Adding a constant V to the wave equation

In summary, the conversation discusses how to show that the wave function picks up a time-dependent phase factor when a constant is added to the potential energy. It is suggested to use the time-independent Hamiltonian and eigenfunctions to show that the new eigenvalues are equal to the old eigenvalues plus the added constant. This leads to the conclusion that the new wave function is equal to the old wave function multiplied by the phase factor. The question of how this affects the expectation values of physical quantities is raised. Another approach is suggested involving the assumption of a time-independent Hamiltonian, which leads to a simpler solution.
  • #1
cyberdeathreaper
46
0
The question says to show that the wave function picks up a time-dependent phase factor,
[tex] e^\left(-i V_0 t / \hbar \right) [/tex],
when you add a constant [itex] V_0 [/itex] to the potential energy. And then it asks what effect does this have on the expecation value of a dynamical variable?

Assuming I only have been given the Schrodinger eqn thus far, and there has not been any discussion thus far about solutions for the equation, where do I start? I was thinking about replacing the V in the equation with [itex] V + V_0 [/itex], but I'm not sure how that would lead to getting the phase factor they talk about.

[tex]
i \hbar \left( \frac {\partial \psi}_{\partial t} \right) = - \left( \frac {\hbar^2}_{2m} \right) \left( \frac {\partial^2 \psi}_{\partial x^2} \right) + V \psi + V_0 \psi
[/tex]

Any ideas?
 
Physics news on Phys.org
  • #2
Just show that if [itex]\psi[/itex] satisfies the original wave equation, [itex]\psi e^{(-i V_0 t/\hbar)}[/itex] solves the modified potential equation.
 
  • #3
Not to break your confidence,but I'm sure this problem has been solved on these forums.But I'm not sure that somone indicated the real elegant solution.Assume a time-independent Hamiltonian (that way,the Dyson series can be restrained to the exponential).

SE:[tex] \frac{d|\psi (t)\rangle}{dt}=\frac{1}{i\hbar} \hat{H}|\psi\rangle [/tex]

had the simple solution (:wink:)

[tex] |\psi (t)\rangle = \hat{U}(t,t_{0})|\psi (t_{0})\rangle [/tex]

,where (here comes the time-independence of the Hamiltonian)

[tex] \hat{U}=\mbox{exp} \left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right) [/tex]

Okay?

Good

Now.Make the addition of a constant to the Hamiltonian.Then the new Hamiltonian is

[tex] \hat{H}+V_{0}\hat{1} [/tex]

It's not difficult to show that the new evolution operator [itex] \hat{\bar{U}} [/itex] is

[tex] \hat{\bar{U}}=\exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\left(\hat{H}+V_{0}\hat{1}\right)\right) [/tex]

which can be written (using the property of the exponential of operators and the fact that the old hamiltonian commutes with the unit operator)

[tex] \hat{\bar{U}}=\exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right)\exp\left(\frac{V_{0}}{i\hbar}\left(t-t_{0}\right) \hat{1}\right) [/tex]

Okay?

Good.

Therefore the new state vector (solution of the new SE) is

[tex] |\bar{\psi} (t)\rangle = \hat{\bar{U}}(t,t_{0})|\psi (t_{0})\rangle
=\left[\exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right)\exp\left(\frac{V_{0}}{i\hbar}\left(t-t_{0}\right) \hat{1}\right)\right]|\psi (t_{0}\rangle [/tex]

Now,use the fact that

[tex] \exp \left(C\hat{1}\right) |\psi (t_{0}\rangle =\exp (C) |\psi (t_{0}\rangle [/tex]

to write

[tex] |\bar{\psi} (t)\rangle =\exp \left(\frac{V_{0}\left(t-t_{0}\right)}{i\hbar}\right) \exp\left(\frac{1}{i\hbar}\left(t-t_{0}\right)\hat{H}\right) |\psi (t_{0}\rangle =\exp \left(\frac{V_{0}\left(t-t_{0}\right)}{i\hbar}\right) |\psi (t)\rangle [/tex]

which proves that the state vector picks up a time-dependent phase factor when a constant is added to the hamiltonian.

Your business is to show what happens in this case with the physical quantities (quantum observables).

Daniel.
 
Last edited:
  • #4
dextercioby:

The Hamiltonian hasn't even been formally introduced yet for us, and neither has the bra-ket notation. This leaves me with just calculus to try to solve this with (any way to translate your approach to this constraint?).

If this has been solved without that notation somewhere else, could someone point me in that direction? The search function seems to glitch out if I try searching for more than 3 terms.

