1. Quantum One

3. The Harmonic Oscillator, continued

4. In the last lecture, we began a study of the quantum mechanical harmonic oscillator.
To treat this system using the algebraic method, we introduced dimensionless position and momentum operators, and then traded those and the Hamiltonian in, for a number operator, and raising and lowering operators, the latter two of which are not Hermitian, and are not even normal.
The product formed from the raising and lowering operators defines the number operator, which shares eigenstates with the harmonic oscillator Hamiltonian.
Using only the commutation relations among this new set of three operators, we deduced the spectrum of the number operator, and therefore completely determined the energy spectrum of the quantum harmonic oscillator without ever having solved any differential equations.

5. Spectrum and Degeneracy of the Harmonic Oscillator Hamiltonian
From the last lecture we saw that
Spectrum 𝑁 = 𝑛 ={0,1,2,⋯}
and that eigenvalues
of the 1D harmonic oscillator form a “ladder” of equally spaced energy levels.
Q: How many eigenvectors for each eigenvalue are there? What is the degeneracy of the 𝑛th energy level?
𝑛=4
𝑛=3
𝑛=2
𝑛=1
𝑛=0 ℏ𝜔
1 2 ℏ𝜔

6. Spectrum and Degeneracy of the Harmonic Oscillator Hamiltonian
From the last lecture we saw that
Spectrum 𝑁 = 𝑛 ={0,1,2,⋯}
and that eigenvalues
of the 1D harmonic oscillator form a “ladder” of equally spaced energy levels.
Q: How many eigenvectors for each eigenvalue are there? What is the degeneracy of the 𝑛th energy level?
𝑛=4
𝑛=3
𝑛=2
𝑛=1
𝑛=0 ℏ𝜔
1 2 ℏ𝜔

7. Spectrum and Degeneracy of the Harmonic Oscillator Hamiltonian
A: It turns out that the eigenvalues of 𝑁 (and therefore 𝐻) are completely nondegenerate.
There exists one and only one linearly independent eigenstate for each distinct eigenvalue.
To prove this we first show the following: If energy level 𝑛 is nondegenerate, then so is energy level 𝑛+1.
𝑛=4
𝑛=3
𝑛=2
𝑛=1
𝑛=0 ℏ𝜔
1 2 ℏ𝜔

8. Spectrum and Degeneracy of the Harmonic Oscillator Hamiltonian
A: It turns out that the eigenvalues of 𝑁 (and therefore 𝐻) are completely nondegenerate.
There exists one and only one linearly independent eigenstate for each distinct eigenvalue.
To prove this we first show the following: If energy level 𝑛 is nondegenerate, then so is energy level 𝑛+1.
𝑛=4
𝑛=3
𝑛=2
𝑛=1
𝑛=0 ℏ𝜔
1 2 ℏ𝜔

9. Spectrum and Degeneracy of the Harmonic Oscillator Hamiltonian
A: It turns out that the eigenvalues of 𝑁 (and therefore 𝐻) are completely nondegenerate.
There exists one and only one linearly independent eigenstate for each distinct eigenvalue.
To prove this we first show the following: If energy level 𝑛 is nondegenerate, then so is energy level 𝑛+1.
𝑛=4
𝑛=3
𝑛=2
𝑛=1
𝑛=0 ℏ𝜔
1 2 ℏ𝜔

10. Spectrum and Degeneracy of the Harmonic Oscillator Hamiltonian
A: It turns out that the eigenvalues of 𝑁 (and therefore 𝐻) are completely nondegenerate.
There exists one and only one linearly independent eigenstate for each distinct eigenvalue.
To prove this we first show the following: If energy level 𝑛 is nondegenerate, then so is energy level 𝑛+1.
𝑛=4
𝑛=3
𝑛=2
𝑛=1
𝑛=0 ℏ𝜔
1 2 ℏ𝜔

11. To see this, assume that level 𝑛 is nondegenerate, and let
be two arbitrary eigenstates of 𝑁 having eigenvalue 𝑛+1.
From these states we can then produce the states and
which would have to be eigenstates of 𝑁 associated with the (assumed) nondegenerate eigenvalue n, and so are linearly dependent.
For two vectors, linear dependence implies proportionality, so there exists a constant 𝜆 such that

12. To see this, assume that level 𝑛 is nondegenerate, and let
be two arbitrary eigenstates of 𝑁 having eigenvalue 𝑛+1.
From these states we can then produce the states and
which would have to be eigenstates of 𝑁 associated with the (assumed) nondegenerate eigenvalue n, and so are linearly dependent.
For two vectors, linear dependence implies proportionality, so there exists a constant 𝜆 such that

13. To see this, assume that level 𝑛 is nondegenerate, and let
be two arbitrary eigenstates of 𝑁 having eigenvalue 𝑛+1.
From these states we can then produce the states and
which would have to be eigenstates of 𝑁 associated with the (assumed) nondegenerate eigenvalue 𝑛, and so are linearly dependent.
For two vectors, linear dependence implies proportionality, so there exists a constant 𝜆 such that

14. To see this, assume that level 𝑛 is nondegenerate, and let
be two arbitrary eigenstates of 𝑁 having eigenvalue 𝑛+1.
From these states we can then produce the states and
which would have to be eigenstates of 𝑁 associated with the (assumed) nondegenerate eigenvalue 𝑛, and so are linearly dependent.
For two vectors, linear dependence implies proportionality, so there exists a constant 𝜆 such that . . .

15. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

16. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

17. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

18. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

19. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

20. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

21. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

22. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

23. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

24. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

25. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

26. Acting with the raising operator 𝑎⁺ then reveals that
Equating the last two expressions in each line above, we deduce that
This shows that and are necessarily linearly dependent.
There is at most one linearly independent eigenvector of 𝑁 with eigenvalue 𝑛 Hence, if eigenvalue 𝑛 is nondegenerate, so is 𝑛+1.

27. To complete the argument, we now show that there exists, in fact, exactly one linearly independent eigenvector |0〉 of 𝑁 with eigenvalue 𝑛=0, from which it follows that all the eigenvalues of 𝑁 are nondegenerate.
To do this we explicitly construct the corresponding eigenfunction
in the position representation.
This is facilitated by the fact, shown above, that any eigenstate |0〉 of 𝑁 with eigenvalue 0 is annihilated by the lowering operator, i.e., it obeys the equation
𝑎|0〉=0.
Using the relation
this implies that . . .

28. To complete the argument, we now show that there exists, in fact, exactly one linearly independent eigenvector |0〉 of 𝑁 with eigenvalue 𝑛=0, from which it follows that all the eigenvalues of 𝑁 are nondegenerate.
To do this we explicitly construct the corresponding eigenfunction
in the position representation.
This is facilitated by the fact, shown above, that any eigenstate |0〉 of 𝑁 with eigenvalue 0 is annihilated by the lowering operator, i.e., it obeys the equation
𝑎|0〉=0.
Using the relation
this implies that . . .

29. To complete the argument, we now show that there exists, in fact, exactly one linearly independent eigenvector |0〉 of 𝑁 with eigenvalue 𝑛=0, from which it follows that all the eigenvalues of 𝑁 are nondegenerate.
To do this we explicitly construct the corresponding eigenfunction
in the position representation.
This is facilitated by the fact, shown above, that any eigenstate |0〉 of 𝑁 with eigenvalue 0 is annihilated by the lowering operator, i.e., it obeys the equation
𝑎|0〉=0.
Using the relation
this implies that . . .

30. To complete the argument, we now show that there exists, in fact, exactly one linearly independent eigenvector |0〉 of 𝑁 with eigenvalue 𝑛=0, from which it follows that all the eigenvalues of 𝑁 are nondegenerate.
To do this we explicitly construct the corresponding eigenfunction
in the position representation.
This is facilitated by the fact, shown above, that any eigenstate |0〉 of 𝑁 with eigenvalue 0 is annihilated by the lowering operator, i.e., it obeys the equation
𝑎|0〉=0.
Using the relation
this implies that . . .

31. in which we have used the differential form taken by the operator p in the position representation.
This first order differential equation leads to the relation
which can be integrated from 𝑞=0 to obtain
ln[φ₀(q)/φ₀(0)]=-(1/2)q² or
Thus, the eigenfunctions of 𝑁 with 𝑛=0 differ from one another only through an overall multiplicative constant

32. in which we have used the differential form taken by the operator 𝑝 in the position representation.
This first order differential equation leads to the relation
which can be integrated from 𝑞=0 to obtain
ln[φ₀(q)/φ₀(0)]=-(1/2)q² or
Thus, the eigenfunctions of 𝑁 with 𝑛=0 differ from one another only through an overall multiplicative constant

33. in which we have used the differential form taken by the operator 𝑝 in the position representation.
This first order differential equation leads to the relation
which can be integrated from 𝑞=0 to obtain
ln[φ₀(q)/φ₀(0)]=-(1/2)q² or
Thus, the eigenfunctions of 𝑁 with 𝑛=0 differ from one another only through an overall multiplicative constant

34. in which we have used the differential form taken by the operator 𝑝 in the position representation.
This first order differential equation leads to the relation
which can be integrated from 𝑞=0 to obtain
ln[φ₀(q)/φ₀(0)]=-(1/2)q² or
Thus, the eigenfunctions of 𝑁 with 𝑛=0 differ from one another only through an overall multiplicative constant

35. in which we have used the differential form taken by the operator 𝑝 in the position representation.
This first order differential equation leads to the relation
which can be integrated from 𝑞=0 to obtain
ln[φ₀(q)/φ₀(0)]=-(1/2)q² or
Thus, the eigenfunctions of 𝑁 with 𝑛=0 differ from one another only through an overall multiplicative constant

