Figuring out if the state x is an eigen state of the hamiltonian

  • #1
I had encountered a problem for which I need to know how to proceed in order to solve it.
Taking a particle m with box potential (one dimensional) where V(x) = 0 when mod(x) <=a and V(x) = infinity when mod(x) > a and where wave function phi(x) = A (phi1(x) + ph2(x)) where phi1(x) and phi2(x) are normalized wave functions of the ground state and first excited state

How do we determine if the state phi an eigenstate of the hamiltonian.

Is my approach correct.

If the state phi has to be an eigenstate of the hamiltonian it has to satisfy two conditions.

a) H phi = E phi -> schrondinger equation.

b) phi (x) tends to 0 when mod(x) tends to infinity.

If this is correct how can I prove that this satisfies a and b.


if it satisfies condition (a) then

(-h^2/dm d^2/dx^2 + V(x)) phi (x) = E phi(x)

when I insert phi(x) = A (phi1(x) + ph2(x)) where phi1(x) = AcosKx1 + BsinKx1

and phi2(x) = AcosKx2 + BsinKx2

phi(x) = A(cosKx1 + coskx2) + B(sinKx1+sinKx2)

phi(x) = 2A*cos(kx1+kx2)/2*cos(kx1-kx2)/2 + 2B*sin(kx1+kx2)/2*cos (kx1-kx2)/2

How do I proceed further ???

How do I test for condition (b)

Regards.
QP
 

Answers and Replies

  • #2
Meir Achuz
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In your case, phi must equal zero at x=+/- A.
 
  • #3
could you please elaborate more on this.You mean it need not satisfy condition a and b or what is to be done.
 
  • #4
nrqed
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I had encountered a problem for which I need to know how to proceed in order to solve it.
Taking a particle m with box potential (one dimensional) where V(x) = 0 when mod(x) <=a and V(x) = infinity when mod(x) > a and where wave function phi(x) = A (phi1(x) + ph2(x)) where phi1(x) and phi2(x) are normalized wave functions of the ground state and first excited state

How do we determine if the state phi an eigenstate of the hamiltonian.

Is my approach correct.

If the state phi has to be an eigenstate of the hamiltonian it has to satisfy two conditions.

a) H phi = E phi -> schrondinger equation.

b) phi (x) tends to 0 when mod(x) tends to infinity.

If this is correct how can I prove that this satisfies a and b.


if it satisfies condition (a) then

(-h^2/dm d^2/dx^2 + V(x)) phi (x) = E phi(x)

when I insert phi(x) = A (phi1(x) + ph2(x)) where phi1(x) = AcosKx1 + BsinKx1

and phi2(x) = AcosKx2 + BsinKx2

phi(x) = A(cosKx1 + coskx2) + B(sinKx1+sinKx2)

phi(x) = 2A*cos(kx1+kx2)/2*cos(kx1-kx2)/2 + 2B*sin(kx1+kx2)/2*cos (kx1-kx2)/2

How do I proceed further ???

How do I test for condition (b)

Regards.
QP

It's much simpler than that.

you have to check whether H applied on the wavefunction gives a constant times the wavefunction .

So you need to calculate [tex] H (\psi_1 + \psi_2) = H \psi_1 + H \psi_2 [/tex]
But since these are solutions of the Schrodinger equation (which is simply an eigenvalue equation for H) you already know what happens when you apply H on psi_1 and on psi_2, right?
 
  • #5
So you mean we need need not see if it satisfied a or b then all that we need to check is this H(psi_1 + psi_2) = hpsi_1 + h_psi2
 
  • #6
I mean condition a) or b) as mentioned in the problem.
 
  • #7
Yes I know psi_1(x) and psi_2(x) it is as follows.

psi_1(x) = 1/sqrt(a) cos pi*x/2a
psi_2(x) = 1/sqrt(a) sin pi*x/a


How do I find H(psi_1(x))
 
  • #8
As I understand H = -h sqr /2m * d(square)/dx(square) + V(x)
So do you mean we have do double differentiate psi_1(x) with respect to x then what happens to V(x) part.How do we calculate that.
 
  • #9
nrqed
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So you mean we need need not see if it satisfied a or b then all that we need to check is this H(psi_1 + psi_2) = hpsi_1 + h_psi2

That's not what I meant. Because of the linearity of H, I know[] that H(psi_1 + psi_2) = H psi_1 + H psi_2.

I meant that now you simply have to calculate (H psi_1) and (H psi_2) and then verify if

H(psi_1 + psi_2) = a constant times (psi_1 + psi_2)
 
  • #10
nrqed
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As I understand H = -h sqr /2m * d(square)/dx(square) + V(x)
So do you mean we have do double differentiate psi_1(x) with respect to x then what happens to V(x) part.How do we calculate that.

You could do the full calculation but you should already know the answer. The time independent schrodinger equation is H psi = E psi. So if you know that a wavefunction is a solution of the TISE with some energy "E", you know what happens when you apply the hamiltonian to it.

(to answer your question about V(x), what is the value of V(x) inside the well?)
 
  • #11
Hφ = Eφ, so Eφ = A(E1 φ1 + E2 φ2)

φ is not an eigenstate of the hamiltonian because E1 is not equal to E2.
In this case how do we find the expectation value of hamiltonian H and its relation to E1 and E2.V(x) is zero in well.I thought this approach is better way of attacking the problem instead of trying to solve it.
 
  • #12
nrqed
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Hφ = Eφ, so Eφ = A(E1 φ1 + E2 φ2)

φ is not an eigenstate of the hamiltonian because E1 is not equal to E2.
Right! You see that this was much more efficient than actually taking second derivatives of the wavefunction and so on!!
In this case how do we find the expectation value of hamiltonian H and its relation to E1 and E2.V(x) is zero in well.I thought this approach is better way of attacking the problem instead of trying to solve it.
the expectation value is simply
[tex] <H> = \int dx (\psi_1^* + \psi_2^*) H (\psi_1 + \psi_2) [/tex]
which is, using the result you gave above,
[tex] \int dx (\psi_1^* + \psi_2^*) (E_1 \psi_1 + E_2 \psi_2) [/tex]
Do you see what this gives? Do you know about the orthonormality properties of the wavefunctions [itex] \psi_n(x) [/itex] of the infinite square well?
 
  • #13
Dont we need to add -infinity to infinity to this integral by definition for the definition of H.I have gone through orthonormality.
 
  • #14
Assume that the wave function (x, t) at time t = 0 is given by
[tex] \psi (x,0) = \phi (x) [/tex] as defined above. Find [tex] \psi (x, t) and \psi (x, t)^2 [/tex]. We need to express the latter in terms of sine or cosine functions, eliminating the
exponentials with the help of Euler’s formula.

Abbreviation to be used is:

[tex] w = \pi ^2 h/8 m a^2 [/tex]

How do I proceed in this regard.

I know that

ih dow W /dow t = -h^2/2m dow^2/dow x^2 + V(x) psi

Since V(x)=0 that term disappears so we have

[tex] ih dow \psi / dow t = -h^2 /2m dow^2/dow x^2 [/tex]




Euler formula.

as I understand [tex] e^i[\theta] = \cos\theta + i \sin\theta[/tex]

Thanks.
Regards.
QP
 
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