-

StatusX:

To clearify your suggestion, your saying let:

[tex]
\psi_2 = \psi_1 e^{(-i V_0 t/\hbar)}
[/tex]

in the equation

[tex]
i \hbar \left( \frac {\partial \psi_2}_{\partial t} \right) = - \left( \frac {\hbar^2}_{2m} \right) \left( \frac {\partial^2 \psi_2}_{\partial x^2} \right) + V \psi_2 + V_0 \psi_2
[/tex]

and prove that each side of the equation is equivalent, right?
 
Last edited:
  • #5
That equation doesn't make sense as it stands. Make sure to distinguish between the new and the old wave functions. But yes, I think you know what to do. This works if all you have to do is verify this new function solves the equation. To actually derive it, you'll need to do something similar to what dextercioby suggested.
 
  • #6
Made a quick edit to those equations... Does that look correct now?
 
  • #7
https://www.physicsforums.com/showthread.php?t=61864&highlight=constant+Schrodinger

That was the thread that came into my mind,but it didn't have the solution.

SOLUTION #2 (sucks compared to the first :yuck: ).

Assume time independent hamiltonian,then the wave-function

[tex] \psi_{1}(t)=e^{\frac{1}{i\hbar} E_{1} t} \psi_{1} [/tex] (1)

is the solution of the time-dep.SE and [itex] \psi_{1} [/itex] is the solution of the time-indep Hamiltonian's spectral eq. (forget about degeneration,this is bastardizing QM)

[tex] H_{1}\psi_{1}=E_{1}\psi_{1} [/tex] (2)

Now,the new Hamiltonian

[tex] H_{2}=:H_{1}+V_{0} [/tex] (3)

commutes with the old one (u can check) and is time-independent as well.Therefore,it has a commune set of eigenfunctions [itex] \psi_{2} [/itex] with the old one

[tex] \psi_{2}=\psi_{1} [/tex] (4)

What about the spectrum...??Well,the new eigenvalues [itex] E_{2} [/itex] will be

[tex] H_{2}\psi_{2}=E_{2}\psi_{2}\Rightarrow (H_{1}+V_{0})\psi_{2}=E_{2}\psi_{2} [/tex] (5)

Now use (4) and (2) to write

[tex] (E_{1}+V_{0})\psi_{1}=E_{2}\psi_{1} [/tex] (6)

Therefore

[tex] E_{2}=E_{1}+V_{0} [/tex] (7)

The solution to the new time-dependent SE will be (cf.(1))

[tex] \psi_{2}(t)=e^{\frac{1}{i\hbar} E_{2}t} \psi_{2} [/tex] (8)

which can be written,according to (7) and (1)

[tex] \psi_{2}(t)=e^{\frac{V_{0}t}{i\hbar}} \psi_{1} [/tex] (9)

Q.e.d.

:yuck: I'd vote for the first solution with my eyes closed... :wink:


Daniel.
 
Last edited:
  • #8
StatusX:

I tried your approach, but I couldn't get the sides to become equivalent. I assume I have my equation mis-labeled in some manner.

Moreover, it is my assumption that the book (the same one mentioned in the other post linked by dextercioby) is looking for a derivation. Is there any other approaches? If not, at this point, I would suffice with a clearified way of proving that is a solution.

-

dextercioby:

Okay, I'm going to try to recap your approach without the use of the Hamiltonian - please correct or clearify what I get wrong:

Assume that

[tex] \psi_{1}(t)=e^{\frac{1}{i\hbar} E_{1} t} \psi_{1} [/tex] (1)

is a solution to the wave function. Moreover, assume

[tex] \psi_{1} [/tex]
and
[tex] \psi_{2} [/tex]

are both solutions to the Schrodinger eqn. Moreover, assume that the potential energy for those solutions are

[tex] E_{1} = V [/tex]
and
[tex] E_{2} = V + V_{0} [/tex]

respectively. Therefore,

[tex] E_{2} = E_{1} + V_{0} [/tex]

The solution for the 2nd wave function is therefore:

[tex] \psi_{2}(t)=e^{\frac{1}{i\hbar} E_{2}t} \psi_{2} [/tex]

Which is simplified to simply:

[tex] \psi_{2}(t)=e^{\frac{V_{0}t}{i\hbar}} \psi_{1} [/tex]

-----

My only question for that approach then is how can I assume (1) if that has not been suggested thus far in the text? I mean, the time-independent schrodinger eqn hasn't even been introduced, and yet it is assumed that this problem is answerable with the material presented thus far (which is simply the schrodinger eqn, how to normalize it, the expectation values of x and p, and the rate of change of the expecation values over time)?
 