36. in which we have used the differential form taken by the operator 𝑝 in the position representation.
This first order differential equation leads to the relation
which can be integrated from 𝑞=0 to obtain
ln[φ₀(q)/φ₀(0)]=-(1/2)q² or
Thus, the eigenfunctions of 𝑁 with 𝑛=0 differ from one another only through an overall multiplicative constant

37. Thus there is only one linearly independent solution with this eigenvalue.
The eigenvalue 𝑛=0 is, therefore, nondegenerate as are all the eigenvalues of 𝑁 and 𝐻.
Thus, we may continue to label the eigenstates |𝑛〉 of 𝑁 and 𝐻 by the eigenvalues 𝑛, we do not need an additional index, since there is no degeneracy.
We now proceed to explicitly construct an orthonormal basis of energy eigenstates |𝑛〉 of 𝐻.
We have already constructed the eigenstate of 𝑁 with 𝑛=0 by deriving the form of the wave function
that represents this state in the position representation.
To complete the picture we need to specify the normalization constant 𝐴.

38. Thus there is only one linearly independent solution with this eigenvalue.
The eigenvalue 𝑛=0 is, therefore, nondegenerate as are all the eigenvalues of 𝑁 and 𝐻.
Thus, we may continue to label the eigenstates |𝑛〉 of 𝑁 and 𝐻 by the eigenvalues 𝑛, we do not need an additional index, since there is no degeneracy.
We now proceed to explicitly construct an orthonormal basis of energy eigenstates |𝑛〉 of 𝐻.
We have already constructed the eigenstate of 𝑁 with 𝑛=0 by deriving the form of the wave function
that represents this state in the position representation.
To complete the picture we need to specify the normalization constant 𝐴.

39. Thus there is only one linearly independent solution with this eigenvalue.
The eigenvalue 𝑛=0 is, therefore, nondegenerate as are all the eigenvalues of 𝑁 and 𝐻.
Thus, we may continue to label the eigenstates |𝑛〉 of 𝑁 and 𝐻 by the eigenvalues 𝑛, we do not need an additional index, since there is no degeneracy.
We now proceed to explicitly construct an orthonormal basis of energy eigenstates |𝑛〉 of 𝐻.
We have already constructed the eigenstate of 𝑁 with 𝑛=0 by deriving the form of the wave function
that represents this state in the position representation.
To complete the picture we need to specify the normalization constant 𝐴.

40. Thus there is only one linearly independent solution with this eigenvalue.
The eigenvalue 𝑛=0 is, therefore, nondegenerate as are all the eigenvalues of 𝑁 and 𝐻.
Thus, we may continue to label the eigenstates |𝑛〉 of 𝑁 and 𝐻 by the eigenvalues 𝑛, we do not need an additional index, since there is no degeneracy.
We now proceed to explicitly construct an orthonormal basis of energy eigenstates |𝑛〉 of 𝐻.
We have already constructed the eigenstate of 𝑁 with 𝑛=0 by deriving the form of the wave function
that represents this state in the position representation.
To complete the picture we need to specify the normalization constant 𝐴.

41. Thus there is only one linearly independent solution with this eigenvalue.
The eigenvalue 𝑛=0 is, therefore, nondegenerate as are all the eigenvalues of 𝑁 and 𝐻.
Thus, we may continue to label the eigenstates |𝑛〉 of 𝑁 and 𝐻 by the eigenvalues 𝑛, we do not need an additional index, since there is no degeneracy.
We now proceed to explicitly construct an orthonormal basis of energy eigenstates |𝑛〉 of 𝐻.
We have already constructed the eigenstate of 𝑁 with 𝑛=0 by deriving the form of the wave function
that represents this state in the position representation.
To complete the picture we need to specify the normalization constant 𝐴.

42. Thus there is only one linearly independent solution with this eigenvalue.
The eigenvalue 𝑛=0 is, therefore, nondegenerate as are all the eigenvalues of 𝑁 and 𝐻.
Thus, we may continue to label the eigenstates |𝑛〉 of 𝑁 and 𝐻 by the eigenvalues 𝑛, we do not need an additional index, since there is no degeneracy.
We now proceed to explicitly construct an orthonormal basis of energy eigenstates |𝑛〉 of 𝐻.
We have already constructed the eigenstate of 𝑁 with 𝑛=0 by deriving the form of the wave function
that represents this state in the position representation.
To complete the picture we need to specify the normalization constant 𝐴.

