QM: Arbitrary operators and their eigenstates

[Niles]

Homework Statement

Hi all.

When riding home today from school on my bike, I was thinking about some QM. The general solution to the Schrödinger equation for t=0 is given by:

$$\Psi(x,0)=\sum_n c_n\psi_n(x),$$

where $\psi_n(x)$ are the eigenfunctions of the Hamiltonian. We know that the probability of $\Psi(x)$ collapsing to one of the eigenstates is given by the absolute square in front of that particular eigenstate.

Now my question comes: Let's say we have an arbitrary operator Q: $Q\psi_n = q_n\psi_n$, where q_n are the eigenvalues. Now what do I do if I want to find the probability of $\Psi(x)$ collapsing to one of the eigenstates of the operator Q upon measurement of the observable Q?

My own attempt is that we write $\Psi(x)$ as a new linear combination of the eigenstates of Q. But to me this seems very difficult.

I hope you can shed some light on this. Thanks in advance.


Niles.

 

[meopemuk]

My own attempt is that we write $\Psi(x)$ as a new linear combination of the eigenstates of Q.

That's exactly what you should do.

 

[Niles]

Great, that's really good.

I have a final question. Let's say we wish to find the expectation value of the operator Q, when our particle is in the state $\Psi$. The expectation value of Q is given by:

$$\left\langle Q \right\rangle = \left\langle \Psi \right|Q\left| \Psi \right\rangle.$$

But does this imply that $\Psi$ must be given in the same basis as the operator Q?

Thanks in advance.

 

[nrqed]

Great, that's really good.

I have a final question. Let's say we wish to find the expectation value of the operator Q, when our particle is in the state $\Psi$. The expectation value of Q is given by:

$$\left\langle Q \right\rangle = \left\langle \Psi \right|Q\left| \Psi \right\rangle.$$

But does this imply that $\Psi$ must be given in the same basis as the operator Q?

Thanks in advance.

This calculation can be done in any basis one desires. The result is independent of the basis used. If the basis is not made of the eigenstates of Q then the representation of Q will not be diagonal, but that does not matter for the final result.

 

[Niles]

This calculation can be done in any basis one desires.

But the only demand is that both $\Psi$ and Q are given in the same basis, right?

 

[nrqed]

But the only demand is that both $\Psi$ and Q are given in the same basis, right?

Yes, to carry out the explicit calculation, Q and Psi must be written using the same basis. That's right.

 

[Niles]

Thanks for all your help, both of you.

 

[Niles]

I have one final question, which is related to what I wrote in my first post:

The general solution to the Schrödinger equation for t=0 is given by:

$$\Psi(x,0)=\sum_n c_n\psi_n(x),$$

where $\psi_n(x)$ are the eigenfunctions of the Hamiltonian.

Why is it that we choose to write it as a linear combination of the stationary states of the Hamiltonian, and not some other operator? Of course, as we talked about in this thread, we can express it as a linear combination of eigenfunctions of an arbitrary observable operator, but why have we chosen the Hamiltonian as the "default" operator?

Thanks in advance.

 

[meopemuk]

I have one final question, which is related to what I wrote in my first post:



Why is it that we choose to write it as a linear combination of the stationary states of the Hamiltonian, and not some other operator? Of course, as we talked about in this thread, we can express it as a linear combination of eigenfunctions of an arbitrary observable operator, but why have we chosen the Hamiltonian as the "default" operator?

Thanks in advance.

The choice of basis is a matter of convenience. For many problems (in atomic or nuclear physics) the use of Hamiltonian eigenfunctions makes calculations simpler than in other bases.

 

[Niles]

The choice of basis is a matter of convenience. For many problems (in atomic or nuclear physics) the use of Hamiltonian eigenfunctions makes calculations simpler than in other bases.

Ahh, I see.

In my book, the expectation value of momentum is defined as:

$$\left\langle p \right\rangle = \int {\psi ^* (x)\left( {\frac{\hbar }{i}\frac{d}{{dx}}} \right)} \,\psi (x)\,dx,$$

where * denotes the complex conjugate. I have written the wavefunction as a function of x to emphasize that it is the wavefunction expressed in the x-basis (i.e. the basis of the position operator).

This must imply (according to our above discussion) that the momentum-operator in the x-basis is given as:

$$\hat p = {\frac{\hbar }{i}\frac{d}{{dx}}}.$$

Questions:

1) Am I correct that it really is the momentum operator given in the x-basis?

2) If yes, then how does one find this result?

I hope I am not annoying you with my questions. Thanks in advance.


Niles.

 

[nrqed]

I have one final question, which is related to what I wrote in my first post:



Why is it that we choose to write it as a linear combination of the stationary states of the Hamiltonian, and not some other operator? Of course, as we talked about in this thread, we can express it as a linear combination of eigenfunctions of an arbitrary observable operator, but why have we chosen the Hamiltonian as the "default" operator?

Thanks in advance.

As meopemuk said, one can use any basis in principle. However, there is something special about using the eigenstates of the Hamiltonian. The hamiltonian is the generator of time translation. What this means is that it is the eigenstates of the Hamiltonian that have simpel time evolutions. They simply evolve as $\psi_n(x) exp(-iE_n t / \hbar)$. Eigenstates of other operators do not have simple time evolutions (unless they commute with the Hamiltonian and hence share common eigenstates). So if we want to know how a wavefunction evolves with time (let's say we are given it at t =0 and want to know what it wil be at a later time), we need to expand it over the eigenstates of the Hamiltonian in order to be able to write down the time dependence.

This is what makes the eigenstates of the Hamiltonian "special" .

 

[nrqed]

1) Yes, absolutely!

2) In Quantum Mechanics, momentum is defined as the generator of translations. An infinitesimal translation shifts the system by a small amount. So we describe the action of an infinitesimal translation operator as : $$\hat T(dx) \psi (x) = \psi (x+dx)$$

Now the generator of any operator $$\hat T$$ can be written as

$$\hat T(dx) = 1-\i \hat K dx$$

Again by Taylor expansion to the first order,
$$\psi(x+dx) = \psi(x) + dx \partial_x \psi(x)$$

So you can see that K is just $$\frac{1 }{i} \frac{d}{{dx}}$$ . The $$\hbar$$ is multiplied to the generator of translator to get the dimensions right for the units we have chosen to work with.

I fixed the TeX

 

[xboy]

Thanks a lot, nrged [:

 

[xboy]

I should add that in the second equation, $$\hat K$$ is the generator.

 

[Niles]

Great, thanks for taking the time to answer all my question to all three of you. It is very nice to get answers and derivations to these questions. I really appreciate it.