I Can a wave-function be a combination of states with different energies?

[dRic2]

Hi everyone, I was wondering if mu thinking was correct abut this argument, here is what I was thinking abut.

A wave-function that is an eigenfunction of the Hamiltonian always describes a state of definite energy. Ok now let's say the eigenvalue is discrete so the wave-function belongs to Hilbert's space. Therefore I can say that the wave-function can be described as a linear combination of other wave-functions. So, in other words, a wave-function that describes a definite state of energy can be expressed as a combination of wave-functions that describes other states of energy meaning a state of energy is the result of the probability combination of other states of energy, right? So if a particle is in a state with energy E_0, when I make a measurement I don't get E_0 because the wave-function collapses in one of its 'base' (linear-combination); then if Ido the average of all the values of E I measured (with infinite number of measurements) I get E_0. Am I correct?

 

[QuantumQuest]

If we have an orthonormal set of energy eigenfunctions ${u_n(x), n = 1, 2, 3\cdots}$, $Hu_n(x) = E_nu_n(x)$ for some Hamiltonian $H$. Then for some arbitrary wave function we can write $$\psi(x) = \sum_{n}^{}c_nu_n(x)$$ i.e. as a linear combination of the functions $u_n(x)$, where the (expansion) coefficients $c_n$ are given by $$c_n = \int_{}^{}dxu_n(x)^{*}\psi(x)$$ This sequence of coefficients is unique for the state of the system and if it is normalized the probability of making a measurement of energy is $|c_n|^2$. Now, if the wavefunction $\psi(x)$ belongs to Hilbert space, its norm $$ \int_{}^{}dx|\psi(x)|^2$$ which is the sum to $n$ of the probability $|c_n|^2$, is finite. So, the probability is conserved.

 

[dRic2]

isn't the same of what I said?

 

[PeroK]

Hi everyone, I was wondering if mu thinking was correct abut this argument, here is what I was thinking abut.

A wave-function that is an eigenfunction of the Hamiltonian always describes a state of definite energy. Ok now let's say the eigenvalue is discrete so the wave-function belongs to Hilbert's space. Therefore I can say that the wave-function can be described as a linear combination of other wave-functions. So, in other words, a wave-function that describes a definite state of energy can be expressed as a combination of wave-functions that describes other states of energy meaning a state of energy is the result of the probability combination of other states of energy, right? So if a particle is in a state with energy E_0, when I make a measurement I don't get E_0 because the wave-function collapses in one of its 'base' (linear-combination); then if Ido the average of all the values of E I measured (with infinite number of measurements) I get E_0. Am I correct?

No. If you have an energy eigenstate, then every measurement of energy gives the same result. The wavefunction does not change as a result of the measurement in this case.

That you can decompose an energy eigenstate into other states is irrelevant here.

 

[PeroK]

If we have an orthonormal set of energy eigenfunctions ${u_n(x), n = 1, 2, 3\cdots}$, $Hu_n(x) = E_nu_n(x)$ for some Hamiltonian $H$. Then for some arbitrary wave function we can write $$\psi(x) = \sum_{n}^{}c_nu_n(x)$$ i.e. as a linear combination of the functions $u_n(x)$, where the (expansion) coefficients $c_n$ are given by $$c_n = \int_{}^{}dxu_n(x)^{*}\psi(x)$$ This sequence of coefficients is unique for the state of the system and if it is normalized the probability of making a measurement of energy is $|c_n|^2$. Now, if the wavefunction $\psi(x)$ belongs to Hilbert space its norm $$ \int_{}^{}dx|\psi(x)|^2$$ which is the sum to $n$ of the probability $|c_n|^2$, is finite. So, the probability is conserved.

This doesn't address the question, which is about an energy eigenstate.

 

[QuantumQuest]

This doesn't address the question, which is about an energy eigenstate.

This one in the OP

So if a particle is in a state with energy E_0, when I make a measurement I don't get E_0 because the wave-function collapses in one of its 'base' (linear-combination); then if Ido the average of all the values of E I measured (with infinite number of measurements) I get E_0. Am I correct?

was added after my post.

 

[dRic2]

No, if you have an energy eigenstate, then every measurement of energy gives the same result. The wavefunction does not change as a result of the measurement in this case.

But, I can derivate that $$ <E> = ∑e_n*c_n $$. It seems to me that $$c_n$$ tells me how much $$u_n$$ "weights"" in the linear combination ( hope the english my is acceptable).

Ps: I used the notation of QuantumQuest because I'm not practical with LaTex

 

[dRic2]

This one in the OP
was added after my post.

