Question:
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Given ϕ(x,t)=∑∞n=1anψn(x)exp(−iEnt/ℏ), what must be true?
Solution:
This is the Parseval's theorem. Since ϕ(x,t)∗ϕ(x,t) represents a probability, we have ∫−∞∞ϕ(x,t)∗ϕ(x,t)dx=1. Substitute in the definition, we have:
1=∞∫−∞ϕ(x,t)∗ϕ(x,t)dx=∞∫−∞(∞∑n=1a∗nψn(x)exp(iEnt/ℏ))(∞∑n=1anψn(x)exp(−iEnt/ℏ))dx=∞∫−∞∞∑p=1∞∑q=1a∗paqexp(iEpt/ℏ)exp(−iEqt/ℏ)ψp(x)ψq(x)dx=∞∑p=1∞∑q=1a∗paqexp(iEpt/ℏ)exp(−iEqt/ℏ)∞∫−∞ψp(x)ψq(x)dx=∞∑p=1∞∑q=1a∗paqexp(iEpt/ℏ)exp(−iEqt/ℏ)δpq=∞∑p=1|ap|2exp(iEpt/ℏ)exp(−iEpt/ℏ)=∞∑p=1|ap|2
Therefore the answer is ∑∞p=1|ap|2=1.
... snip ...
Given ϕ(x,t)=∑∞n=1anψn(x)exp(−iEnt/ℏ), what must be true?
Solution:
This is the Parseval's theorem. Since ϕ(x,t)∗ϕ(x,t) represents a probability, we have ∫−∞∞ϕ(x,t)∗ϕ(x,t)dx=1. Substitute in the definition, we have:
1=∞∫−∞ϕ(x,t)∗ϕ(x,t)dx=∞∫−∞(∞∑n=1a∗nψn(x)exp(iEnt/ℏ))(∞∑n=1anψn(x)exp(−iEnt/ℏ))dx=∞∫−∞∞∑p=1∞∑q=1a∗paqexp(iEpt/ℏ)exp(−iEqt/ℏ)ψp(x)ψq(x)dx=∞∑p=1∞∑q=1a∗paqexp(iEpt/ℏ)exp(−iEqt/ℏ)∞∫−∞ψp(x)ψq(x)dx=∞∑p=1∞∑q=1a∗paqexp(iEpt/ℏ)exp(−iEqt/ℏ)δpq=∞∑p=1|ap|2exp(iEpt/ℏ)exp(−iEpt/ℏ)=∞∑p=1|ap|2
Therefore the answer is ∑∞p=1|ap|2=1.
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