43. Correct normalization requires that
The integral appearing in this condition is well known and has the value 𝜋 ,
from which we deduce that the correctly normalized ground state wave function has the form
It is also possible to express this in terms of the "real" position variable 𝑥, rather
than the dimensionless variable

44. Correct normalization requires that
The integral appearing in this condition is well known and has the value 𝜋 ,
from which we deduce that the correctly normalized ground state wave function has the form
It is also possible to express this in terms of the "real" position variable 𝑥, rather
than the dimensionless variable

45. Correct normalization requires that
The integral appearing in this condition is well known and has the value 𝜋 ,
from which we deduce that the correctly normalized ground state wave function has the form
It is also possible to express this in terms of the "real" position variable 𝑥, rather
than the dimensionless variable

46. Correct normalization requires that
The integral appearing in this condition is well known and has the value 𝜋 ,
from which we deduce that the correctly normalized ground state wave function has the form
It is also possible to express this in terms of the "real" position variable 𝑥, rather
than the dimensionless variable

47. This is most easily done by noting that, in general, normalization requires that
so that
Making the appropriate substitution gives the ground state wave function

48. This is most easily done by noting that, in general, normalization requires that
so that
Making the appropriate substitution gives the ground state wave function

49. This is most easily done by noting that, in general, normalization requires that
so that
Making the appropriate substitution gives the ground state wave function

50. This is most easily done by noting that, in general, normalization requires that
so that
Making the appropriate substitution gives the ground state wave function

51. This is most easily done by noting that, in general, normalization requires that
so that
Making the appropriate substitution gives the ground state wave function

52. The remaining eigenstates can be generated from the ground state by repeated application of the raising operator 𝑎⁺.
Unfortunately, repeated application of 𝑎⁺to the ground state does not generate normalized eigenstates.
To see this, let us denote by |𝑛〉 and |𝑛+1〉 the square-normalized eigenstates of 𝑁 with eigenvalues 𝑛 and 𝑛+1, respectively.
Our earlier argument show that 𝑎⁺ |𝑛〉 is also an eigenstate of 𝑁 with eigenvalue 𝑛+1, and so must be proportional to |𝑛+1〉, since eigenstates of 𝑁 are nondegenerate. Thus, there exists a constant 𝜆 such that 𝑎⁺|𝑛〉=𝜆|𝑛+1〉
Taking the norm of this vector reveals that 〈𝑛|𝑎𝑎⁺|𝑛〉=|𝜆|²
Uusing the fact that 𝑎𝑎⁺=𝑁+1, we see that |𝜆|²=𝑛+1.

53. The remaining eigenstates can be generated from the ground state by repeated application of the raising operator 𝑎⁺.
Unfortunately, repeated application of 𝑎⁺to the ground state does not generate normalized eigenstates.
To see this, let us denote by |𝑛〉 and |𝑛+1〉 the square-normalized eigenstates of 𝑁 with eigenvalues 𝑛 and 𝑛+1, respectively.
Our earlier argument show that 𝑎⁺ |𝑛〉 is also an eigenstate of 𝑁 with eigenvalue 𝑛+1, and so must be proportional to |𝑛+1〉, since eigenstates of 𝑁 are nondegenerate. Thus, there exists a constant 𝜆 such that 𝑎⁺|𝑛〉=𝜆|𝑛+1〉
Taking the norm of this vector reveals that 〈𝑛|𝑎𝑎⁺|𝑛〉=|𝜆|²
Uusing the fact that 𝑎𝑎⁺=𝑁+1, we see that |𝜆|²=𝑛+1.

54. The remaining eigenstates can be generated from the ground state by repeated application of the raising operator 𝑎⁺.
Unfortunately, repeated application of 𝑎⁺to the ground state does not generate normalized eigenstates.
To see this, let us denote by |𝑛〉 and |𝑛+1〉 the square-normalized eigenstates of 𝑁 with eigenvalues 𝑛 and 𝑛+1, respectively.
Our earlier argument show that 𝑎⁺ |𝑛〉 is also an eigenstate of 𝑁 with eigenvalue 𝑛+1, and so must be proportional to |𝑛+1〉, since eigenstates of 𝑁 are nondegenerate. Thus, there exists a constant 𝜆 such that 𝑎⁺|𝑛〉=𝜆|𝑛+1〉
Taking the norm of this vector reveals that 〈𝑛|𝑎𝑎⁺|𝑛〉=|𝜆|²
Uusing the fact that 𝑎𝑎⁺=𝑁+1, we see that |𝜆|²=𝑛+1.

55. The remaining eigenstates can be generated from the ground state by repeated application of the raising operator 𝑎⁺.
Unfortunately, repeated application of 𝑎⁺to the ground state does not generate normalized eigenstates.
To see this, let us denote by |𝑛〉 and |𝑛+1〉 the square-normalized eigenstates of 𝑁 with eigenvalues 𝑛 and 𝑛+1, respectively.
Our earlier argument show that 𝑎⁺ |𝑛〉 is also an eigenstate of 𝑁 with eigenvalue 𝑛+1, and so must be proportional to |𝑛+1〉, since eigenstates of 𝑁 are nondegenerate. Thus, there exists a constant 𝜆 such that 𝑎⁺|𝑛〉=𝜆|𝑛+1〉
Taking the norm of this vector reveals that 〈𝑛|𝑎𝑎⁺|𝑛〉=|𝜆|²
Uusing the fact that 𝑎𝑎⁺=𝑁+1, we see that |𝜆|²=𝑛+1.

