# 1D Infinite Square Well: Box Suddenly Increases in Size. How treat this?

I am currently working my way through John S. Townsend book "A Fundamental Approach to Modern Physics" (ISBN: 978-1-891389-62-7). Exercise 3.12 (p.111) is about the 1D infinite square well. The box has the potential barriers at $x=0$ and $x=L$. $$V\left(x\right) = \begin{cases} \infty & x < 0 \\ 0 & 0 \leq x \leq L \\ \infty & x > L \end{cases}$$

The text states the following:

A particle of mass $m$ is in the lowest energy (ground) state of the infinite potential energy well. At time $t=0$ the wall located at $x=L$ is suddenly pulled back to a position at $x=2L$. This change occurs so rapidly that instantaneously the wave function does not change. ($a$) Calculate the probability that a measurement of the energy will yield the ground-state energy of the new well.

The answer to this question involves re-solving the T.I.S.E., applying new boundary conditions, and re-normalizing the wave function. Then we project the new wave-function onto the old one, and calculate the overlap of the two wave-functions (initial and final wave-function).

I cannot find any discussion about this particular problem that would make me able to reason my way to calculate an overlap between the initial and final wave-function. No way. Unless I have a background in linear algebra, and would be able to foresee the need to project the one onto the other, and sort of find my way logically to the answer, I see no way any student without any experience with quantum physics would be able to solve this on their own. I had to use Google until I found the solution to a similar problem.

Or, maybe QM just does not come naturally to me. A lot of problem-solving is required in these types of courses to for us to build up some "intuition" about what to expect as results, what to do, etc. I still do not fully understand the logic behind the solution to the problem.

Phenomenologically, there is a particle inside the box. Then, suddenly, the box widens to twice its size. However, the wave-function does not change. Even though the particle has a bigger box in which to move around, the initial wave-function does not take this extra room into account, and would equal zero when $x>L$. Hence, I would not expect for the particle to venture out to the new region much, if we used only the initial wave-function. Or, is the problem that the particle will indeed venture out there (since $V(x)=0$, and it is physically allowed to), but we have not considered this in our normalizing of the function? Hence, it makes no sense to calculate any property of the particle with the "initial" wave-function, since this is simply the incorrect wave-function for the new well?

New ideas and thoughts

Okay, so I have to evaluate the integral

$$c_1 = \int\limits_{- \infty}^{\infty} \psi_1 (x) \Psi (x) dx$$

where $\psi_1$ is the theoretical wave-function for a particle in the entire box, that is from $0$ to $2L$, and $\Psi$ is the wave-function for the actual particle, that is from $0$ to $L$.

Now, I can split this integral into two parts:

$$c_1 = \int\limits_{- \infty}^{\infty} \psi_1 (x) \Psi (x) dx = \int\limits_{0}^{L} \psi_1 (x) \Psi (x) dx + \int\limits_{L}^{2L} \psi_1 (x) \Psi (x) dx$$

We see that since the actual particle's wave-function is not defined when $x>L$, the second term will be zero - that is, the wave-functino is normalized for $0<x<L$, and the boundary conditions ensures that $\Psi \rightarrow 0$ when $x=0$ and $x=L$. Thus, the the integral over is reduced to

\begin{align*} c_1 &= \int\limits_{0}^{L} \psi_1 (x) \Psi (x) dx \\ &= \int\limits_{0}^{L} \sqrt{\frac{1}{L}} \sin \left( \frac{\pi x}{2L} \right) \sqrt{\frac{2}{L}} \sin \left( \frac{\pi x}{L} \right) dx \\ &= \frac{\sqrt{2}}{L} \int\limits_{0}^{L} \sin \left( \frac{\pi x}{2L} \right) \sin \left( \frac{\pi x}{L} \right) dx \\ &= \frac{4 \sqrt{2}}{3 \pi} \\ \Rightarrow c_1^2 &= \frac{32}{9 \pi ^2} \\ c_1^2 & \approx 0.36 \end{align*}

I think maybe my difficulty with "visualzing" the problem is that I do not fully understand the expression for $c_n^2$, and how this gives the probability for measuring the energy level $n$.

