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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 $$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_1(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_1(x,0))=U(t)f(x).$$ 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's zerofor 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.

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.

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 $$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_1(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_1(x,0))=U(t)f(x).$$ 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 ![enter image description here][1] [1]: http://img17.imageshack.us/img17/5292/bild1ow.png This expression is the only $x$-dependent part and it's zerofor all integers $n$.

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 $$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_1(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_1(x,0))=U(t)f(x).$$ 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's zerofor 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.

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 $$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_1(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_1(x,0))=U(t)f(x).$$ 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.

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 $$\Phi(x,t)=c_1\Phi_1(x,t)+c_2\Phi_1(x,t)=U(t)(c_1\Phi_1(x,0)+c_2\Phi_1(x,0))=U(t)f(x).$$ 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 ![enter image description here][1] [1]: http://img17.imageshack.us/img17/5292/bild1ow.png This expression is the only $x$-dependent part and it's zerofor all integers $n$.