂ 2 + 2 1 x 2 = 2 1 ( P 2 + Q 2 ) where [ Q , P ] = i I . If we define the so-called lowering and raising operators A = 2 1 ( P − i Q ) A ⋆ = 2 1 ( P + i Q ) then H 0 = A ⋆ A + 2 1 I = A A ⋆ − 2 1 I [ A , A ⋆ ] = I [ H 0 , A ] = − A [ H 0 , A ⋆ ] = A
If ψ is an eigenstate of H 0 with energy level E + 2 1 , then we see that H 0 A ψ ∥ A ψ ∥ 2 = = ( E − 2 1 ) A ψ E ∥ ψ ∥ 2 H 0 A ⋆ ψ ∥ A ⋆ ψ ∥ 2 = = ( E + 2 3 ) A ⋆ ψ ( E + 1 ) ∥ ψ ∥ 2 and hence A ⋆ ψ is another eigenstate of H 0 with energy level E + 2 3 , while A ψ is another eigenstate of H 0 with energy level E − 2 1 provided that E > 0 . If E ∈ N ∪ { 0 } , then A r ψ will be a eigenstate with energy level E − r + 2 1 for all r ∈ N , and hence E − r ≥ 0 for all r ∈ N . This is impossible, and hence E ∈ N ∪ { 0 } . Thus the possible values of E are nonnegative integers only, and so the energy levels of H 0 are E n = n + 2 1 for n ∈ N ∪ { 0 } .
Returning to the problem, our Hamiltonian is H = 2 1 P 2 + 2 1 Q 2 + 2 1 Q = H 0 + 2 1 Q More usefully, we see that H = 2 1 P 2 + 2 1 ( Q + 2 1 I ) 2 − 4 1 I and it is much simpler to compare H with the translated version of H 0 obtained from Q ^ = Q + 2 1 I and P ^ = P . Thus the energy levels of H are E n = E n − 4 1 = 4 1 ( 4 n + 1 ) for n ∈ N ∪ { 0 } .
Thus n = 0 ∑ ∞ E n 2 1 = n = 0 ∑ ∞ ( 4 n + 1 ) 2 1 6 = π 2 + 8 G where G is the Catalan constant. Thus a = 2 , b = 8 , and so the answer is 2 + 8 = 1 0 .