56. The remaining eigenstates can be generated from the ground state by repeated application of the raising operator 𝑎⁺.
Unfortunately, repeated application of 𝑎⁺to the ground state does not generate normalized eigenstates.
To see this, let us denote by |𝑛〉 and |𝑛+1〉 the square-normalized eigenstates of 𝑁 with eigenvalues 𝑛 and 𝑛+1, respectively.
Our earlier argument show that 𝑎⁺ |𝑛〉 is also an eigenstate of 𝑁 with eigenvalue 𝑛+1, and so must be proportional to |𝑛+1〉, since eigenstates of 𝑁 are nondegenerate. Thus, there exists a constant 𝜆 such that 𝑎⁺|𝑛〉=𝜆|𝑛+1〉
Taking the norm of this vector reveals that 〈𝑛|𝑎𝑎⁺|𝑛〉=|𝜆|²
Uusing the fact that 𝑎𝑎⁺=𝑁+1, we see that |𝜆|²=𝑛+1.

57. The remaining eigenstates can be generated from the ground state by repeated application of the raising operator 𝑎⁺.
Unfortunately, repeated application of 𝑎⁺to the ground state does not generate normalized eigenstates.
To see this, let us denote by |𝑛〉 and |𝑛+1〉 the square-normalized eigenstates of 𝑁 with eigenvalues 𝑛 and 𝑛+1, respectively.
Our earlier argument show that 𝑎⁺ |𝑛〉 is also an eigenstate of 𝑁 with eigenvalue 𝑛+1, and so must be proportional to |𝑛+1〉, since eigenstates of 𝑁 are nondegenerate. Thus, there exists a constant 𝜆 such that 𝑎⁺|𝑛〉=𝜆|𝑛+1〉
Taking the norm of this vector reveals that 〈𝑛|𝑎𝑎⁺|𝑛〉=|𝜆|²
Using the fact that 𝑎𝑎⁺=𝑁+1, we see that |𝜆|²=𝑛+1.

58. Fixing the relative phase of our basis vectors such that 𝜆 is real and positive, we obtain
𝜆= 𝑛+1
from which we deduce the basic action
𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉
of 𝑎 + between neighboring basis vectors .
For the purpose of constructing these states it is useful to write this relation in the equivalent form
By recursion, this allows us to express the state |𝑛〉 in terms of the ground state, i.e., . . .

59. Fixing the relative phase of our basis vectors such that 𝜆 is real and positive, we obtain
𝜆= 𝑛+1
from which we deduce the basic action
𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉
of 𝑎 + between neighboring basis vectors .
For the purpose of constructing these states it is useful to write this relation in the equivalent form
By recursion, this allows us to express the state |𝑛〉 in terms of the ground state, i.e., . . .

60. Fixing the relative phase of our basis vectors such that 𝜆 is real and positive, we obtain
𝜆= 𝑛+1
from which we deduce the basic action
𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉
of 𝑎 + between neighboring basis vectors .
For the purpose of constructing these states it is useful to write this relation in the equivalent form
By recursion, this allows us to express the state |𝑛〉 in terms of the ground state, i.e., . . .

61. Fixing the relative phase of our basis vectors such that 𝜆 is real and positive, we obtain
𝜆= 𝑛+1
from which we deduce the basic action
𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉
of 𝑎 + between neighboring basis vectors .
For the purpose of constructing these states it is useful to write this relation in the equivalent form
By recursion, this allows us to express the state |𝑛〉 in terms of the ground state, i.e., . . .

62. or
To find the wave functions which represent the eigenstates in the position representation we project this onto the basis vectors |𝑞〉 of that representation

63. or
To find the wave functions which represent the eigenstates in the position representation we project this onto the basis vectors |𝑞〉 of that representation

64. or
To find the wave functions which represent the eigenstates in the position representation we project this onto the basis vectors |𝑞〉 of that representation

65. or
To find the wave functions which represent the eigenstates in the position representation we project this onto the basis vectors |𝑞〉 of that representation

66. or
To find the wave functions which represent the eigenstates in the position representation we project this onto the basis vectors |𝑞〉 of that representation

67. Using the explicit form for the normalized ground state wave function we find that
Another useful form of this follows from the relation 𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉 derived earlier, from which it follows that
giving, in the position representation, the recursion relation
that allows all the wave functions to be generated from the bottom up.

68. Using the explicit form for the normalized ground state wave function we find that
Another useful form of this follows from the relation 𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉 derived earlier, from which it follows that
giving, in the position representation, the recursion relation
that allows all the wave functions to be generated from the bottom up.

69. Using the explicit form for the normalized ground state wave function we find that
Another useful form of this follows from the relation 𝑎 + 𝑛 = 𝑛+1 |𝑛+1〉 derived earlier, from which it follows that
giving, in the position representation, the recursion relation
that allows all the wave functions to be generated from the bottom up.

