Probability of measuring specific energy of particle (in a box)

You have a sum of complex numbers, which is fine, but then you're multiplying the sum by something else, and you're not making it clear which part of the sum is being multiplied by that something else.In summary, the conversation discusses a particle-in-a-box problem in which a particle of mass m is subject to a potential with infinite barriers at the endpoints of the box and zero potential in between. The eigenstates of the Hamiltonian H and their corresponding eigenvalues are given, along with the state of the particle at time t=0 and the condition that it is normalized. The conversation also presents questions about the probability of finding a certain energy value, the mean
  • #1
doorknob
3
0

Homework Statement


Consider a particle-in-a-box problem, involving a
particle of mass m subject to a potential:
V(x) = +∞ for X≤0
V(x) = 0 for 0<X<L
V(x) = +∞ for X≥L

|ϕn> are the eigenstates of the Hamiltonian H with the corresponding eigenvalues:
En = (n^2)*(hbar^2)*(Pi^2) / 2mL^2

The state of the particle at the instant t = 0 is:
|Ψ(0)> = c1 |ϕ1> + c2 |ϕ2> + c3 |ϕ3> + c4 |ϕ4>

where ci are complex coefficients that satisfy Σ|ci|^2 = 1 (i.e., the state is normalized).

a) When the energy of the particle in the state |Ψ(0)> is measured what is the probability of
finding a value smaller than
3*(hbar^2)*(Pi^2) /mL^2?

b) What is the mean energy <H> and the uncertainty ΔE = [<H^2> − <H>^2] in the energy of the particle in the state |Ψ(0)> ?

c) Give the expression for the state vector |Ψ(t)> at time t. Do the results found in parts (a) and
(b) for t = 0 remain valid at an arbitrary time t? Make sure to show how you arrived at your conclusion.

d) The energy was measured at time t and the result
8*(hbar^2)*(Pi^2) /mL^2?
was found. After this measurement,
what is the state of the system? What is the result if the energy is measured again?

The Attempt at a Solution



a) the probability of obtaining an eigenvalue corresponding to the eigenstate |Ψ(t)> is:

|<ϕi|Ψ(t)>|^2

However, how do I compare the value to 3*(hbar^2)*(Pi^2) /mL^2? I do not know what the complex coefficients (c1,c2..c4) are.

b) Expectation energy <H> is defined as:
<H> = <Ψ|H|Ψ>

=> <H> = c1*E1<ϕ1|ϕ1> + c2*E2<ϕ2|ϕ2> + c3*E3<ϕ3|ϕ3> + c4*E4<ϕ4|ϕ4>
since normalized, => <ϕi|ϕi> = 1
So <H> = c1*E1 + c2*E2 + c3*E3 +c4*E4
But I do not know what the complex coefficients (c1,c2..c4) are.

c) don't know since have not completed part (a) or (b)

d) I would guess 3nd excited state (I'm assuming n=1 is the ground state)? Since if n=4, then 4^2 = 16.

16*(hbar^2)*(Pi^2) /2mL^2
reduces to 8*(hbar^2)*(Pi^2) /mL^2

After a measurement has been made, the system |Ψ> collapses to one of its eigenstates where it remains forever. If the energy were measured again, it should remain the same value as when you first made the measurement (since the system collapsed).

Any comments, feedback, criticism, would be helpful. I am more interested in the concept and understanding the material. Thank you in advance!
 
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  • #2
doorknob said:

Homework Statement


Consider a particle-in-a-box problem, involving a
particle of mass m subject to a potential:
V(x) = +∞ for X≤0
V(x) = 0 for 0<X<L
V(x) = +∞ for X≥L

|ϕn> are the eigenstates of the Hamiltonian H with the corresponding eigenvalues:
En = (n^2)*(hbar^2)*(Pi^2) / 2mL^2

The state of the particle at the instant t = 0 is:
|Ψ(0)> = c1 |ϕ1> + c2 |ϕ2> + c3 |ϕ3> + c4 |ϕ4>

where ci are complex coefficients that satisfy Σ|ci|^2 = 1 (i.e., the state is normalized).

a) When the energy of the particle in the state |Ψ(0)> is measured what is the probability of
finding a value smaller than
3*(hbar^2)*(Pi^2) /mL^2?

b) What is the mean energy <H> and the uncertainty ΔE = [<H^2> − <H>^2] in the energy of the particle in the state |Ψ(0)> ?

c) Give the expression for the state vector |Ψ(t)> at time t. Do the results found in parts (a) and
(b) for t = 0 remain valid at an arbitrary time t? Make sure to show how you arrived at your conclusion.

d) The energy was measured at time t and the result
8*(hbar^2)*(Pi^2) /mL^2?
was found. After this measurement,
what is the state of the system? What is the result if the energy is measured again?

