Find Error in Calculating Energy Eigenstates of Particle in Box

[dimensionless]

I have a particle of mass m in a box of length L. The energy eigenstates of this particle have wave functions
$$\phi_{n}(x)=\sqrt{2/L}sin(n \pi x/L)$$
and energies
$$E_n = n^{2}\pi^{2}\hbar^{2}/2mL^{2}$$
where n=1, 2, 3,... At time t=0, the particle is in a state described as follows.
$$\Psi(t=0)=\frac{1}{\sqrt{14}}[\phi_1 + 2\phi_2 + 3\phi_3]$$
To find the energy for state $$\Psi$$ I did the following.
$$\sum_{1, 2, 3} E_n = (1^2 + 2^2 +3^2) \frac{\pi^2\hbar^2}{2mL^2}=14\frac{\pi^2\hbar^2}{2mL^2}= 14E_1$$
where $$E_1=\frac{\pi^2\hbar^2}{2mL^2}$$
I have made a mistake somewhere because the actual answer is $$9 E_1$$. Does anyone know where my error is?

 

[marlon]

Well just apply the Hamiltonian onto the state psi. this state is the sum of three phi-terms. So each phi-state yields :
$$H \phi_1 = E_1 \phi_1$$
$$2H \phi_2 = 2E_2 \phi_2$$
$$3H \phi_3 = 3E_3 \phi_3$$

Just add up everything and you get :

$$H \Psi =\frac {1}{\sqrt14} (E_1 \phi_1 + 2E_2 \phi_2 + 3E_3 \phi_3)$$

The clue is to write down each energy term as a function of $$E_1$$. You have a formula given to do that. Keep in mind that the coefficients of the psi-wave-function denote the possible energy values for the system



Good Luck

marlon

 

[marlon]

Ps, as an additional question : can you give me the probability that the psi-system has energy value 9E_1 ?

regards
marlon