  • #9
If u don't use that essential commutation relation between the 2 hamiltonians,u can't make the connection between the 2 wavefunctions...That's because the 2 spectral equations are different,because the operators are different.

U assumed that [itex]\psi_{1}[/itex] and [itex] \psi_{2} [/itex] are eigenfunction of the same operator,which is incorrect.

I find the two solutions provided enough to get full mark.Especially the first,which is elegant.

Daniel.
 
  • #10
You don't know that the new spectral equation has the same eigenstates [itex] \psi_{n} [/itex] as the the old one,unless proving those 2 hamiltonians commute...Which is basically my second solution.

Daniel.

EDIT:Status X,why did u delete your post??
 
  • #11
I think this is the best you can do given you haven't gotten into operators or commutators or anything. I assume you have covered separation of variables and the time-independent schrodinger equation:

[tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x) \psi_n(x) = E_n \psi_n(x) [/tex]*

Now, assume every possible solution can be expressed as a linear combination of these solutions, with the appropriate time factor tagged on:

[tex] \Psi_1(x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-i E_n t/\hbar} [/tex]

Now look at the new equation:

[tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +(V(x)+V_0) \psi_n(x) = (E_n)' \psi_n(x) [/tex]

which can be rewritten as:

[tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x)\psi_n(x) = ((E_n)' -V_0)\psi_n(x)[/tex]*

Now, since the nth eigenstate for the first potential had an eigenvalue of En, the En' here will be related to the old eigenvalues by:

[tex](E_n)' = E_n +V_0[/tex]

So:

[tex] \Psi_2(x,t) = \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-i((E_n)') t/\hbar} = \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-i(E_n+V_0) t/\hbar} =e^{(-i V_0 t/\hbar)} \sum_{n=1}^{\infty} c_n \psi_n(x) e^{-iE_n t/\hbar}[/tex]

[tex] \Psi_2(x,t) = e^{(-i V_0 t/\hbar)} \Psi_1(x,t) [/tex]

EDIT: Sorry, I realized I made a mistake and I wasn't sure how long it would take to fix it. It should be right now. As for your comment, the operator on the left sides of the starred equations above are identical, so they have identical eignfunctions and eigenvalues.
 
Last edited:
  • #12
StatusX:

Unfortunately, both separation of variables and the time-independent Schrodinger equation have both not been formally covered...

Sheesh; I mean, if there are all these methods of solving this problem using those concepts, then why was the question asked before they are even presented? Is there any way to solve it without them, or is this author making seriously skewed assumptions about what can be answered with the presented concepts thus far? Am I just misreading this or something? Here is the word-for-word question, just like in the link:

"Suppose you add a constant [itex] V_0 [/itex] to the potential energy (by "constant" I mean independent of x as well as t). In classical mechanics this doesn't change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor: [itex] exp(-i V_{0} t / \hbar ) [/itex]. What effect does this have on the expectation value of a dynamical variable?"

dextercioby:

I appreciate your efforts - it's clear that by Ch 2, I could use the approach presented by your second solution. Until then, I'm not sure what this author is thinking.

--------

In general, if there is no way to solve this without using some of the techniques you two have presented, I'll just go ahead and use one of them, explaining that there was no other way to approach the problem.
 
  • #13
If you don't know any ways to solve the equation, I can't think of a way to derive the result. You can check that it works, though, by plugging the new solution into the new equation. Just be careful to label the new and old wavefunctions, and use the fact that the left and right sides of the first equation are equal when you plug in the old solution.
 
  • #14
Nope,Status X.They're different operators acting on different (a priori) wavefunctions...I can say that your third equation is incorrect,as it is not justified...


Daniel.
 
  • #15
cyberdeathreaper said:
StatusX:

Unfortunately, both separation of variables and the time-independent Schrodinger equation have both not been formally covered...

Sheesh; I mean, if there are all these methods of solving this problem using those concepts, then why was the question asked before they are even presented? Is there any way to solve it without them, or is this author making seriously skewed assumptions about what can be answered with the presented concepts thus far? Am I just misreading this or something? Here is the word-for-word question, just like in the link:

"Suppose you add a constant [itex] V_0 [/itex] to the potential energy (by "constant" I mean independent of x as well as t). In classical mechanics this doesn't change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor: [itex] exp(-i V_{0} t / \hbar ) [/itex]. What effect does this have on the expectation value of a dynamical variable?"

dextercioby:

I appreciate your efforts - it's clear that by Ch 2, I could use the approach presented by your second solution. Until then, I'm not sure what this author is thinking.