70. We derive below the first three harmonic oscillator wave functions
For 𝑛=0 we have graphed as solid lines, with the probability densities indicated as dashed lines

71. The first three harmonic oscillator wave functions
For 𝑛=1

72. The first three harmonic oscillator wave functions
For 𝑛=1

73. The first three harmonic oscillator wave functions
For 𝑛=2

74. The first three harmonic oscillator wave functions
For 𝑛=2

75. The wave function is customarily expressed in terms of the 𝑛th order Hermite polynomial 𝐻 𝑛 (𝑧), defined through the relation
With this definition
Since 𝐻 commutes with the parity operator, and because its spectrum is nondegenerate, the eigenstates of 𝐻 are also eigenstates of the parity operator.
The parity eigenvalue of the 𝑛th harmonic oscillator state is (−1)ⁿ

76. The wave function is customarily expressed in terms of the 𝑛th order Hermite polynomial 𝐻 𝑛 (𝑧), defined through the relation
With this definition
Since 𝐻 commutes with the parity operator, and because its spectrum is nondegenerate, the eigenstates of 𝐻 are also eigenstates of the parity operator.
The parity eigenvalue of the 𝑛th harmonic oscillator state is (−1)ⁿ

77. The wave function is customarily expressed in terms of the 𝑛th order Hermite polynomial 𝐻 𝑛 (𝑧), defined through the relation
With this definition
Since 𝐻 commutes with the parity operator, and because its spectrum is nondegenerate, the eigenstates of 𝐻 are also eigenstates of the parity operator.
The parity eigenvalue of the 𝑛th harmonic oscillator state is (−1)ⁿ

78. The wave function is customarily expressed in terms of the 𝑛th order Hermite polynomial 𝐻 𝑛 (𝑧), defined through the relation
With this definition
Since 𝐻 commutes with the parity operator, and because its spectrum is nondegenerate, the eigenstates of 𝐻 are also eigenstates of the parity operator.
The parity eigenvalue of the 𝑛th harmonic oscillator state is (−1)ⁿ

79. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
And so

80. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
And so

81. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
And so

82. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
And so

83. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
And so

84. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
And so

85. Action of Operators in the Energy Representation We now consider the action and expansion of various operators in the basis of energy eigenstates.
For example, the action of the number operator in this representation is particularly simple,
Thus, 𝑁|𝑛〉=𝑛|𝑛〉 so that 𝑛 ′ 𝑁 𝑛 =𝑛 𝑛 ′ 𝑛 =𝑛 𝛿 𝑛′,𝑛
Hence it follows that
Similarly, the harmonic oscillator Hamiltonian has the action
so that
and so

86. Action of Operators in the Energy Representation The action of the raising and lowering operators 𝑎 and 𝑎⁺ are also easily deduced.
We have already derived the relation
from which we deduce the matrix elements
and so
which is clearly not diagonal, even though it is represented by a single index Taking the adjoint of this last relation gives an expansion for the annihilation operator

87. Action of Operators in the Energy Representation The action of the raising and lowering operators 𝑎 and 𝑎⁺ are also easily deduced.
We have already derived the relation
from which we deduce the matrix elements
and so
which is clearly not diagonal, even though it is represented by a single index Taking the adjoint of this last relation gives an expansion for the annihilation operator

88. Action of Operators in the Energy Representation The action of the raising and lowering operators 𝑎 and 𝑎⁺ are also easily deduced.
We have already derived the relation
from which we deduce the matrix elements
and so
which is clearly not diagonal, even though it is represented by a single index Taking the adjoint of this last relation gives an expansion for the annihilation operator

89. Action of Operators in the Energy Representation The action of the raising and lowering operators 𝑎 and 𝑎⁺ are also easily deduced.
We have already derived the relation
from which we deduce the matrix elements
and so
which is clearly not diagonal, even though it is represented by a single index Taking the adjoint of this last relation gives an expansion for the annihilation operator

90. Action of Operators in the Energy Representation The action of the raising and lowering operators 𝑎 and 𝑎⁺ are also easily deduced.
We have already derived the relation
from which we deduce the matrix elements
and so
which is clearly not diagonal, even though it is represented by a single index Taking the adjoint of this last relation gives an expansion for the lowering operator

91. Action of Operators in the Energy Representation The action of the raising and lowering operators 𝑎 and 𝑎⁺ are also easily deduced.
We have already derived the relation
from which we deduce the matrix elements
and so
which is clearly not diagonal, even though it is represented by a single index Taking the adjoint of this last relation gives an expansion for the lowering operator

92. Action of Operators in the Energy Representation which implies that
From these we can derive relations for the position and momentum operators
and
Thus, for example, we deduce that
so that
and thus

93. Action of Operators in the Energy Representation which implies that
From these we can derive relations for the position and momentum operators
and
Thus, for example, we deduce that
so that
and thus

94. Action of Operators in the Energy Representation which implies that
From these we can derive relations for the position and momentum operators
and
Thus, for example, we deduce that
so that
and thus

95. Action of Operators in the Energy Representation which implies that
From these we can derive relations for the position and momentum operators
and
Thus, for example, we deduce that
so that
and thus

