What Do Coefficients and Expectation Values Mean in Quantum Mechanics?

In summary: 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. 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 <H><H>\langle \hat{H} \rangle to get the energy of the system.So does this mean that the Hamiltonian represents the total energy of the system, while the wave function represents the state of the system at a particular instance (which can be a superposition of eigenstates)?
  • #1
WWCY
479
12

Homework Statement



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Screen Shot 2017-09-06 at 11.53.03 PM.png


I have a few questions I'd like to ask about this example. (C1 was already derived before the second part)

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.
 
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  • #2
(Note: This is from Griffiths, Introduction to Quantum Mechanics).

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

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

WWCY said:
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.

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

WWCY said:
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.

WWCY said:
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}##.

WWCY said:
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?

WWCY said:
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.
 
  • #3
Thank you so much for the reply,

DrClaude said:
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?

DrClaude said:
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.

DrClaude said:
^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?

DrClaude said:
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.
 
  • #4
Lets start with
WWCY said:
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

WWCY said:
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]
 
  • #5
Apologies for the late reply, I have been busy with school lately!

DrClaude said:
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?

DrClaude said:
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?

Thank you for your assistance.
 
  • #6
WWCY said:
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##.)

WWCY said:
Is it right to interpret the initial wave-function then, as some superposition of ##\psi_1## and some other "higher-level" states?
Yes.
 
  • #7
DrClaude said:
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.

Thank you for your assistance.
 
  • #8
WWCY said:
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.

WWCY said:
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.
 
  • #9
DrClaude said:
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!
 

Related to What Do Coefficients and Expectation Values Mean in Quantum Mechanics?

What is an infinite potential well in quantum mechanics?

An infinite potential well is a theoretical model used in quantum mechanics to study the behavior of a particle confined to a certain region. It is often used as a simplified representation of a particle trapped inside a box with infinitely high walls.

What is the significance of the infinite potential well in quantum mechanics?

The infinite potential well provides a simple yet powerful tool for understanding the fundamental principles of quantum mechanics, such as wave-particle duality and quantization of energy levels. It also serves as a building block for more complex quantum systems.

How is the wave function of a particle in an infinite potential well described?

The wave function of a particle in an infinite potential well is described by a piecewise function, with a value of zero outside the well and a non-zero value within the well. It follows the Schrödinger equation and can be represented as a standing wave with discrete energy levels.

What are the boundary conditions for the wave function in an infinite potential well?

The boundary conditions for the wave function in an infinite potential well require that the wave function must be continuous and have a derivative of zero at the boundaries of the well. This ensures that the particle is confined within the well and cannot escape.

Can the infinite potential well model be applied to real-world systems?

While the infinite potential well is a simplified model, it can still provide insights into the behavior of particles in more complex systems. It has been used to describe the behavior of electrons in atoms, as well as the vibrations of molecules. However, it is important to note that in reality, there is no such thing as an infinitely high potential well.

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