# QM: Infinite Potential Well

## Homework Statement

1. What does the line "The rest of the coefficients make up the difference" actually mean?

2. What does "As one might expect....because of the admixture of the excited states" mean?

3. Does the <H> in this case represent the expectation value of total energy in the system?

Questions slightly unrelated to the example above:

4. If a Hamiltonian is written as such: Hψn = Enψn, does it now represent the total allowed energy of the wavefunction with quantum number n?

5. If I derived an equation for ψn(x) in the case of a infinite potential well, and derived an expression for its corresponding coefficient, before putting it all together into a linear combination for the general solution for the time-dependent Schrodinger equation, what exactly am I doing?

2. Homework Equations

## The Attempt at a Solution

1 and 2) These lines, to me, suggest that there are other Ψn's in the system apart from that of the ground state. If true, how is this possible? Do excited states not have different "shapes" (different n's different number of nodes)?
If not, what should my interpretation of these statements be?

3 and 4) I have a feeling that H can both describe the total energy in a system and of a particle in a specific state n.

5) My interpretation was that the general solution would also tell me the possible measurement outcomes of energy and their corresponding probabilities. If this is true, is it also true that there is a possibility of measuring more than 1 energy level despite there being only a single particle in the box?

Many, many thanks to anyone who takes the time to help clear up my misconceptions.

Last edited:

DrClaude
Mentor
(Note: This is from Griffiths, Introduction to Quantum Mechanics).

1. What does the line "The rest of the coefficients make up the difference" actually mean?

2. What does "As one might expect....because of the admixture of the excited states" mean?

1 and 2) These lines, to me, suggest that there are other Ψn's in the system apart from that of the ground state. If true, how is this possible? Do excited states not have different "shapes" (different n's different number of nodes)?
The eigenfunctions are given by eq. (2.28). As you can see, the ground state is a sine, not a parabola, hence ##\Psi(x,0)## is not any particular eigenstate of the well, but a superposition of different eigenstates, although it is mostly like the ground state. Yes, the excited states have correspond to higher frequency sines, with nodes, but that doesn't matter. The eigenfunctions form a complete basis, such that any function that goes to zero at 0 and a can be represented as a sum of those eigenfunctions.

3. Does the <H> in this case represent the expectation value of total energy in the system?
Yes.

4. If a Hamiltonian is written as such: Hψn = Enψn, does it now represent the total allowed energy of the wavefunction with quantum number n?
The Hamiltonian corresponds to the physical situation, namely in this particular case a particle in a 1D box with infinite walls. You have to distinguish that from the wave function, which corresponds to the state of a particular instance of this quantum system. If the system is in state ##\psi_n##, then it has a definite energy that is given directly by the energy eigenvalue. If the system is not in a single eigenstate, but in a superposition of such states, as in the case above, then one must calculate ##\langle \hat{H} \rangle## to get the energy of the system.

3 and 4) I have a feeling that H can both describe the total energy in a system and of a particle in a specific state n.
##\hat{H}## is the energy operator. That operator has eigenstates ##\psi_n##. The state actual physical system is specified by its wave function ##\Psi##, which can be expressed in terms of the eigenfunctions of ##\hat{H}##.

5. If I derived an equation for ψn(x) in the case of a infinite potential well, and derived an expression for its corresponding coefficient, before putting it all together into a linear combination for the general solution for the time-dependent Schrodinger equation, what exactly am I doing?

5) My interpretation was that the general solution would also tell me the possible measurement outcomes of energy and their corresponding probabilities. If this is true, is it also true that there is a possibility of measuring more than 1 energy level despite there being only a single particle in the box?
You have to distinguish between an expectation value and the result of a measurement. When a measurement of the energy is made, then after measurement the system will be found in an eigenstate of the Hamiltonian, and the value of the energy measured will be the corresponding eigenvalue. On the other hand, the expectation value gives you the average energy that would be measured if many identical quantum systems were prepared and each one measured.

Thank you so much for the reply,

The eigenfunctions are given by eq. (2.28). As you can see, the ground state is a sine, not a parabola, hence Ψ(x,0)Ψ(x,0)\Psi(x,0) is not any particular eigenstate of the well, but a superposition of different eigenstates, although it is mostly like the ground state. Yes, the excited states have correspond to higher frequency sines, with nodes, but that doesn't matter. The eigenfunctions form a complete basis, such that any function that goes to zero at 0 and a can be represented as a sum of those eigenfunctions.

Most of the wave-functions in an infinite potential well I have seen thus far (in my limited experience) seem to "stack" on top of each other from one level to another, this graph however, plots them as a superposition. Is this down to the fact that the y-axis does not correspond to energy levels?

The Hamiltonian corresponds to the physical situation, namely in this particular case a particle in a 1D box with infinite walls. You have to distinguish that from the wave function, which corresponds to the state of a particular instance of this quantum system. If the system is in state ψnψn\psi_n, then it has a definite energy that is given directly by the energy eigenvalue.

^HH^\hat{H} is the energy operator. That operator has eigenstates ψnψn\psi_n. The state actual physical system is specified by its wave function ΨΨ\Psi, which can be expressed in terms of the eigenfunctions of ^H

I'm afraid I still don't quite understand the meaning of the Hamiltonian. The text introduced it as a sort of operator that in classical-mechanics, described the total energy of a system. I've not reached the part of the text talking about eigenfunctions and eigenvalues yet, so I'm kind of struggling to understand your explanation. Is there a) another way to explain what a Hamiltonian is with respect to the 2 situations I've raised and b) somewhere I can get a beginner's crash course in the eigen-stuff?

