Quantum Mechanics Question

In summary, the conversation discusses two questions regarding quantum mechanics, specifically the relationship between an arbitrary function and its position and momenta, as well as the change of expectation values in a time dependent state. The first question involves deriving general functions using commutation rules, while the second question delves into the evolution of a general state with a time-independent Hamiltonian. The conversation also reveals the speaker's lack of understanding in quantum mechanics and their struggle to grasp the concepts being discussed.
  • #1
guruoleg
15
0
Hey all-
I am new to quantum mechanics so these questions will be elementary and don't be afraid to go over simple concepts. There are two questions on the homework that I barely even know how to approach (I don't know how to type p sub k properly so that is how I will be writing it):

1. Let f(p,x) be an integer function of operators p sub k,x sub k.
Derive general functions (df/dx sub k) = -[f,p sub k] and (df/dp sub k) = [f,x sub k]
With abbreviation [f,g] = i/h bar(fg - gf) from the commutation rules for p sub k and x sub k.
----------------------------------------------------------------------------------
Firstly, I thought that [A,B] = i h bar (A,B) Poisson Bracket where (A,B) Poisson Bracket has partial derivatives and a summation. Are the two representations the same?
Anyway, I absolutely cannot see how to begin proving a relationship between a completely arbitrary function f and its position and momenta. Since I cannot take the partial derivative of an arbitrary function I tried to work backwards and compute [f, p] but by the definition given above (fp - pf) = 0. In fact, it seems like fp-pf will be zero for any values of p and f. However, p and f are not variables, right? It is difficult to see but should they be operators expressed in the bra-ket notation?
Anyway, I can prove the above relationships using [x,B] = i h bar dB/dp but that seems inadequate because that formula is essentially the same as the problem itself and the book does not really derive it.

------------------------------------------------------------------------------------

2. Let <A> be the expectation value of an operator not explicitly dependent on time in a time dependent state Psi. How does <A> change with time? What follows for <x sub k> and <p sub k>?
-------------------------------------------------------------------------------
At first glance it seems that if an operator does not depend on time neither will its expectation value nor will x and p. The only relevant equation I know is <A>=<Psi|A|Psi>. This may just be a ridiculously easy question and what I said before is the answer so please tell me if it's more involved than that.

Thank you for your help and attention.
 
Physics news on Phys.org
  • #2
guruoleg said:
Hey all-
I am new to quantum mechanics so these questions will be elementary and don't be afraid to go over simple concepts. There are two questions on the homework that I barely even know how to approach (I don't know how to type p sub k properly so that is how I will be writing it):

1. Let f(p,x) be an integer function of operators p sub k,x sub k.
Derive general functions (df/dx sub k) = -[f,p sub k] and (df/dp sub k) = [f,x sub k]
With abbreviation [f,g] = i/h bar(fg - gf) from the commutation rules for p sub k and x sub k.
----------------------------------------------------------------------------------
Firstly, I thought that [A,B] = i h bar (A,B) Poisson Bracket where (A,B) Poisson Bracket has partial derivatives and a summation. Are the two representations the same?
Anyway, I absolutely cannot see how to begin proving a relationship between a completely arbitrary function f and its position and momenta. Since I cannot take the partial derivative of an arbitrary function I tried to work backwards and compute [f, p] but by the definition given above (fp - pf) = 0. In fact, it seems like fp-pf will be zero for any values of p and f. However, p and f are not variables, right?
Yes, they are not scalars. Do you know how to write the p operator in the x-basis? [Hint: you have probably seen this in the time independent Schrodinger equation.]

2. Let <A> be the expectation value of an operator not explicitly dependent on time in a time dependent state Psi. How does <A> change with time? What follows for <x sub k> and <p sub k>?
-------------------------------------------------------------------------------
At first glance it seems that if an operator does not depend on time neither will its expectation value nor will x and p. The only relevant equation I know is <A>=<Psi|A|Psi>. This may just be a ridiculously easy question and what I said before is the answer so please tell me if it's more involved than that.

Thank you for your help and attention.
Yes, it is more involved. Notice the part I bolded. [itex]| \psi (t) \rangle \neq | \psi (0) \rangle [/itex]

How does a general state (of a system, with a time-independent Hamitonian H) evolve in time?
 
  • #3
Gokul43201 said:
Yes, they are not scalars. Do you know how to write the p operator in the x-basis? [Hint: you have probably seen this in the time independent Schrodinger equation.]