I apologize because I have a bad internet connection

 

[PeroK]

But, I can derivate that $$ <E> = ∑e_n*c_n $$. It seems to me that $$c_n$$ tells me how much $$u_n$$ "weights"" in the linear combination ( hope the english my is acceptable).

Ps: I used the notation of QuantumQuest because I'm not practical with LaTex

Yes, but in an energy eigenstate, the wavefunction has $c_n = 0$ for all but the one eigenfunction. E.g. in the $E_0$ state, you have $c_0 = 1$ and $c_n = 0$ otherwise. In other words, this state is 100% weighted towards $E_0$.

 

[jtbell]

So, in other words, a wave-function that describes a definite state of energy can be expressed as a combination of wave-functions that describes other states of energy meaning a state of energy is the result of the probability combination of other states of energy, right?

Actually, no. A system’s energy eigenstates are orthogonal to each other. You cannot represent one of them as a superposition (linear combination) of the others.

It’s exactly analogous to the fact that you cannot express one of the Cartesian unit vectors (e.g. $\hat x$) as a linear combination of the others ($\hat y$ and $\hat z$).

 

[dRic2]

Actually, no. A system’s energy eigenstates are orthogonal to each other. You cannot represent one of them as a superposition (linear combination) of the others.

I forgot that. Does it come from the fact that eigenvalues of Energy are discrete?

So if I change operator, will my question make more sense? For example, taking the position operator I should reach the conclusion that $|c_n|^2 = |ψ(x)|^2$.

 

[PeroK]

I forgot that. Does it come from the fact that eigenvalues of Energy are discrete?

So if I change operator, will my question make more sense? For example, taking the position operator I should reach the conclusion that $|c_n|^2 = |ψ(x)|^2$.

For the position operator, what is $c_n$?

 

[dRic2]

$c_n = ψ(x)$ ?

 

[PeroK]

$c_n = ψ(x)$ ?

That makes no sense. For an operator with a discrete spectrum you have a sequence of eigenfunctions (for that operator). The sequence can be indexed by a whole numbers $n$. Every state is a (finite or infinite) linear combination of these eigenstates.

The position operator has a continuous spectrum, which means two things. First, the eigenstates are not physically realisable. Second, each state is not a sum of these eigenstates but a continuous distribution of them. There are, therefore, no coefficients ($c_n$) in this case.

 

[dRic2]

That makes no sense.

You are right. I'm a bit confused right now. I need to think about it. Tomorrow i'll be back

 

[dRic2]

Ok, I think I figured this out. Let's say I want to know a property (for example the energy) of a particle. I have $<E> = ∫ψHψ dr$ (H is the operator).
Now I need to find $ψ$ so I need to solve the Schrodinger equation. As a solution I get different orthonormal functions $ψ_n$ and I combine them to find $ψ$
like this:
$$ ψ = ∑c_n*ψ_n $$
Every $ψ_n$ represent a specific state of Energy $E_n$ and their linear combination gives me the expected values of E (<E>). When I make a measurement I don't get exactly E, but the Schrodinger equation collapse into one of the $ψ_n$ states thus the energy I measure will be $E_n$ (the eigenvalue of $ψ_n$). This brings me to the conclusion that $c_n$ is linked to the concept of probability. Since the wave function must be normalized I get $∑|c_n|^2 = 1$.

So, I was wrong when I wrote this:

But, I can derivate that

<E>=∑en∗cn<E>=∑en∗cn​

= ∑e_n*c_n

because $$<E> = ∑e_n*|c_n|^2$$

So $|c_n|^2$ tells me the probability to find the value $E_n$ after a measurement.

I have just one question: How can I find $c_n$? Everywhere I find that
$$ c_n = ∫ψ*ψ_n dr $$
but I can't understand one thing: since $ψ = ∑c_n*ψ_n$ how am I supposte to find the value $c_n$? I need $c_n$ to find $ψ$ so how can I use $ψ$ to find $c_n$?

 

[PeroK]

The wavefunction is a vector and $c_n$ are its components - in this case in the basis of energy eigenfunctions.

Just like a vector, if you know the wavefuction you can calculate the components; and, if you know the components, you know the vector.

But, if you don't know either, then you don't know the wavefunction.

 

[dRic2]

So, solving the Schrodinger equation is not enough to find the wave function of a particle ?

 

[PeroK]

So, solving the Schrodinger equation is not enough to find the wave function of a particle ?

The Schrödinger equation tells you how the wavefuction evolves over time. The wavefunction at any given time can be any square integrable function.

There is a unique solution to the Schrödinger equation given an initial wavefunction.