I know this seems like I am rambling, and maybe I am, but I hope you understand my confusion. I appreciate any help!

• I might be able to help you out here, but I don't understand your phrase 'So, the new "available" space for the particle is not "taken care of" in the normalizing of the particle's wave-function.' Can you restate that some other way? By the way, are you working through this book on your own, or as part of a course? – garyp May 18 '14 at 13:35
• It is all part of an introductory course to quantum physics. I am a master student in inorganic chemical analysis, and wanted more theory. I will restate the phrase; I hope it became clearer. – Yoda May 18 '14 at 13:46
• I updated my question with some new thoughts. – Yoda May 18 '14 at 15:04
• What are the constraints of the wave function? That the ends of the wave at the boundary have zero amplitude. If you double the length of the wave, you see twice the wavelength (with everything else being equal). In essence, you would have twice the number of eigenfunctions when the distance is doubled. – LDC3 May 18 '14 at 16:14
• @LCD I think you need to rephrase what you are trying to say, because as it is written it is not correct. Both before and after the number of eigenstates is infinite. Also, I think your comment does not address the OP. – garyp May 18 '14 at 22:05

I would forget about the movement of the wall. The potential is the infinite square well of width $2L$ (potential is $\infty$ aside from the region $0 < x < 2L$, where it is $0$), and the wavefunction is $$\Psi\left(x,t\right) = \sum_{n=1}^\infty c_n \psi_n\left(x\right) \exp\left(-\frac{iE_n t}{\hbar}\right),$$ where $\psi_n\left(x\right) = \sqrt{1/L} \sin\left(n \pi x / 2L\right)$ is the $n$-th stationary state, and $E_n = n^2 \pi^2 \hbar^2 / \left(8 m L^2\right)$ is its energy.

To determine the coefficients $c_n$, we multiply $\Psi\left(x,0\right)$ by $\psi_m^*\left(x\right)$, integrate, and use the orthonormality of the stationary states: $$\int_0^{2L} dx \ \psi^*_m\left(x\right) \Psi\left(x,0\right) = \sum_{n=1}^\infty c_n \int_0^{2L} dx \ \psi^*_m\left(x\right) \psi_n\left(x\right) = \sum_{n=1}^\infty c_n\delta_{mn} = c_m$$ If that last step is confusing, remember that it is equivalent to using the orthonormality of Cartesian unit vectors $\hat{\bf e}_i \bullet \hat{\bf e}_j = \delta_{ij}$ to determine the components $V_i$ of a 3 dimensional vector $\bf V$ by taking the dot product with $\hat{\bf e}_i$: ${\bf V} \bullet \hat{\bf e}_i = \left(\sum_j V_j \hat{\bf e}_j\right) \bullet \hat{\bf e}_i = \sum_j V_j \delta_{ji} = V_i$.

Anyway, now we use the fact that the initial wavefunction $\Psi\left(x,0\right)$ is the ground state of the infinite square well of width $L$, which is $\sqrt{2/L} \sin\left(\pi x/L\right)$ for $0 < x < L$ and $0$ otherwise. So (note the change in integration range), $$c_n = \frac{\sqrt{2}}{L} \int_0^{L} dx \ \sin\left(\frac{n \pi x}{ 2L}\right) \sin\left(\frac{\pi x}{L}\right) = \frac{\sqrt{2}}{\pi} \int_0^{\pi} du \ \sin\left(\frac{n u}{2}\right) \sin\left(u\right),$$ or $$c_1 = \frac{\sqrt{2}}{\pi} \int_0^{\pi} du \ \sin\left(\frac{u}{2}\right) \sin\left(u\right),$$ as you have.