96. Action of Operators in the Energy Representation
Similarly,
so that
and thus

97. Action of Operators in the Energy Representation
Similarly,
so that
and thus

98. Action of Operators in the Energy Representation
Similarly,
so that
and thus

99. Action of Operators in the Energy Representation
Thus, the operators 𝑁 and 𝐻 are diagonal in the energy representation, while the operators 𝑎, 𝑎⁺, 𝑞 , and 𝑞 connect each energy eigenstate to the states immediately above or below it.
The matrices representing these operators are straightforward to construct, and appear below

100. Action of Operators in the Energy Representation
Thus, the operators 𝑁 and 𝐻 are diagonal in the energy representation, while the operators 𝑎, 𝑎⁺, 𝑞 , and 𝑞 connect each energy eigenstate to the states immediately above or below it.
The matrices representing these operators are straightforward to construct, and appear below

101. Action of Operators in the Energy Representation
Thus, the operators 𝑁 and 𝐻 are diagonal in the energy representation, while the operators 𝑎, 𝑎⁺, 𝑞 , and 𝑞 connect each energy eigenstate to the states immediately above or below it.
The matrices representing these operators are straightforward to construct, and appear below

102. Action of Operators in the Energy Representation
Thus, the operators 𝑁 and 𝐻 are diagonal in the energy representation, while the operators 𝑎, 𝑎⁺, 𝑞 , and 𝑞 connect each energy eigenstate to the states immediately above or below it.
The matrices representing these operators are straightforward to construct, and appear below

103. Action of Operators in the Energy Representation

104. Action of Operators in the Energy Representation

105. Action of Operators in the Energy Representation

106. Action of Operators in the Energy Representation

107. Action of Operators in the Energy Representation
It is clear from the structure of these last two matrices that the mean position and momentum associated with any eigenstate vanishes, i.e.,
a fact that also follows from the symmetry (i.e., parity) of the wave functions.
It is also interesting to consider the mean value
Note that in any expression involving the expectation value of a product of 𝑎 's and 𝑎⁺ 's with respect to an energy eigenstate, the only terms that can survive are those with an equal number of 𝑎's and 𝑎⁺'s

108. Action of Operators in the Energy Representation
It is clear from the structure of these last two matrices that the mean position and momentum associated with any eigenstate vanishes, i.e.,
a fact that also follows from the symmetry (i.e., parity) of the wave functions.
It is also interesting to consider the mean value
Note that in any expression involving the expectation value of a product of 𝑎 's and 𝑎⁺ 's with respect to an energy eigenstate, the only terms that can survive are those with an equal number of 𝑎's and 𝑎⁺'s

109. Action of Operators in the Energy Representation
It is clear from the structure of these last two matrices that the mean position and momentum associated with any eigenstate vanishes, i.e.,
a fact that also follows from the symmetry (i.e., parity) of the wave functions.
It is also interesting to consider the mean value
Note that in any expression involving the expectation value of a product of 𝑎 's and 𝑎⁺ 's with respect to an energy eigenstate, the only terms that can survive are those with an equal number of 𝑎's and 𝑎⁺'s

110. Action of Operators in the Energy Representation
Thus, we find that
so the uncertainty in position
increases with the quantum number 𝑛.

111. Action of Operators in the Energy Representation
Thus, we find that
so the uncertainty in position
increases with the quantum number 𝑛.

112. Action of Operators in the Energy Representation
Thus, we find that
so the uncertainty in position
increases with the quantum number 𝑛.

113. Action of Operators in the Energy Representation
Thus, we find that
so the uncertainty in position
increases with the quantum number 𝑛.

114. Action of Operators in the Energy Representation
Thus, we find that
so the uncertainty in position
increases with the quantum number 𝑛.

115. Action of Operators in the Energy Representation
Thus, we find that
so the uncertainty in position
increases with the quantum number 𝑛.

116. Action of Operators in the Energy Representation
Similarly, we can consider the second moment of the momentum
so that

117. Action of Operators in the Energy Representation
Similarly, we can consider the second moment of the momentum
so that

118. Action of Operators in the Energy Representation
Similarly, we can consider the second moment of the momentum
so that

119. Action of Operators in the Energy Representation
Similarly, we can consider the second moment of the momentum −
so that

120. Action of Operators in the Energy Representation
The uncertainty product is therefore
which clearly satisfies the uncertainty theorem.

121. Action of Operators in the Energy Representation
The uncertainty product is therefore
which clearly satisfies the uncertainty theorem.