--------

In general, if there is no way to solve this without using some of the techniques you two have presented, I'll just go ahead and use one of them, explaining that there was no other way to approach the problem.


I have the book and looked through it.Ha hasn't covered the material to account for a proof.One bad thing for the book... :yuck:

Daniel.
 
  • #16
Alright, thanks for your help. I may try mimicking the separation of variable technique in the next chapter to reach a point where I can use this approach anyway. If not, I'll just plug in like you suggested.

well, after spending w-a-y too much time on this problem, I'm off to bed... thanks to both of you for all your help
 
  • #17
dextercioby said:
I have the book and looked through it.Ha hasn't covered the material to account for a proof.One bad thing for the book... :yuck:

Daniel.

Yeah, really. This book is terrible about correlating the material with the questions. Even Prob 1.3 took me forever (my professor, a M.A. in theoretical physics couldn't even initially figure it out). Hopefully it gets better as the chapters progress, but so far I'm not digging the approach. I have had other overviews before that were better organized than this.
 
  • #18
What can i say,"good luck" with & post all the problems u encounter.I'm sure someone around here would know a bit about QM & would be willing to help.


Daniel.
 
  • #19
dextercioby said:
Nope,Status X.They're different operators acting on different (a priori) wavefunctions...I can say that your third equation is incorrect,as it is not justified...

My third equation (by which I assume you mean the third line of latex) is just the time independent schroedinger's equation for the new potential. The En' are just the energies for this new hamiltonian, and are not initially connected to any of the previous equations. Moving V0 to the other side, the left side equation obviously has the same eigenfunctions and eigenvalues as the left side of the first equation, since they are one in the same. I really don't see any problems here.
 
  • #20
Nope,that equation should HAVE OTHER WAVEFUNCTIONS,becasue the operator is different...A priori.Then,u can equate those sets of eigenfunctions,by proving the 2 hamiltnians commute,ergo form a csco...

Daniel.
 
  • #21
An operator and the same operator plus a constant trivially commute, and this is probably implictly used when I bring the constant over to the other side. But in any case, I'm not working with operators here, I'm working with the schroedinger equation. For this eigenvalue equation:

[tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x) \psi_n=\lambda_n \psi_n(x)[/tex]

There will be a certain set of eigenfunctions [itex]\psi_n[/itex] and eigenvalues [itex]\lambda_n[/itex]. Then for the two wave equations:

[tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x) \psi_n(x)=E_n \psi_n(x)[/tex]

[tex]\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \psi_n(x) +V(x) \psi_n(x)=((E_n)'-V_0)\psi_n(x)[/tex]

We can equate the constants on the left hand side with the [itex]\lambda_n[/itex] and the eigenfunctions from above solve both. The En and En' are determined by these eigenvalues.
 
Last edited:
  • #22
Voilà.So you make use of the commutation of the 2 op-s...You have to.There's no other way.

Daniel.
 

1. What is the purpose of adding a constant V to the wave equation?

The constant V is added to the wave equation to account for the presence of a medium through which the wave is traveling. This medium can affect the speed and direction of the wave, and the constant V helps to incorporate those effects into the equation.

2. How does the addition of a constant V affect the shape of a wave?

Depending on the value of the constant V, it can affect the amplitude, wavelength, and frequency of a wave. A larger constant V may result in a larger amplitude and shorter wavelength, while a smaller constant V may result in a smaller amplitude and longer wavelength.

3. Is the addition of a constant V necessary for all types of waves?

No, the addition of a constant V is not necessary for all types of waves. It is most commonly used for mechanical waves that travel through a medium, such as sound waves and water waves. Electromagnetic waves, on the other hand, do not require the use of a constant V in their wave equations.

4. Can the constant V change over time?

Yes, the constant V can change over time if the medium through which the wave is traveling changes. For example, if the temperature or pressure of the medium changes, the constant V may also change, resulting in a different wave behavior.

5. How does adding a constant V affect the speed of a wave?

The constant V directly affects the speed of a wave. In fact, the speed of a wave can be calculated by dividing the wavelength by the constant V. This means that a larger constant V will result in a slower wave speed, while a smaller constant V will result in a faster wave speed.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
778
  • Introductory Physics Homework Help
Replies
25
Views
461
  • Introductory Physics Homework Help
Replies
6
Views
733
  • Introductory Physics Homework Help
Replies
3
Views
805
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
768
  • Introductory Physics Homework Help
Replies
11
Views
1K
Replies
7
Views
558
  • Introductory Physics Homework Help
Replies
4
Views
902
  • Introductory Physics Homework Help
Replies
6
Views
1K
Back
Top