# One dimensional Schrödinger equation equation with initial condition, finding the probability of the particle's future position

A particle of mass $m$ moves freely in the interval $[0,a]$ on the $x$ axis. Initially the wave function is: $$f(x)=\frac{1}{\sqrt{3}}\operatorname{sin}\Big( \frac{\pi x}{a} \Big)\Big[1+2\operatorname{cos}\Big( \frac{\pi x}{a} \Big) \Big]$$

I let the normalised wave function in one dimension be:

$$\Phi_n(x,t)=\sqrt{\frac{2}{a}}\operatorname{sin}\Big( \frac{n\pi x}{a} \Big)\operatorname{exp} \Big( \frac{-in^2\pi^2 h t}{2ma^2} \Big)$$ Where $h$ is $2\pi$ Planck's constant. So here I have said that as $$f(x)=\frac{1}{\sqrt{2}}[\Phi_1(x,0)+\Phi_2(x,0)]$$ That for later $t$ the wave function of the particle is given by: $$\Phi(x,t)=\frac{1}{\sqrt{2}}[\Phi_1(x,t)+\Phi_2(x,t)]$$ Could anyone help me go about normalizing this? (or showing it is already normalized, however I am guessing it is not). And help me find the probability that the particle is in the interval $[0,a/2]$.

A problem I am encountering is:

$$|\Phi|^2=\frac{1}{2}|\Phi_1+\Phi_2|^2=\frac{1}{2}(\Phi_1+\Phi_2)\overline{(\Phi_1+\Phi_2)}=$$ $$\Rightarrow |\Phi|^2=\frac{1}{2}(|\Phi_1|^2+|\Phi_2|^2+[\overline{(\Phi_1)}{(\Phi_1)}+\overline{(\Phi_1)}{(\Phi_1)}])$$ $$\Rightarrow \int_0^a|\Phi|^2=\frac{1}{2}\int_0^a|\Phi_1|^2+|\Phi_2|^2\operatorname{d}x+ \frac{1}{2}\int_0^a\overline{(\Phi_1)}{(\Phi_2)}+\overline{(\Phi_2)}{(\Phi_1)}$$ $$\Rightarrow \int_0^a|\Phi|^2=1+ \frac{1}{2}\int_0^a\overline{(\Phi_1)}{(\Phi_2)}+\overline{(\Phi_2)}{(\Phi_1)}$$

From here I can't see how to proceed to normalize the wave funtion.

• Is this a homework or self-study question, by any chance? It kind of sounds like one (not that there's anything wrong with that, just that it should have the homework tag if that's appropriate). Jan 29, 2012 at 22:24

The integrals you have to compute are not that difficult. First of all, the spatial parts of your wavefunctions are real (this is wise and always possible in one dimension), and therefore complex conjugates just do not matter. So, the remaining integral you have on the rhs is (up to some expression dependent on t (S(t)) which we will leave for now):

$$I(t) = S(t) \int_0^a\Phi_n(x)\Phi_m(x) dx = S(t) \frac{2}{a} \int_0^a \operatorname{sin}\Big( \frac{n\pi x}{a} \Big) \operatorname{sin}\Big( \frac{m\pi x}{a} \Big)dx$$

Now, you have to make a substitution $t = \frac{\pi x}{a}$ to get

$$I(t) = S(t) \frac{2}{\pi} \int_0^\pi \operatorname{sin} ( n x ) \operatorname{sin} ( m x )dx$$

First of all, if either n or m are 0 then I(t) = 0 and your initial state is normalized.

Now you integrate by parts (or by mathematica if you can't by parts) to obtain

$$I(t) = S(t) \frac{2}{\pi} [ -(\frac{1}{m} \sin{n x} \cos{m x}) |_0^\pi + \frac{n}{m} \int_0^\pi \cos{n x} \cos({m x}) dx]$$

The first term is 0 (cause sin is 0 in all multiplicities of $\pi$). Now we can proceed with integration by parts of the second term.

$$I(t) = S(t) \frac{2}{\pi} ( \frac{n}{m} \int_0^\pi \cos{n x} \cos({m x}) dx) = S(t) \frac{2}{\pi} [ (\frac{n}{m^2} \cos{n x} \sin{m x}) |_0^\pi$$ $$+ \frac{n^2}{m^2} \int_0^\pi \sin{n x} \sin({m x}) dx] = \frac{n^2}{m^2} I(t)$$

So, there are two possibilities - either n=m or I(t)=0 (n=-m is excluded as negative labels are not in the basis because sin(-nx) = -sin(nx) - linear dependency). What you get is ORTHOGONALITY RELATION - if eigenvectors have different labels (e.g. 1 and 2 as in your case) they are orthogonal in $L^2([0,a])$ which happens to be your Hilbert space. There are theorems which state that this is always true for Hamiltonian eigenstates - corresponding theory of differental operators is known as Strum-Liouville theory (for finite dimensional Hilbert spaces this is a trivial property of self-adjoined operators).