Yes! p = -i h bar d/dq. That's awesome! So, for the first one, -[f,p] = (-1/i h bar) (f p - p f) = (-1/i h bar)(f -i h bar d/dx - (-i h bar df/dx)) = f d/dx + df/dx! The second part of that is the answer, but what is f d/dx and how does it go away? Should I perhaps make all this stuff act on x? If so, should the x be taken out later (I've only seen this done once)?

Gokul43201 said:
Yes, it is more involved. Notice the part I bolded. [itex]| \psi (t) \rangle \neq | \psi (0) \rangle [/itex]

How does a general state (of a system, with a time-independent Hamitonian H) evolve in time?

Good grief... I have no idea - this is my first time seeing quantum mechanics mathematically and the class is quantum optics so the prof. assumes I know all this (and there is no textbook, just his notes). Come to think of it, I completely understand next to none of this so let's see if I know anything at all; A is an operator. <A> = <Psi|A|Psi> means the expectation value of A on Psi: so if A were to operate on state Psi, <A> would be the probabilistic distribution of the result. What is a state Psi anyway? Is it a snapshot of a system? So it contains all information about a system at a certain time?
I cannot think of a time dependent (t.d.) operator but an example of a time independent (t.i.d.) one might be Q^ - the translation operator (because it has no t in it?). So does the question ask what will happen when t.i.d. operator acts upon t.d. system? If so I have no idea.

To re-iterate, I haven't the foggiest how the general state evolves in time.

Also, I would appreciate a reference to an introductory QM text that uses bra-ket notation. I'm not sure if the Dirac notation is only useful for QM with systems with infinite degrees of freedom but that was the only book I could find that uses that notation.

Thank you.
 
  • #4
guruoleg said:
Yes! p = -i h bar d/dq. That's awesome! So, for the first one, -[f,p] = (-1/i h bar) (f p - p f) = (-1/i h bar)(f -i h bar d/dx - (-i h bar df/dx)) = f d/dx + df/dx! The second part of that is the answer, but what is f d/dx and how does it go away? Should I perhaps make all this stuff act on x? If so, should the x be taken out later (I've only seen this done once)?
Start with the commutator and operate it on some arbitrary position-space wavefunction, [itex] \psi (x) [/itex]:

[tex][f,p] \psi = f \left( -i\hbar \frac{\partial \psi}{\partial x} \right) - (-i \hbar) \frac {\partial}{\partial x} (f \psi) [/tex]

You know how to find the derivative of a product. Use it to expand the second term on the RHS and see what you are left with.

Similarly, try the other commutator.
 
  • #5
Brilliant... that's very helpful... thank you very much.

Now for the other question; I think the state of a system evolves with
[tex]i\hbar \frac{\partial }{\partial t} \right)|\psi(t)> = H|\psi(t)>[/tex]
That LaTeX code stuff is helpful, so I'm learning it as I go along... Anyway, I still don't know what is being asked. The original question was "Let <A> be the expectation value of an operator not explicitly dependent on time in a time dependent state Psi." What is the thing that is "in a time dependent state Psi?" The time independent operator? The expectation value of an operator is in a state? What does that mean?

I am not pretending; I have never seen any of this before. Should I perhaps somehow use the above time evolution equation with the expectation value equation [tex]<A>=<\psi|A|\psi>[/tex]? Maybe I should then take the derivative of something to find the time variance.

Perhaps I should solve the above differential equation to get [tex]\psi=e^{H/i\hbar}[/tex]. This doesn't appear helpful and I'm not sure it even makes sense for an operator (H) to be used that way in a solution.

Last guess... and this one may actually be partially coherent: From the very first equation we know that [tex]\frac{\partial }{\partial t} \right)|\psi(t)> = \frac{H}{i\hbar}|\psi(t)>[/tex]. This is the time derivative of [tex]\psi[/tex] and may be what we want in the first place. So is that the answer? Except we want to know how the expectation change with time so... ummm... [tex]<\frac{\partial A(t)}{\partial t} \right)> = <\psi|\frac{H}{i\hbar}\psi(t)|\psi>[/tex]?

That looks totally made up and I have no idea how to bring x and p into this. Please help soon...
 
  • #6
I think this might have to do with perturbation theory. This homework is due in a few hours but even of you don't get around to answering by then I'd still like to know. Thanks.
 
  • #7
Check out wikipedia at http://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics [Broken]). Under Time-dependent perturbation theory there is some helpful looking stuff. I think I might know how to put A into all this but I totally don't understand the step that get you to d cn/d t. It says something about a partial differential equation but I don't know how to get there and I don't see where the V(t) comes from in the first place.
 