• Passing question: does that mean that the particle is in a superposition of the different states of the new well? When the particle realises that the well has been expanded, its wavefunction will inevitably adjust to fit the new well. Assuming the new wavefunction is the ground state of the new well, where did the extra energy go? – user120404 Jan 6 '18 at 19:14
• @user120404 Yes, the state is a superposition of the stationary states of the well (width $2L$), with coefficients $c_n$ determined from the initial wave function equal to the ground state of a well of width $L$. The way I've chosen to approach this problem, the well has width $2L$ at $t=0$, so there is no expansion of the well. Finally, I'm not sure what you mean when you say "assuming the new wavefunction is the ground state of the new well." Do you mean "assuming a measurement of the energy yields the ground-state energy of the well of width $2L$?" – Eric Angle Jan 8 '18 at 16:03
• Yes, assuming that a measurement collapses the state in the new well to be in the ground state, where does the energy difference between the ground state in the old well and the ground state in the new well go? – user120404 Jan 9 '18 at 16:57
• @user120404 to the measurement apparatus. – Ruslan Jan 10 '18 at 14:26
• @user120404 Keep in mind that while $\Psi\left(x,0\right)$ would be an energy eigenstate of a well of width $L$, it is not an energy eigenstate of the actual well of width $2L$. Upon measuring the energy, the probability of observing $E_n = n^2 \pi^2 \hbar^2 / \left(8 m L^2\right)$ is $\left|c_n\right|^2$. However, note that the expectation value of the energy $\left<E\right> = \sum_{n=1}^{\infty} \left|c_n\right|^2 E_n = \pi^2 \hbar^2 / \left(2mL^2\right)$, which is the ground state energy of a well of width $L$. That sum takes a bit of work, but isn't too hard to work out. – Eric Angle Feb 8 '18 at 13:42

...it makes no sense to calculate any property of the particle with the "initial" wave-function, since this is simply the incorrect wave-function for the new well?

The wavefunction can't be "incorrect for the well". Your wavefunction is just an initial condition for time-dependent Schrödinger equation. Here's how it would evolve if you solve the time-dependent equation (I ignore normalization here):

We see that since the actual particle's wave-function is not defined when $x>L$...

The wavefunction is defined for all $x\in\mathbb R$. It's just zero outside of $(0,L)$ because the potential is infinite there.

the second term will be zero

This remains true, however, because of what I said above.

the wave-functino is normalized for $0<x<L$

Actually, again, the wavefunction is normalized period. It's defined for the whole real line, and zero outside the well, so when you normalized using the integral over the well, it's the same as if you integrated over $\mathbb R$. If it were not this way, your "normalization for some domain" would not make any sense, i.e. it'd not be a normalization at all.

Your further calculation looks OK to me.

I think maybe my difficulty with "visualzing" the problem is that I do not fully understand the expression for $c^2_n$, and how this gives the probability for measuring the energy level $n$.

That this gives the probability for measuring energy level $n$ is known as Born rule. You find projection of your actual wave function onto the eigenstate of energy, namely on state $n$. By Born rule, its magnitude squared is the probability of measuring the system to appear in that eigenstate.

The fact that you used only the unchanged original wavefunction for calculations, even though it almost immediately drastically changes as time goes on, is that despite its change in shape, its coefficients $c_n$ actually only change their phase like $$c_n\propto\exp\left(-\frac i\hbar E_nt\right),$$

but remain the same in magnitude — because the potential is independent of time. Thus you can measure the energy after some time passes, and will still get the same result.

It seems like the thing you are missing is a very basic ingredient of quantum mechanics -- the Born rule. When we make a measurement of some observable $A$, we will find that $A$ takes values in the eigenvalues $a_i$ of the corresponding operator $A$. The Born rule tells us that, for a system initially in the state $\left|\psi\right>$, the probability of finding the system in the eigenstate $\left|a_i\right>$ corresponding to eigenvalue $a_i$ is $\left|\left<a_i|\psi\right>\right|^2$. In standard mathematics notation, this is $\left|(a_i,\psi)\right|^2$, the inner product between the the two states. In the position basis, the states are wavefunctions, and we can write their inner product explicitly as $\int\,dx\, \phi_i(x)\psi(x)$.

The probability that a measurement made on a particle will yield the ground state energy is proportional to the overlap between the initial wave function and the ground state eigenfunction. So you just need to calculate the overlap integral between the initial state and the final one. Whether you integrate over the entire expanded well or just its original unexpanded extent will make no difference to the result, because the original wave function is zero everywhere in the expanded region.