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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 ⟩|𝑆𝑆(𝑡𝑡) that is a vector that lives “out there in Hilbert space”.

When necessary, we can “project” 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 Ψ 𝑥𝑥, 𝑡𝑡 = ⟨𝑥𝑥 ⟩|𝑆𝑆(𝑡𝑡) , which is what we call “the wavefunction” representation of the quantum state.

However, we have also expressed the quantum state in terms of the eigenfunctions of the momentum operator as Φ 𝑝𝑝, 𝑡𝑡 = ⟨𝑝𝑝 ⟩|𝑆𝑆(𝑡𝑡) .

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The effect of an operator is to “transform” one vector (quantum state) into another.

Operators as Rotations in Hilbert Space

A quantum state ⟩|𝛼𝛼 is transformed into state ⟩|𝛽𝛽 by means of a linear transformation brought on by application of an operator �𝑄𝑄 as follows: ⟩|𝛽𝛽 = �𝑄𝑄 ⟩|𝛼𝛼 .

Introduce an orthonormal basis ⟩|𝑒𝑒𝑛𝑛 with 𝑛𝑛 = 1, 2, 3, … that spans the Hilbert space. This could be a set of eigenfunctions of any Hermitian operator. They obey ��𝑒𝑒𝑚𝑚|𝑒𝑒𝑛𝑛 = 𝛿𝛿𝑛𝑛,𝑚𝑚

We will express the operator in terms of its “matrix elements” in this particular basis as 𝑄𝑄𝑚𝑚𝑛𝑛 ≡ ��𝑒𝑒𝑚𝑚| �𝑄𝑄|𝑒𝑒𝑛𝑛 , which is an infinite set of complex numbers since 𝑛𝑛 = 1, 2, 3, … and 𝑚𝑚 = 1, 2, 3, …

The transformation can now be written as ⟩|𝛽𝛽 = �𝑄𝑄 ⟩|𝛼𝛼 , or ∑𝑛𝑛=1∞ 𝑏𝑏𝑛𝑛 ⟩|𝑒𝑒𝑛𝑛 = ∑𝑛𝑛=1∞ 𝑎𝑎𝑛𝑛 �𝑄𝑄 ⟩|𝑒𝑒𝑛𝑛 .

Take the inner product of this expression with state ⟩|𝑒𝑒𝑚𝑚 to find 𝑏𝑏𝑚𝑚 = ∑𝑛𝑛=1∞ 𝑄𝑄𝑚𝑚𝑛𝑛𝑎𝑎𝑛𝑛.

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⟩|𝟎𝟎 ⟩|𝟏𝟏

Double Well Potential Without Tunneling

𝝐𝝐 𝝐𝝐

Consider two identical finite square wells, widely separated in space. Suppose they each has a single bound state, of identical energy 𝝐𝝐.

5https://www.math.ru.nl/~landsman/Robin.pdf

⟩|𝒔𝒔+ ⟩|𝒔𝒔−

𝝐𝝐 − 𝜟𝜟 𝝐𝝐 + 𝜟𝜟

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 × 2 matrix: �ℋ = ℋ00 ℋ01ℋ10 ℋ11

, where ℋ01 = �⟨0| �ℋ|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 𝒊𝒊𝒊 𝒅𝒅

𝒅𝒅𝒅𝒅⟩|𝑺𝑺(𝒅𝒅) = �𝓗𝓗 ⟩|𝑺𝑺(𝒅𝒅) , for the time evolution of state ⟩|𝑺𝑺(𝒅𝒅) = 𝒂𝒂(𝒅𝒅)

𝒃𝒃(𝒅𝒅) .

This is actually two differential equations written in vector form: 𝑖𝑖𝒊 𝑑𝑑𝑑𝑑𝑑𝑑

𝑎𝑎(𝑑𝑑)𝑏𝑏(𝑑𝑑) = �ℋ 𝑎𝑎(𝑑𝑑)

𝑏𝑏(𝑑𝑑) = 𝜖𝜖 −∆−∆ 𝜖𝜖

𝑎𝑎(𝑑𝑑)𝑏𝑏(𝑑𝑑) .