Last edited by a moderator:
  • #8
no, perturbation theory is when your system is perturbed..

Now you have a state that evolves with time, You should instead find info about "Quantum dynamics". Sakurai "Modern Quantum mechanics" has a good chapter about it.

In shrödinger picture you have time indepentend operators, and time depending states (ket's). And to evolve a state in time, you apply the time evolution operator to it.

And you could have an operator like this: [tex] e^{-iHt/\hbar} [/tex] Which is the time-evol-operator. The exponential of an operator is defined as the power series expansion of e^x; i.e you have 1 + x + X^2/2 + .. etc; but with your operator instead of x here.
 
  • #9
guruoleg said:
Brilliant... that's very helpful... thank you very much.

Now for the other question; I think the state of a system evolves with
[tex]i\hbar \frac{\partial }{\partial t} \right)|\psi(t)> = H|\psi(t)>[/tex]
That LaTeX code stuff is helpful, so I'm learning it as I go along... Anyway, I still don't know what is being asked. The original question was "Let <A> be the expectation value of an operator not explicitly dependent on time in a time dependent state Psi." What is the thing that is "in a time dependent state Psi?" The time independent operator? The expectation value of an operator is in a state? What does that mean?

I am not pretending; I have never seen any of this before. Should I perhaps somehow use the above time evolution equation with the expectation value equation [tex]<A>=<\psi|A|\psi>[/tex]? Maybe I should then take the derivative of something to find the time variance.

Perhaps I should solve the above differential equation to get [tex]\psi=e^{H/i\hbar}[/tex]. This doesn't appear helpful and I'm not sure it even makes sense for an operator (H) to be used that way in a solution.

Last guess... and this one may actually be partially coherent: From the very first equation we know that [tex]\frac{\partial }{\partial t} \right)|\psi(t)> = \frac{H}{i\hbar}|\psi(t)>[/tex]. This is the time derivative of [tex]\psi[/tex] and may be what we want in the first place. So is that the answer? Except we want to know how the expectation change with time so... ummm... [tex]<\frac{\partial A(t)}{\partial t} \right)> = <\psi|\frac{H}{i\hbar}\psi(t)|\psi>[/tex]?

That looks totally made up and I have no idea how to bring x and p into this. Please help soon...


never mind my other post, given the way the question is phrased, I thik what they want i sthe following. ( and you were on the right track).

write [tex] \partial_t <A>A = (\partial_t <\Psi|) A |\Psi> + <\Psi| A ~(\partial_t |\Psi>) [/tex]

Then use Schrodinger equation for the bra and the ket. You will end up with the expectation value of a simple expression containing A and H (if you use the notation of commutators)
 

1. What is quantum mechanics?

Quantum mechanics is a branch of physics that studies the behavior of particles on a very small scale, such as atoms and subatomic particles. It describes the fundamental principles that govern the behavior of these particles and their interactions with each other and with energy.

2. How does quantum mechanics differ from classical mechanics?

Classical mechanics describes the behavior of larger objects that we can observe and measure, while quantum mechanics deals with the behavior of particles that are too small to be directly observed. In classical mechanics, objects have definite properties like position and velocity, but in quantum mechanics, particles are described by wave functions that can only predict the probability of finding a particle in a certain state.

3. What are the key principles of quantum mechanics?

The key principles of quantum mechanics include superposition, which states that particles can exist in multiple states simultaneously; wave-particle duality, which describes the dual nature of particles as both waves and particles; and the uncertainty principle, which states that it is impossible to know both the position and momentum of a particle with absolute precision.

4. What are some real-world applications of quantum mechanics?

Quantum mechanics has numerous applications, including in the development of new technologies such as transistors and lasers. It also plays a crucial role in fields such as chemistry, where it helps explain the behavior of atoms and molecules, and in cryptography, where it is used to create secure communication systems.

5. Is quantum mechanics a complete theory?

Quantum mechanics is currently the most complete theory we have for describing the behavior of particles on a small scale. However, it is still being actively studied and refined, and there are some aspects that are still not fully understood, such as the role of gravity in quantum systems. As our understanding of the universe continues to evolve, it is possible that new theories may emerge that build upon or even replace quantum mechanics.

Similar threads

  • Advanced Physics Homework Help
Replies
10
Views
458
  • Advanced Physics Homework Help
Replies
15
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
732
  • Advanced Physics Homework Help
Replies
3
Views
842
  • Advanced Physics Homework Help
Replies
5
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
766
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
5
Views
2K
  • Advanced Physics Homework Help
Replies
24
Views
678
Back
Top