Wave function linear combination

[gfd43tg]

Homework Statement

Homework Equations

The Attempt at a Solution

a) I am not sure exactly what they are looking for, but I said that the operators associate, distribute, and do not necessarily commute.

b) I know by definition ##\langle a' \mid a \rangle = \int a'^{*}a dx = \delta_{a'a}##. This is sort of a weak argument, I am wondering how to give a better answer.

c) I understand the question, but not exactly the form in wish is intended. I put
$$ \mid \psi \rangle = \mid 1 0 0 \frac {1}{2} \rangle + A \mid 2 1 1 - \frac {1}{2} \rangle + A \mid 2 1 -1 - \frac {1}{2} \rangle + \sqrt{7}A \mid 2 1 0 - \frac {1}{2} \rangle $$
Another possible answer I have is
$$ \mid \psi \rangle = R_{10}Y_{0}^{0} \chi_{+} + AR_{21}Y_{1}^{1} \chi_{-} + AR_{21}Y_{1}^{-1} \chi_{-} + \sqrt {7} AR_{21}Y_{1}^{0} \chi_{-}$$

d) I'm not sure how to exploit the orthogonal properties, but I do know ##\int \psi^{*}\ \psi dx = 1##, So I chose one of the components of the wave function with the unknown constant ##A##, and did exactly this.

Choosing ##A \frac {r}{a} e^{-r/2a} Y_{1}^{-1}##,

$$ \frac {3}{8 \pi} \frac {A^2}{a^2} \int_0^{2 \pi} \int_0^{\pi} \int_0^{\infty} r^{3} e^{-r/a} \sin^{2} \theta \hspace {0.02 in} dr d\theta d\phi = 1 $$
$$ \frac {3}{8 \pi} \frac {A^2}{a^2} * 2 \pi * \frac {\pi}{2} \int_0^{\infty} r^{3} e^{-r/a} dr = 1 $$
$$ \frac {3 \pi}{8 a^{2}} A^{2} \frac {3!}{(1/a)^{4}} = 1$$
$$ A = \frac {2}{3a \sqrt{\pi}}$$
Which is not what the answer is supposed to be.

 

[blue_leaf77]

a) This point asks you to mention the properties of operators whose associated quantum numbers is 'good', and second a property that allows us to write a unique eigenstate defined by a collective numbers composed of each of the quantum numbers from the operators A, B, C, ...

b) I know by definition ##\langle a' \mid a \rangle = \int a'^{*}a dx = \delta_{a'a}##.

b) But there is a reason why that identity can be true. For a simpler case with only one quantum number, use the fact that ##\langle a'|A| a \rangle##, where the ##|a \rangle##'s are eigenstates of operator A, can be calculated in two ways.

c) Remember that the modulus square of a certain expansion coefficient in the wavefunction represented in some basis is interpreted as a probability to find that wavefunction to be in that basis state. How will you use this information to calculate A?

d) A is supposed to be already computed in c). For this point, simply use the orthonormality of the eigenstates of H atom and the interpretation of the expansion coefficient as I have mentioned in c).

 

[gfd43tg]

Okay, does it have something to do with the things like ##L^{2} \mid l m \rangle = l(l+1) \hbar^{2} \mid l m \rangle##?

 

[blue_leaf77]

Yes, ##l## is one of good quantum numbers for H atom. Do you know why is ##L^2## called 'good'?

 

[gfd43tg]

I am not certain, but I can give two speculations. One is that it commutes, i.e. ##[L^{2}, L_{i}] = 0## for ##i = x,y,z##, the other that its quantum number is good, ergo its a good operator

 

[blue_leaf77]

What about googling or reading in the book where the problem is? I'm almost sure, that problem is not the first place where the term 'good quantum number' appears. Just for hint, this is related with time evolution operator.

I am not certain, but I can give two speculations. One is that it commutes, i.e. ##[L^{2}, L_{i}] = 0## for ##i = x,y,z##, the other that its quantum number is good, ergo its a good operator

Using commutativity property as well as what I hint you above is the right direction.

 

[gfd43tg]

Does it mean that they have simultaneous eigenfunction, such as ##L^{2}## and ##L_{z}##?

 

[blue_leaf77]

they

Which do you refer by 'they'?
Allright to get you further in this problem, I'll let you know the mathematical property which an operator A whose quantum number is said to begood must satisfy, that is [H,A] = 0. But the problem asks you to interpret the physical implication of that relation.

 

[gfd43tg]

Okay, so that means that the operator is conservative. But how to I answer the first question more generally? That the eigenvalue of the operator is discrete and depends on its good quantum number?

 

[blue_leaf77]

Yes, it must be conserved in time. If it's not, it will be inconvenient for us since if we allow our system to evolve in time, we must always compute the new quantum number for that particular time as our initial quantum number has changed. That's the reason why it's always desireable to seek for the good quantum numbers for a system before being able to represent any eigenstate as a unique collective indices of those quantum numbers.