The two equations are: 𝒊𝒊𝒊 𝒅𝒅𝒂𝒂𝒅𝒅𝒅𝒅

= 𝝐𝝐 𝒂𝒂 − ∆ 𝒃𝒃 and 𝒊𝒊𝒊 𝒅𝒅𝒃𝒃𝒅𝒅𝒅𝒅

= −∆ 𝒂𝒂 + 𝝐𝝐 𝒃𝒃

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If the Hamiltonian is time-independent we can separate variables and write the quantum state as

⟩|𝑆𝑆(𝑡𝑡) = ⟩|𝑠𝑠 𝑒𝑒−𝑖𝑖𝑖𝑖𝑑𝑑/𝒊

This leads to the time-independent Schrodinger equation (TISE):

�ℋ ⟩|𝑠𝑠 = 𝐸𝐸 ⟩|𝑠𝑠 , where 𝐸𝐸 is the energy eigenvalue and ⟩|𝑠𝑠 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:𝐸𝐸 = 𝜖𝜖 + ∆ with associated eigenfunction ⟩|𝑠𝑠− = 1

21−1 , and

𝐸𝐸 = 𝜖𝜖 − ∆ the associated eigenfunction is ⟩|𝑠𝑠+ = 12

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Note that ⟩|𝑠𝑠− and ⟩|𝑠𝑠+ form an orthonormal basis: �𝑠𝑠+ ⟩|𝑠𝑠+ = �𝑠𝑠− ⟩|𝑠𝑠− = 1 and �𝑠𝑠+ ⟩|𝑠𝑠− = �𝑠𝑠− ⟩|𝑠𝑠+ = 0.

8https://www.math.ru.nl/~landsman/Robin.pdf

⟩|𝒔𝒔+ ⟩|𝒔𝒔−

𝝐𝝐 − 𝜟𝜟 𝝐𝝐 + 𝜟𝜟

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 ⟩|0 = 1

0 , i.e. the particle starts initially in the left well.

Calculation in two-dimensional Hilbert Space

�)|𝑆𝑆(0 = ⟩|0 = 10 =

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⟩|𝑠𝑠+ + ⟩|𝑠𝑠−

This can be turned back into a time-dependent quantum state by adding back the separated time dependence: ⟩|𝑆𝑆(𝑡𝑡) = 1

2⟩|𝑠𝑠+ 𝑒𝑒−𝑖𝑖 𝜖𝜖−∆ 𝑑𝑑/𝒊 + ⟩|𝑠𝑠− 𝑒𝑒−𝑖𝑖 𝜖𝜖+∆ 𝑑𝑑/𝒊

�)|𝑆𝑆(𝑡𝑡 = 𝑒𝑒−𝑖𝑖𝜖𝜖 ⁄𝑑𝑑 𝒊 cos ∆ ⁄𝑡𝑡 𝒊−𝑖𝑖 sin ∆ ⁄𝑡𝑡 𝒊

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 “direction in Hilbert space” that corresponds to occupation of the left well, namely the state ⟩|0 :

⟨0 ⟩|𝑆𝑆(𝑡𝑡) = 1 0 𝑒𝑒−𝑖𝑖𝜖𝜖𝑑𝑑/𝒊 cos ∆𝑡𝑡/𝒊−𝑖𝑖 sin ∆𝑡𝑡/𝒊 = 𝑒𝑒−𝑖𝑖𝜖𝜖𝑑𝑑/𝒊 cos ∆𝑡𝑡/𝒊 .

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We can find the probability to be in the left well: 𝑃𝑃 0, 𝑡𝑡 = ⟨0 �)|𝑆𝑆(𝑡𝑡 2 = 𝑐𝑐𝑐𝑐𝑠𝑠2 ∆ ⁄𝑡𝑡 𝒊

Calculation in two-dimensional Hilbert Space

From the last slide: ⟨0 ⟩|𝑆𝑆(𝑡𝑡) = 1 0 𝑒𝑒−𝑖𝑖𝜖𝜖𝑑𝑑/𝒊 cos ∆𝑡𝑡/𝒊−𝑖𝑖 sin ∆𝑡𝑡/𝒊 = 𝑒𝑒−𝑖𝑖𝜖𝜖𝑑𝑑/𝒊 cos ∆𝑡𝑡/𝒊 .

This is an oscillating function of time, starting at 𝑃𝑃 0,0 = 1 at time 𝑡𝑡 = 0, and then decreasing initially.The probability to be in the left well vanishes at time 𝑡𝑡 = 𝜋𝜋𝒊/2∆. It then increases to unity at time 𝑡𝑡 = 𝜋𝜋𝒊/∆and 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 “tunneling strength” ∆.