Now, to the second part of your question. The modulus squared of wavefunction is by definition probability density of finding a particle in an interval $dx$

So, the probability of finding particle in [0,a/2] is simply

$$\int_0^{a/2} |f(x,t)|^2 dx = \frac{1}{2}\int_0^{a/2} |\Phi_1(x)|^2 + |\Phi_2(x)|^2 dx +$$

$$\frac{1}{2}\int_0^{a/2} \Phi_1(x) \Phi_2(x) \operatorname{exp} \Big( (-1^2+2^2)\frac{-i\pi^2 h t}{2ma^2} \Big)+ \Phi_2(x) \Phi_1(x) \operatorname{exp} \Big( (-2^2+1^2) \frac{-i\pi^2 h t}{2ma^2} \Big)$$ $$= \frac{1}{2}\int_0^{a/2} |\Phi_1(x)|^2 + |\Phi_2(x)|^2 dx + \cos{\frac{3\pi^2 h t}{2ma}} \int_0^{a/2} \Phi_1(x) \Phi_2(x) dx$$

First integral is simply 1 because we take half of the area of $sin(x)^2$ and $sin(2x)$, and add it. Second integral can be easily computed using double 'by parts' formula from above and replacing $\pi$ with $\pi/2$ in integral limits. One gets $\frac{4}{3 \pi}$ Therefore finally

$$\int_0^{a/2} |f(x,t)|^2 dx = \frac{1}{2}+\frac{4}{3 \pi} \cos{\frac{3\pi^2 h t}{2ma}}$$

As $\frac{4}{3 \pi} < \frac{1}{2}$ the result makes sense. As the body is in mixture of two states, the probability is no longer constant in time.

• Sorry it took me so long to get back to you. This is extremely useful, however I am only worried as to why we can ignore the complex conjugates, as surely they still have an imaginary part? Jan 31, 2012 at 12:39
• Complex conjugates do not matter for real numbers and all that is complex in your problem is time dependence. Therefore, i have collected all of it in S(t). In the second part of answer i have used them to change signs in $(-1^2+2^2)$ and $(-2^2+1^2)$ Feb 1, 2012 at 12:01

@Nick's answer is correct, but redundant.

A state needs not be normalized to begin with, nor be represented by the eigenstates in order to conserve its norm.

Time evolution is a unitary transformation (indeed, because the Hamiltonian is Hermitian). Therefore the norm of a state, any state, is conserved with time. This is explained in detail here.

If your $\Phi_n(x,t)$ are alreay the energy eigenstates, then they are orthogonal and therefore, if "let the normalised wave function in one dimension be" means you have already normalized it, then $f(x)$ at $t=0$ is normalized.

For later times you have $$\Phi_n(x,t)=e^{-iE_nt}\Phi_n(x,0)=\exp({-iHt})\Phi_n(x,0)=U(t)\Phi_n(x,0)$$ and so $$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_2(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_2(x,0))=U(t)f(x).$$ Since $U(t)$ is unitary, the state will stay normalized.

You have not given us the Hamiltonian to see if these $\Phi_n$ are really the eigenstates, but the point is that if they are, then the states are orthogonal because the Hamiltonian is hermitean and consequently the mixing terms are zero.

The integration (for normalization, checking orthogonality as well as finding the particle in a finite interval) is a purely mathematical problem.

edit: Okay, here you go

http://img15.imageshack.us/img15/6774/bild3nl.png

This expression is the only $x$-dependent part and it is zero for all integers $n\ne m$.

You can also plug "Integrate[Sin[\pi x/a]Sin[2\pi x/a],{x,0,a}]" into Wolfram Alpha.

• Sorry it took me so long to get back to you. This is helpful thanks, I got there in the end! Jan 31, 2012 at 12:39