 

[gfd43tg]

Okay, so then for a) you would say that the operators are conservative in time, and that the eigenvalues of the operator are constants which have discrete values?

 

[blue_leaf77]

that the eigenvalues of the operator are constants which have discrete values?

This one is not really necessary as the problem only asks about the properties of the operators. More importantly, the property that makes those operators have a common set of eigenstates.

 

[gfd43tg]

What property makes them have a common set of eigenstates?

 

[blue_leaf77]

I thought you have known the answer judging from what you said in comment #7.

 

[gfd43tg]

I mean, for an arbitrary operator ##A##, then ##A^{2}## and ##A_{z}## have simultaneous eigenfunctions? I mean, I was using orbital angular momentum as a concrete example. Who is to say that there is a z-component of some arbitrary operator?

 

[blue_leaf77]

Who is to say that there is a z-component of some arbitrary operator?

No, such analogy is unnecessary, you can only define spatial components of an operator if it is a vector operator that satisfies its own requirement.

For H atom the operators H, L², L_z and S_z share a common set of eigenstate commonly written as ##|nlm_lm_s\rangle##. What property do all those 4 operators have in relation with certain mathematical relation?

 

[gfd43tg]

I am not sure what you are trying to get me to say, is it that they have simultaneous eigenstates??

 

[gfd43tg]

for d)

I know just to take the absolute of the square of each coefficient. so the terms are 1/2 for the first, 1/18 for the second and third, and 7/18 for the last term. But what are the energies?

 

[blue_leaf77]

I am not sure what you are trying to get me to say, is it that they have simultaneous eigenstates??

Imagine you have arbitrary state ##|\psi \rangle##, now you apply this operator: ##AB-BA## on to our arbitrary state where A (with quantum number ##a##) and B (with quantum number ##b##) are known to have a common set of eigenstates, which we write as ##|a,b \rangle##. Hence we have relations like ##A|a,b\rangle = a|a,b\rangle## and ##B|a,b\rangle = b|a,b\rangle## . Since ##|a,b \rangle##'s are eigenstates, they must be complete and therefore we can expand ##|\psi \rangle## into the bases ##|a,b \rangle## for all ##a,b##. My question is, given those forementioned conditions, what is the result of ##(AB-BA)|\psi \rangle##?

But what are the energies?

On which quantum number does the energies of hydrogen atom depend?

 

[gfd43tg]

The result is zero. The energies depend on ##n##. I mean, half of them are at ##n = 1## and half at ##n = 2##, but I don't know the numeric value of the energy

EDIT: so then for ##n =1##, it's -13.6 eV (50%), and for ##n=2##, its -3.4 eV for the other 50%??

 

[gfd43tg]

for b)

$$\langle a' | A | a \rangle = a_{1} \langle a' | a \rangle$$
$$\langle a' | A | a \rangle = \langle Aa' | a \rangle = a_{2}\langle a' | a \rangle$$
Subtracting,
$$(a_{1} - a_{2}) \langle a' | a \rangle = 0 $$
$$ \langle a' | a \rangle = 0 $$
But how to show that it is equal to the kronecker delta?

 

[blue_leaf77]

From your calculation, what can you conclude about two observables having the same eigenstates? What if it is generalized to more than two observables?

for b)
$$(a_{1} - a_{2}) \langle a' | a \rangle = 0 $$
$$ \langle a' | a \rangle = 0 $$
But how to show that it is equal to the kronecker delta?

Between those two steps, you have made too quick of an implication. You haven't considered the case when ##a_1 = a_2##, from then you can probably see that you can relate things with kronecker delta assuming ##|a \rangle ## is already normalized.

 

[gfd43tg]

Which calculation are you talking about?

And for b), If I say ##a_{1} = a_{2}##, then ##\langle a' | a \rangle## is nonzero, and we can assume it to be normalized, thus it is equal to one, therefore ##\langle a' | a \rangle = \delta_{a'a}##?

Also, where did I go wrong for (c) as far as my normalization constant A not being correct?

 

[blue_leaf77]

Which calculation are you talking about?

Part a), have you figured out what property the operators A, B, C, ... need to satisfiy in order for them to have common eigenstates?

And for b), If I say ##a_{1} = a_{2}##, then ##\langle a' | a \rangle## is nonzero, and we can assume it to be normalized, thus it is equal to one, therefore ##\langle a' | a \rangle = \delta_{a'a}##?

Yes.

 

[gfd43tg]

Is it that the operators A,B,C must commute with each other?

 

[blue_leaf77]

Yes, the operators A, B, C, and so on need to form the so-called maximal set of commuting observables, so that they can share the same set of eigenstates.