Topic 4: The Finite Potential Well

Outline:

The quantum well

The finite potential well (FPW)

Even parity solutions of the TISE in the FPW

Odd parity solutions of the TISE in the FPW

Tunnelling into classically forbidded regions

Comparison with the IPW

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The AlGaAs-GaAs QuantumWell

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The AlGaAs-GaAs QuantumWellExample of a potential well:

Sandwich of GaAsand AlGaAs layers

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The AlGaAs-GaAs QuantumWellExample of a potential well:

Sandwich of GaAsand AlGaAs layers

Constrained motion along the x-axis; free motion in the y z plane.

V0 : 1-dimensional infinite potential well

V0 < : 1-dimensional finite potential well

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The Finite Potential Well

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The Finite Potential WellConsider a particle in the potential

V (x) =

V0 x < L2

0 L2 x L

2

V0 x >L2

V0 > 0

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The Finite Potential WellConsider a particle in the potential

V (x) =

V0 x < L2

0 L2 x L

2

V0 x >L2

V0 > 0

E > V0 unbound states, total energy E continuous (not quantized)

E < V0 bound states, expect discrete states

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The Finite Potential WellConsider a particle in the potential

V (x) =

V0 x < L2

0 L2 x L

2

V0 x >L2

V0 > 0

E > V0 unbound states, total energy E continuous (not quantized)

E < V0 bound states, expect discrete states

Solve the TISE:

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The Finite Potential WellConsider a particle in the potential

V (x) =

V0 x < L2

0 L2 x L

2

V0 x >L2

V0 > 0

E > V0 unbound states, total energy E continuous (not quantized)

E < V0 bound states, expect discrete states

Solve the TISE:

L2 x L

2(Region I):

!2

2md2

dx2(x) = E(x) (x) = k2(x) k2 = 2mE

!2> 0

solutions: I(x) = A sin kx + B cos kx, (as for the IPW)A, B arbitrary constants

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x > L2(Region II):

Note: in region II E = KE + PE = KE + V0 < V0 KE < 0!

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x > L2(Region II):

Note: in region II E = KE + PE = KE + V0 < V0 KE < 0!

!2

2md2

dx2(x) + V0(x) = E(x)

(x) = 2(x) 2 = 2m(V0E)!2

> 0

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x > L2(Region II):

Note: in region II E = KE + PE = KE + V0 < V0 KE < 0!

!2

2md2

dx2(x) + V0(x) = E(x)

(x) = 2(x) 2 = 2m(V0E)!2

> 0

solutions II(x) = Cex + DexC, D arbitrary constants

put D = 0, otherwise (x) not square integrable (blows up atlarge +ve x)

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x > L2(Region II):

Note: in region II E = KE + PE = KE + V0 < V0 KE < 0!

!2

2md2

dx2(x) + V0(x) = E(x)

(x) = 2(x) 2 = 2m(V0E)!2

> 0

solutions II(x) = Cex + DexC, D arbitrary constants

put D = 0, otherwise (x) not square integrable (blows up atlarge +ve x)

x < L2(Region III):

solutions III(x) = Fex + Gex (like in region II)F,G arbitrary constants

put F = 0, otherwise (x) not square integrable (blows up atlarge -ve x)

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

Apply general conditions on at x = L/2:

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B cos (kL/2) = C exp (L/2)

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B cos (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk sin (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider symmetric solutions only.

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B cos (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk sin (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider symmetric solutions only.

divide them side-by-side: tan kL2

= k

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B cos (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk sin (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider symmetric solutions only.

divide them side-by-side: tan kL2

= k

introduce = kL2

LHS: y() = tan

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The potential is symmetric w.r.t. to x = 0 expect symmetric(even-parity) and antisymmetric (odd-parity) states

Consider even-parity solutions only: I(x) = B cos kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B cos (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk sin (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider symmetric solutions only.

divide them side-by-side: tan kL2

= k

introduce = kL2

LHS: y() = tan

and 0 = k0L2 const. where k20 =

2mV0!2

> 0

RHS: y() = k

=

k20

k2 1 =

V0EE

=

20

2 1

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The even-parity solutions are determined when the curve y = tan intersects the curve y =

k. = kL

2

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The even-parity solutions are determined when the curve y = tan intersects the curve y =

k. = kL

2

The intersection pointsdetermine k and henceE = !