When a measurement of the energy is made, then after measurement the system will be found in an eigenstate of the Hamiltonian, and the value of the energy measured will be the corresponding eigenvalue.

Does the eigenstate refer to a stationary-state solution that was part of the general solution to the time-dependent Schrodinger?

Thank you again for your patience.

DrClaude
Mentor
I'm afraid I still don't quite understand the meaning of the Hamiltonian. The text introduced it as a sort of operator that in classical-mechanics, described the total energy of a system. I've not reached the part of the text talking about eigenfunctions and eigenvalues yet, so I'm kind of struggling to understand your explanation. Is there a) another way to explain what a Hamiltonian is with respect to the 2 situations I've raised and b) somewhere I can get a beginner's crash course in the eigen-stuff?

Does the eigenstate refer to a stationary-state solution that was part of the general solution to the time-dependent Schrodinger?
which I think is the most important point to clarify first.

You are solving the time-independent Schrödinger equation, which is an eigenvalue problem
$$\hat{H} \psi_n = E_n \psi_n$$
with ##\hat{H}## corresponding to the Hamiltonian which, as in classical mechanics, is the total energy of the system. One thing that is new in QM is that it is an operator (instead of a function), and as such has eigenfunctions ##\psi_n##, function which are invariant upon the action of the operator, up to a real multiplicative factor (the eigenvalue).

Observables such as the Hamiltonian act on any quantum state ##\Psi##, and one can use that to calculate the expectation values of that observable, which is energy in the case of the Hamiltonian, see eq. (2.13). If ##\Psi = \psi_n##, then ##\langle H \rangle = E_n##, and ##\Psi## is a stationary state.

Given all the eigenstates ##\psi_n## of an observable, one can always write the state ##\Psi## of a system as
$$\Psi = \sum_n c_n \psi_n$$
with ##c_n \in \mathbb{C}##. If more than one ##c_n \neq 0##, then ##\Psi## is not a stationary state

Most of the wave-functions in an infinite potential well I have seen thus far (in my limited experience) seem to "stack" on top of each other from one level to another, this graph however, plots them as a superposition. Is this down to the fact that the y-axis does not correspond to energy levels?
That "stacking" is just for visualization. What is important is the sum.[/QUOTE]

Apologies for the late reply, I have been busy with school lately!

Observables such as the Hamiltonian act on any quantum state ##\Psi##, and one can use that to calculate the expectation values of that observable, which is energy in the case of the Hamiltonian, see eq. (2.13). If ##\Psi = \psi_n##, then ##\langle H \rangle = E_n##, and ##\Psi## is a stationary state.

Given all the eigenstates ##\psi_n## of an observable, one can always write the state ##\Psi## of a system as
$$\Psi = \sum_n c_n \psi_n$$
with ##c_n \in \mathbb{C}##. If more than one ##c_n \neq 0##, then ##\Psi## is not a stationary state

So if we don't have a stationary state general solution, ##\langle H \rangle## becomes ##\sum_n |c_n|^2E_n##,
what does it now represent?

That "stacking" is just for visualization. What is important is the sum.

Is it right to interpret the initial wave-function then, as some superposition of ##\psi_1## and some other "higher-level" states?

DrClaude
Mentor
So if we don't have a stationary state general solution, ##\langle H \rangle## becomes ##\sum_n |c_n|^2E_n##,
what does it now represent?
It is the expectation value of the energy. As for all expectations values, you should consider them as ensemble averages: identically prepare a huge number of systems, and measure them. The average value of those measurements will be the expectation value. (You can see that directly from the the equation you wrote, since ##|c_n|^2## is the probability of being in state ##n##.)

Is it right to interpret the initial wave-function then, as some superposition of ##\psi_1## and some other "higher-level" states?
Yes.

It is the expectation value of the energy. As for all expectations values, you should consider them as ensemble averages: identically prepare a huge number of systems, and measure them. The average value of those measurements will be the expectation value. (You can see that directly from the the equation you wrote, since ##|c_n|^2## is the probability of being in state ##n##.)

So is it right to say that ##\langle H \rangle## is the average amount of energy observed when we make measurements across an ensemble of particles with state ##\Psi##?

Also, how does one actually physically make such measurements, and how does one prepare an ensemble of particles in a certain state? A lot of the stuff I've read thus far don't really go into these things, which makes it all rather confusing.

DrClaude
Mentor
So is it right to say that ##\langle H \rangle## is the average amount of energy observed when we make measurements across an ensemble of particles with state ##\Psi##?
That is correct.

Also, how does one actually physically make such measurements, and how does one prepare an ensemble of particles in a certain state? A lot of the stuff I've read thus far don't really go into these things, which makes it all rather confusing.
This is too wide a question to be answered in a simple post. To give two examples, one can use filtering, as in the Stern-Gerlach experiment where only one spin polarization is kept (the same can be done for photons with polarizers), or you can prepare the state of an atom using lasers.

That is correct.

This is too wide a question to be answered in a simple post. To give two examples, one can use filtering, as in the Stern-Gerlach experiment where only one spin polarization is kept (the same can be done for photons with polarizers), or you can prepare the state of an atom using lasers.

Thank you so much for your help!