The Attempt at a Solution



a) the probability of obtaining an eigenvalue corresponding to the eigenstate |Ψ(t)> is:

|<ϕi|Ψ(t)>|^2

However, how do I compare the value to 3*(hbar^2)*(Pi^2) /mL^2? I do not know what the complex coefficients (c1,c2..c4) are.
I'd expect your answer is supposed to be written in terms of the ci's.
b) Expectation energy <H> is defined as:
<H> = <Ψ|H|Ψ>

=> <H> = c1*E1<ϕ1|ϕ1> + c2*E2<ϕ2|ϕ2> + c3*E3<ϕ3|ϕ3> + c4*E4<ϕ4|ϕ4>
since normalized, => <ϕi|ϕi> = 1
So <H> = c1*E1 + c2*E2 + c3*E3 +c4*E4
But I do not know what the complex coefficients (c1,c2..c4) are.
The ci's are complex, so your expression for <H> will generally turn out to be a complex quantity. Does that make sense? You need to be a bit more careful in your calculations.
c) don't know since have not completed part (a) or (b)

d) I would guess 3nd excited state (I'm assuming n=1 is the ground state)? Since if n=4, then 4^2 = 16.

16*(hbar^2)*(Pi^2) /2mL^2
reduces to 8*(hbar^2)*(Pi^2) /mL^2

After a measurement has been made, the system |Ψ> collapses to one of its eigenstates where it remains forever. If the energy were measured again, it should remain the same value as when you first made the measurement (since the system collapsed).
That's right.
 
  • #3
Vela,
Thank you for your response. However, can I use the fact that I know:

Σ|ci|^2 = 1 (i.e., the state is normalized)

to obtain an expression without Ci? For example, when calculating mean energy <H>:
Expectation energy <H> is defined as:
<H> = <Ψ|H|Ψ>

*Note, (my mistake) my initial expression for <H> did not include the squared complex coefficients. It is now resolved to:
=> <H> = (c1^2)*E1<ϕ1|ϕ1> + (c2^2)*E2<ϕ2|ϕ2> + (c3^2)*E3<ϕ3|ϕ3> + (c4^2)*E4<ϕ4|ϕ4>

=> <H> = Σci*Ei*<ϕi|ϕi>

and since <ϕi|ϕi> = 1 we get:
Σ(|ci|^2)*ΣEi
= ΣEi

Is my line of reasoning correct?
 
  • #4
doorknob said:
Vela,
Thank you for your response. However, can I use the fact that I know:

Σ|ci|^2 = 1 (i.e., the state is normalized)

to obtain an expression without Ci? For example, when calculating mean energy <H>:
Expectation energy <H> is defined as:
<H> = <Ψ|H|Ψ>

*Note, (my mistake) my initial expression for <H> did not include the squared complex coefficients. It is now resolved to:
=> <H> = (c1^2)*E1<ϕ1|ϕ1> + (c2^2)*E2<ϕ2|ϕ2> + (c3^2)*E3<ϕ3|ϕ3> + (c4^2)*E4<ϕ4|ϕ4>
That's better but still not quite right. I hope you realize that ci2≠|ci|2.
=> <H> = Σci*Ei*<ϕi|ϕi>

and since <ϕi|ϕi> = 1 we get:
Σ(|ci|^2)*ΣEi
= ΣEi

Is my line of reasoning correct?
These remaining lines are all wrong.
 
  • #5
vela said:
That's better but still not quite right. I hope you realize that ci2≠|ci|2.

These remaining lines are all wrong.

Wow...okay, I guess I'll end up with an expression in terms of Ci's. Also, when you said "these remaining lines are all wrong" did you also mean that I was wrong to make the following assumption:

since normalized, => <ϕi|ϕi> = 1
 
  • #6
No, you're correct about the normalization.
 

What is the probability of measuring a specific energy of a particle in a box?

The probability of measuring a specific energy of a particle in a box depends on the energy levels of the particle and the size of the box. If the energy levels are evenly spaced, then the probability is equal for each level. However, if the energy levels are not evenly spaced, then the probability will vary for each level.

How does the size of the box affect the probability of measuring a specific energy?

The size of the box affects the probability of measuring a specific energy because it determines the possible energy levels that the particle can have. A larger box will have more energy levels, resulting in a higher probability of measuring a specific energy.

Can the probability of measuring a specific energy ever be 100%?

In theory, the probability of measuring a specific energy can never be 100%, as there is always a chance for error or uncertainty in the measurement. However, the probability can be very close to 100% if the energy levels are evenly spaced and the box size is large enough.

How does the uncertainty principle affect the probability of measuring a specific energy?

The uncertainty principle states that it is impossible to know both the exact position and momentum of a particle simultaneously. This means that there will always be a level of uncertainty in the measurement of a particle's energy, which can affect the probability of measuring a specific energy.

Does the probability of measuring a specific energy change over time?

The probability of measuring a specific energy can change over time if the particle is in a state of superposition, meaning it has a chance of being in multiple energy levels simultaneously. As time passes, the probability of measuring a specific energy will change as the particle's state changes.

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