2k2

2m .

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The even-parity solutions are determined when the curve y = tan intersects the curve y =

k. = kL

2

The intersection pointsdetermine k and henceE = !

2k2

2m .

When E # V0

k =r

20

2 1 1

(solid line)

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The even-parity solutions are determined when the curve y = tan intersects the curve y =

k. = kL

2

The intersection pointsdetermine k and henceE = !

2k2

2m .

When E # V0

k =r

20

2 1 1

(solid line)

When E V0 0

k =r

20

2 1 0

(dashed line)

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The even-parity solutions are determined when the curve y = tan intersects the curve y =

k. = kL

2

The intersection pointsdetermine k and henceE = !

2k2

2m .

When E # V0

k =r

20

2 1 1

(solid line)

When E V0 0

k =r

20

2 1 0

(dashed line)

bound states with discrete (quantized) energyQM PHY202 p. 22

Comments:

larger V0 more bound states;smaller V0 less bound states

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Comments:

larger V0 more bound states;smaller V0 less bound states

when 0 < V0 < 22!2

mL2

only one symmetric state exists

In the FPW there is always at leastone bound state.

Even in a very shallow well (V0 0).

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Comments:

larger V0 more bound states;smaller V0 less bound states

when 0 < V0 < 22!2

mL2

only one symmetric state exists

In the FPW there is always at leastone bound state.

Even in a very shallow well (V0 0).

IPW: wavenumbers kn = nL , or n =n2

(n = 1,3,5, ... for symmetric states). The wavenumber and energy of the nth state is less than in theIPW.

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The FPW odd-parity solutions

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

Apply general conditions on at x = L/2:

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B sin (kL/2) = C exp (L/2)

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B sin (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk cos (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider antisymmetric solutions only.

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B sin (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk cos (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider antisymmetric solutions only.

divide them side-by-side: cot kL2

= k

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B sin (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk cos (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider antisymmetric solutions only.

divide them side-by-side: cot kL2

= k

(as before) introduce = kL2

LHS: y() = cot

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The FPW odd-parity solutionsConsider odd-parity solutions only: I(x) = B sin kx

Apply general conditions on at x = L/2:

continuous at x = L/2: I(L/2) = II(L/2) B sin (kL/2) = C exp (L/2)

continuous at x = L/2: I(L/2) = II(L/2) Bk cos (kL/2) = C exp (L/2)

No new constraints at x = L/2 since we consider antisymmetric solutions only.

divide them side-by-side: cot kL2

= k

(as before) introduce = kL2

LHS: y() = cot

and 0 = k0L2 const. where k20 =

2mV0!2

> 0

RHS: y() = k

=

k20

k2 1 =

20

2 1

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The odd-parity solutions are determined when the curve y = cot intersects the curve y =

k. = kL

2

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The odd-parity solutions are determined when the curve y = cot intersects the curve y =

k. = kL

2larger (smaller)V0 more (less)bound states

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The odd-parity solutions are determined when the curve y = cot intersects the curve y =

k. = kL

2larger (smaller)V0 more (less)bound states

when 0 0

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Quantum Tunnelling

At x > L/2 the wavefn (x) ex;at x < L/2 the wavefn (x) ex. 2 = 2m(V0E)

!2> 0

when V0 1/ 0

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Quantum Tunnelling

At x > L/2 the wavefn (x) ex;at x < L/2 the wavefn (x) ex. 2 = 2m(V0E)

!2> 0

when V0 1/ 0

when E V0 0 1/

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Quantum Tunnelling

At x > L/2 the wavefn (x) ex;at x < L/2 the wavefn (x) ex. 2 = 2m(V0E)

!2> 0

when V0 1/ 0

when E V0 0 1/

The depth of tunneling is determined by

the penetration depth

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Quantum Tunnelling

At x > L/2 the wavefn (x) ex;at x < L/2 the wavefn (x) ex. 2 = 2m(V0E)

!2> 0

when V0 1/ 0

when E V0 0 1/

The depth of tunneling is determined by

the penetration depthNon-zero wavefunction in classically forbidden regions (KE < 0!) isa purely quantum mechanical effect. It allows tunnelling betweenclassically allowed regions.