122. Time Evolution of the Harmonic Oscillator
Having solved the eigenvalue problem for the time-independent harmonic oscillator Hamiltonian we have essentially solved the time evolution problem as well.
We can immediately construct the evolution operator 𝑈(𝑡)=exp⁡(−𝑖𝐻𝑡/ℏ), which is diagonal in the energy representation and given in the present problem by
where so

123. Time Evolution of the Harmonic Oscillator
Having solved the eigenvalue problem for the time-independent harmonic oscillator Hamiltonian we have essentially solved the time evolution problem as well.
We can immediately construct the evolution operator 𝑈(𝑡)=exp⁡(−𝑖𝐻𝑡/ℏ), which is diagonal in the energy representation and given in the present problem by
where so

124. Time Evolution of the Harmonic Oscillator
Having solved the eigenvalue problem for the time-independent harmonic oscillator Hamiltonian we have essentially solved the time evolution problem as well.
We can immediately construct the evolution operator 𝑈(𝑡)=exp⁡(−𝑖𝐻𝑡/ℏ), which is diagonal in the energy representation and given in the present problem by
where so

125. Time Evolution of the Harmonic Oscillator
If the system is initially in the state |𝜓(0)〉 at time 𝑡=0, associated with the wave function 𝜓(𝑞,0) = 〈𝑞|𝜓(0)〉
then its state at time 𝑡 is represented by the expansion
where
and the wave function can be then be obtained as an expansion . . .

126. Time Evolution of the Harmonic Oscillator
If the system is initially in the state |𝜓(0)〉 at time 𝑡=0, associated with the wave function 𝜓(𝑞,0) = 〈𝑞|𝜓(0)〉
then its state at time 𝑡 is represented by the expansion
where
and the wave function can be then be obtained as an expansion . . .

127. Time Evolution of the Harmonic Oscillator
If the system is initially in the state |𝜓(0)〉 at time 𝑡=0, associated with the wave function 𝜓(𝑞,0) = 〈𝑞|𝜓(0)〉
then its state at time 𝑡 is represented by the expansion
where
and the wave function can be then be obtained as an expansion . . .

128. Time Evolution of the Harmonic Oscillator
If the system is initially in the state |𝜓(0)〉 at time 𝑡=0, associated with the wave function 𝜓(𝑞,0) = 〈𝑞|𝜓(0)〉
then its state at time 𝑡 is represented by the expansion
where
and the wave function can be then be obtained as an expansion . . .

129. Time Evolution of the Harmonic Oscillator
in harmonic oscillator energy eigenfunctions

130. Time Evolution of the Harmonic Oscillator
in harmonic oscillator energy eigenfunctions

131. In this lecture, we completed our study of the quantum mechanical harmonic oscillator.
Having previously determined the spectrum of the Hamiltonian, we showed that each eigenvalue is non-degenerate, and developed the tools to contruct and ONB of energy eigenstates, that we could express through a recursion relation giving the corresponding energy eigenfunctions.
We went on to determine the action of operators of interest in the energy representation, and computed their matrix elements, which we used to generate ket-bra expansions, and explicit matrices representing them in this basis.
Finally, we considered the evolution of the system, seeing how to compute the state of the system, or its wave function, at an arbitrary time following an arbitrary initial condition.
With that, we bring to a close our study of the first semester quantum mechanics course. I hope to see you for Quantum Two.

132. In this lecture, we completed our study of the quantum mechanical harmonic oscillator.
Having previously determined the spectrum of the Hamiltonian, we showed that each eigenvalue is non-degenerate, and developed the tools to contruct and ONB of energy eigenstates, that we could express through a recursion relation giving the corresponding energy eigenfunctions.
We went on to determine the action of operators of interest in the energy representation, and computed their matrix elements, which we used to generate ket-bra expansions, and explicit matrices representing them in this basis.
Finally, we considered the evolution of the system, seeing how to compute the state of the system, or its wave function, at an arbitrary time following an arbitrary initial condition.
With that, we bring to a close our study of the first semester quantum mechanics course. I hope to see you for Quantum Two.

133. In this lecture, we completed our study of the quantum mechanical harmonic oscillator.
Having previously determined the spectrum of the Hamiltonian, we showed that each eigenvalue is non-degenerate, and developed the tools to contruct and ONB of energy eigenstates, that we could express through a recursion relation giving the corresponding energy eigenfunctions.
We went on to determine the action of operators of interest in the energy representation, and computed their matrix elements, which we used to generate ket-bra expansions, and explicit matrices representing them in this basis.
Finally, we considered the evolution of the system, seeing how to compute the state of the system, or its wave function, at an arbitrary time following an arbitrary initial condition.
With that, we bring to a close our study of the first semester quantum mechanics course. I hope to see you for Quantum Two.

134. In this lecture, we completed our study of the quantum mechanical harmonic oscillator.
Having previously determined the spectrum of the Hamiltonian, we showed that each eigenvalue is non-degenerate, and developed the tools to contruct and ONB of energy eigenstates, that we could express through a recursion relation giving the corresponding energy eigenfunctions.
We went on to determine the action of operators of interest in the energy representation, and computed their matrix elements, which we used to generate ket-bra expansions, and explicit matrices representing them in this basis.
Finally, we considered the evolution of the system, seeing how to compute the state of the system, or its wave function, at an arbitrary time following an arbitrary initial condition.
With that, we bring to a close our study of the first semester quantum mechanics course. I hope to see you for Quantum Two.