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My doubt was that if a linear combination of separable solutions is also a solution of the Schrodinger's equation, but the linear combination doesn't necessarily satisfy the time independent part, it implies that the linear combination might not be an energy eigen function. This means that it will not have a definite value of energy, even though the wavefunctions out of which it is made of do have a definite value of energy.

Moreover, since the linear combination is also a separable solution of the Schrodinger's equation, it should have had a definite value of energy (as per Introduction to Quantum Mechanics by Griffiths). Where am I going wrong?

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    $\begingroup$ A superposition of solutions with different eigenvalues ( = different values of the energy) does not have a well-defined energy. One can talk about the energy of the state in the sense of the expectation value of the Energy, will give a weighted average of the contributions of each component of the superposition. But the state is not an eigenstate of the Hamiltonian and does not a single energy value. $\endgroup$
    – user245141
    Oct 27, 2019 at 8:47
  • $\begingroup$ @yu-v Please post answers as answers, not comments. $\endgroup$ Oct 27, 2019 at 9:23

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My doubt was that if a linear combination of separable solutions is also a solution of the Schrodinger's equation, but the linear combination doesn't necessarily satisfy the time independent part, it implies that the linear combination might not be an energy eigen function. This means that it will not have a definite value of energy, even though the wavefunctions out of which it is made of do have a definite value of energy.

Yes.

Moreover, since the linear combination is also a separable solution of the Schrodinger's equation

No, it isn't. The fact that the Schrödinger equation is linear means that any linear combination of solutions will also be a solution. However, not all solutions are separable, and the linearity does not extend to separability - a linear combination of separable solutions need not be (and, in general, it won't be) separable.

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  • $\begingroup$ Oh right, I understood where I was going wrong. Thanks! $\endgroup$
    – Janak Ruia
    Oct 29, 2019 at 5:36

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