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Quantum Tunnelling

At x > L/2 the wavefn (x) ex;at x < L/2 the wavefn (x) ex. 2 = 2m(V0E)

!2> 0

when V0 1/ 0

when E V0 0 1/

The depth of tunneling is determined by

the penetration depthNon-zero wavefunction in classically forbidden regions (KE < 0!) isa purely quantum mechanical effect. It allows tunnelling betweenclassically allowed regions.

It follows from requiring that both (x) and (x) are continuous!

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Quantum Tunnelling

At x > L/2 the wavefn (x) ex;at x < L/2 the wavefn (x) ex. 2 = 2m(V0E)

!2> 0

when V0 1/ 0

when E V0 0 1/

The depth of tunneling is determined by

the penetration depthNon-zero wavefunction in classically forbidden regions (KE < 0!) isa purely quantum mechanical effect. It allows tunnelling betweenclassically allowed regions.

It follows from requiring that both (x) and (x) are continuous! Requiring a reasonable behaviour of the wavefunction leads toa (classically) crazy phenomenon of tunnelling

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Quantum states in potential wells

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Quantum states in potential wellsSome general properties:

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Quantum states in potential wellsSome general properties:

quantum (discrete) energy states are a typical property of anywell-type potential

the corresponding wavefunctions (and probability) are mostlyconfined inside the potential but exhibit non-zero tails in theclassically forbidden regions of KE < 0

Except when V (x) where the tails are not allowed.

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Quantum states in potential wellsSome general properties:

quantum (discrete) energy states are a typical property of anywell-type potential

the corresponding wavefunctions (and probability) are mostlyconfined inside the potential but exhibit non-zero tails in theclassically forbidden regions of KE < 0

Except when V (x) where the tails are not allowed.

both properties result from requiring the wavefunction (x) and itsderivative (x) to be continuous everywhere

Except when V (x) where (x) is not continuous.

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Quantum states in potential wellsSome general properties:

quantum (discrete) energy states are a typical property of anywell-type potential

the corresponding wavefunctions (and probability) are mostlyconfined inside the potential but exhibit non-zero tails in theclassically forbidden regions of KE < 0

Except when V (x) where the tails are not allowed.

both properties result from requiring the wavefunction (x) and itsderivative (x) to be continuous everywhere

Except when V (x) where (x) is not continuous.

quantum states in symmetric potentials (w.r.t. reflections x x)are either symmetric (i.e., even parity), with an even number ofnodes, or else antisymmetric (i.e, odd parity), with an odd number ofnodes

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the lowest energy state (ground state) is always above the bottom ofthe potential and is symmetric

A consequence of the HUP.

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the lowest energy state (ground state) is always above the bottom ofthe potential and is symmetric

A consequence of the HUP.

the wider and/or more shallow the potential, the lower the energies ofthe quantum states

A consequence of the HUP.

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the lowest energy state (ground state) is always above the bottom ofthe potential and is symmetric

A consequence of the HUP.

the wider and/or more shallow the potential, the lower the energies ofthe quantum states

A consequence of the HUP.

inside FPW-type of potentials the number of quantum states is finite

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the lowest energy state (ground state) is always above the bottom ofthe potential and is symmetric

A consequence of the HUP.

the wider and/or more shallow the potential, the lower the energies ofthe quantum states

A consequence of the HUP.

inside FPW-type of potentials the number of quantum states is finite

when the total energy E is larger than the height of the potential, theenergy becomes continuous, i.e., we have continuous states

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the lowest energy state (ground state) is always above the bottom ofthe potential and is symmetric

A consequence of the HUP.

the wider and/or more shallow the potential, the lower the energies ofthe quantum states

A consequence of the HUP.

inside FPW-type of potentials the number of quantum states is finite

when the total energy E is larger than the height of the potential, theenergy becomes continuous, i.e., we have continuous states

when V = V (x), both bound and continuous states are stationary,i.e, the time-dependent wavefunctions are of the form(x, t) = (x) exp

(

i!Et

)

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