Phys 401 Spring 2020 Lecture # 23 1 April, 2020â¬ÂŠÂ · The probability to be in the left well...
Transcript of Phys 401 Spring 2020 Lecture # 23 1 April, 2020â¬ÂŠÂ · The probability to be in the left well...
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Phys 401Spring 2020Lecture #231 April, 2020
Welcome Back Everybody!
The slides for this lecture are posted in Supplementary Material
Today:Brief Review of Lecture 22New Material:
Conclusion of the two-dimensional Hilbert space calculation
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Change of Basis in Hilbert Space
We think about a general quantum state â©|ðð(ð¡ð¡) that is a vector that lives âout there in Hilbert spaceâ.
When necessary, we can âprojectâ this quantum state into the set of all eigenfunctions of any Hermitian operator.
Up to this point we have mainly considered the projection of the quantum state in the direction of the position eigenfunctions as Κ ð¥ð¥, ð¡ð¡ = âšð¥ð¥ â©|ðð(ð¡ð¡) , which is what we call âthe wavefunctionâ representation of the quantum state.
However, we have also expressed the quantum state in terms of the eigenfunctions of the momentum operator as Ί ðð, ð¡ð¡ = âšðð â©|ðð(ð¡ð¡) .
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The effect of an operator is to âtransformâ one vector (quantum state) into another.
Operators as Rotations in Hilbert Space
A quantum state â©|ðŒðŒ is transformed into state â©|ðœðœ by means of a linear transformation brought on by application of an operator ï¿œðð as follows: â©|ðœðœ = ï¿œðð â©|ðŒðŒ .
Introduce an orthonormal basis â©|ðððð with ðð = 1, 2, 3, ⊠that spans the Hilbert space. This could be a set of eigenfunctions of any Hermitian operator. They obey ᅵᅵðððð|ðððð = ð¿ð¿ðð,ðð
We will express the operator in terms of its âmatrix elementsâ in this particular basis as ðððððð ⡠ᅵᅵðððð| ï¿œðð|ðððð , which is an infinite set of complex numbers since ðð = 1, 2, 3, ⊠and ðð = 1, 2, 3, âŠ
The transformation can now be written as â©|ðœðœ = ï¿œðð â©|ðŒðŒ , or âðð=1â ðððð â©|ðððð = âðð=1â ðððð ï¿œðð â©|ðððð .
Take the inner product of this expression with state â©|ðððð to find ðððð = âðð=1â ðððððððððð.
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â©|ðð â©|ðð
Double Well Potential Without Tunneling
ðð ðð
Consider two identical finite square wells, widely separated in space. Suppose they each has a single bound state, of identical energy ðð.
5https://www.math.ru.nl/~landsman/Robin.pdf
â©|ðð+ â©|ððâ
ðð â ðð ðð + ðð
Double Well Potential With TunnelingNow suppose that the two finite square wells are brought close together such that a finite height and finite width barrier separates the wells. As we saw in our earlier study of 1D quantum mechanics, at some point there will be tunneling between the wells such that a particle initially placed in the left well will have a finite probability of finding itself in theright well.
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The Hamiltonian is now a 2 à 2 matrix: ï¿œâ = â00 â01â10 â11
, where â01 = ï¿œâš0| ï¿œâ|1 , etc.
Calculation in two-dimensional Hilbert Space
To understand this Hamiltonian it is good to see it in action. The time-dependent Schrodinger equation is ððð ð ð
ð ð ð ð â©|ðºðº(ð ð ) = ï¿œðð â©|ðºðº(ð ð ) , for the time evolution of state â©|ðºðº(ð ð ) = ðð(ð ð )
ðð(ð ð ) .
This is actually two differential equations written in vector form: ððð ðððððð
ðð(ðð)ðð(ðð) = ï¿œâ ðð(ðð)
ðð(ðð) = ðð ââââ ðð
ðð(ðð)ðð(ðð) .
The two equations are: ððð ð ð ððð ð ð ð
= ðð ðð â â ðð and ððð ð ð ððð ð ð ð
= ââ ðð + ðð ðð
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If the Hamiltonian is time-independent we can separate variables and write the quantum state as
â©|ðð(ð¡ð¡) = â©|ð ð ððâðððððð/ð
This leads to the time-independent Schrodinger equation (TISE):
ï¿œâ â©|ð ð = ðžðž â©|ð ð , where ðžðž is the energy eigenvalue and â©|ð ð is the energy eigenstate.
Calculation in two-dimensional Hilbert Space
The next step is to solve the TISE for the energy eigenvalues and eigenfunctions of the tunneling system. This is a standard linear algebra problem involving 2x2 matrices and vectors.
Answer:ðžðž = ðð + â with associated eigenfunction â©|ð ð â = 1
21â1 , and
ðžðž = ðð â â the associated eigenfunction is â©|ð ð + = 12
11 .
Note that â©|ð ð â and â©|ð ð + form an orthonormal basis: ï¿œð ð + â©|ð ð + = ï¿œð ð â â©|ð ð â = 1 and ï¿œð ð + â©|ð ð â = ï¿œð ð â â©|ð ð + = 0.
8https://www.math.ru.nl/~landsman/Robin.pdf
â©|ðð+ â©|ððâ
ðð â ðð ðð + ðð
Double Well Potential With Tunneling
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Now go back to solve the time-dependent Schrodinger equation with an initial condition that the particle starts in state â©|0 = 1
0 , i.e. the particle starts initially in the left well.
Calculation in two-dimensional Hilbert Space
ï¿œ)|ðð(0 = â©|0 = 10 =
12
â©|ð ð + + â©|ð ð â
This can be turned back into a time-dependent quantum state by adding back the separated time dependence: â©|ðð(ð¡ð¡) = 1
2â©|ð ð + ððâðð ððââ ðð/ð + â©|ð ð â ððâðð ðð+â ðð/ð
ï¿œ)|ðð(ð¡ð¡ = ððâðððð âðð ð cos â âð¡ð¡ ðâðð sin â âð¡ð¡ ð
What is the probability that the particle is still in the left well? To answer this question we can project the quantum state onto the âdirection in Hilbert spaceâ that corresponds to occupation of the left well, namely the state â©|0 :
âš0 â©|ðð(ð¡ð¡) = 1 0 ððâðððððð/ð cos âð¡ð¡/ðâðð sin âð¡ð¡/ð = ððâðððððð/ð cos âð¡ð¡/ð .
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We can find the probability to be in the left well: ðð 0, ð¡ð¡ = âš0 ï¿œ)|ðð(ð¡ð¡ 2 = ððððð ð 2 â âð¡ð¡ ð
Calculation in two-dimensional Hilbert Space
From the last slide: âš0 â©|ðð(ð¡ð¡) = 1 0 ððâðððððð/ð cos âð¡ð¡/ðâðð sin âð¡ð¡/ð = ððâðððððð/ð cos âð¡ð¡/ð .
This is an oscillating function of time, starting at ðð 0,0 = 1 at time ð¡ð¡ = 0, and then decreasing initially.The probability to be in the left well vanishes at time ð¡ð¡ = ððð/2â. It then increases to unity at time ð¡ð¡ = ððð/âand continues to oscillate. The tunneling process allows the particle to move back and forth between the twowells at a rate that is inversely proportional to the âtunneling strengthâ â.