Quantum Simple Harmonic Oscillator

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Homework Statement



We know that a particle in SHM is in a state such that measurements of the energy will yield either [itex]E_0[/itex] or [itex]E_1[/itex] (and nothing else), each with equal probability. Show that the state must be of the form

[tex] \psi = \frac{1}{\sqrt2} \psi_0 + \frac{e^{i \phi}}{\sqrt2} \psi_1 [/tex]

where [itex] \psi_ [/itex] and [itex] \psi_1 [/itex] are the ground and first excited state, respectively.

Homework Equations



For a Hamiltonian with discrete energy spectrum, the probability of measuring the particular eigenvalue associated with the orthonormalized eigenfunction [itex]f_n[/itex] is [itex] \mid c_n \mid ^2 [/itex].

The Attempt at a Solution



Since we are just as likely to measure [itex]E_0[/itex] as we are to measure [itex]E_1[/itex], we know that the wave function must look like

[tex] \psi = c_1 \psi_0 + c_2 \psi_1 [/tex]

where

[tex] \mid c_1 \mid ^2 + \mid c_2 \mid ^2 =1 \rightarrow \mid c_n \mid ^2 = \frac{1}{2} [/tex]

I have no idea where the factor of [itex]e^{i \phi}[/itex] comes from in the final answer.
 
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Answers and Replies

  • #2
ideasrule
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In the equation, c2=[tex]
\frac{e^{i \phi}}{\sqrt2}
[/tex]. What's the square of the absolute value of c2?
 
  • #3
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In the equation, c2=[tex]
\frac{e^{i \phi}}{\sqrt2}
[/tex]. What's the square of the absolute value of c2?

[tex] \left| \frac{e^{i \phi}}{\sqrt{2}} \right| ^2 = \frac{e^{-i \phi}}{\sqrt{2}} \frac{e^{i \phi}}{\sqrt{2}} = \frac{1}{2} [/tex]

But we are supposed to derive the equation.
 
  • #4
fzero
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[tex] \mid c_1 \mid ^2 + \mid c_2 \mid ^2 =1 \rightarrow \mid c_n \mid ^2 = \frac{1}{2} [/tex]

It doesn't follow from the normalization that

[tex]\mid c_n \mid ^2 = \frac{1}{2},[/tex]

only that

[tex] |c_2|^2 = 1-|c_1|^2.[/tex]

Instead you should be able to conclude that [tex]\mid c_n \mid ^2 =1/2[/tex] from the requirement that [tex]\psi_1[/tex] and [tex]\psi_2[/tex] apply from equal probabilities. Can you write down an expression for the probability that a measurement on [tex]\psi[/tex] finds the state [tex]\psi_1[/tex]?

As for the single phase, you need to use the fact that rescaling a wavefunction represents the same physical state. Namely that

[tex]\alpha \psi~~\text{and} ~~\psi [/tex]

are equivalent. This lets us reduce the number of phases to one.
 
  • #5
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Right.

The sum of the [itex]\mid c_n \mid ^2[/itex]s equals 1 by normalization:

[tex]\mid c_1 \mid ^2 + \mid c_2 \mid ^2 =1[/tex]

Since [itex]\mid c_n \mid ^2[/itex] represents the probability that a measurement of energy would yield the corresponding eigenvalue, we know that (since [itex]E_0[/itex] and [itex]E_1[/itex] are equally likely) [itex] \mid c_1 \mid ^2 = \mid c_2 \mid ^2 [/itex]. Or, making a substitution, that

[tex] \mid c_1 \mid ^2 = \frac{1}{2} [/tex]

and

[tex] \mid c_2 \mid ^2 = \frac{1}{2} [/tex]

What does it mean to take the square root of these quantities, since they are potentially complex? Intuitively, I feel like "an expression for the probability that a measurement on [itex]\psi[/itex] finds the state [itex]\psi_1[/itex]" is:

[tex] \psi = \frac{1}{\sqrt2} \psi_1 + \frac{1}{\sqrt2} \psi_2 [/tex]

Could you explain more what you mean when you say "rescaling a wavefunction represents the same physical state"? Thanks!
 
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  • #6
fzero
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What does it mean to take the square root of these quantities, since they are potentially complex? Intuitively, I feel like "an expression for the probability that a measurement on [itex]\psi[/itex] finds the state [itex]\psi_1[/itex]" is:

[tex] \psi = \frac{1}{\sqrt2} \psi_1 + \frac{1}{\sqrt2} \psi_2 [/tex]

That's not the most general form that satisfies the requirements, precisely because the [tex]c_i[/tex] are complex. The point is that specifying [tex]|c_i|=1/\sqrt{2}[/tex] only tells you that

[tex]c_i = \frac{1}{\sqrt{2}} e^{i\theta_i},[/tex]

where [tex]\theta_i[/tex] is an unspecified angle. This is because we can write any complex number as [tex]z=r e^{i\theta}[/tex], where [tex]r[/tex] is real and [tex]\theta[/tex] is an angle, and [tex]|e^{i\theta}|=1[/tex] for any [tex]\theta[/tex].

Could you explain more what you mean when you say "rescaling a wavefunction represents the same physical state"? Thanks!

Yes, the reason is that we really define the expectation value of an observable (in the state [tex]|\psi\rangle[/tex]) as

[tex] \langle A \rangle_\psi = \frac{ \langle \psi | A | \psi \rangle}{ \langle \psi | \psi \rangle }.[/tex]

When [tex]|\psi\rangle[/tex] is normalized, we ignore the denominator because it's just equal to one. But in general we can use the complete formula to show that in the state [tex]\alpha |\psi\rangle[/tex]

[tex] \langle A \rangle_{\alpha \psi} = \frac{ \langle \psi | \alpha^* A \alpha | \psi \rangle}{ \langle \psi | \alpha^* \alpha | \psi \rangle } = \frac{ |\alpha|^2}{|\alpha|^2} \frac{ \langle \psi | A | \psi \rangle}{ \langle \psi | \psi \rangle } = \langle A \rangle_\psi .[/tex]

Since this formula is true for all operators [tex]A[/tex], we find that any observable measured in the state [tex]\alpha |\psi\rangle[/tex] is the same as the measurement in the state [tex] |\psi\rangle[/tex]. Therefore we conclude that we have the freedom to rescale a quantum state by a number without changing the physical state that it corresponds to.

An important thing to note is that this property is precisely what allows us to take an arbitrary state and normalize it without changing the physics.
 
  • #7
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Nice! That all makes sense. Thank you.
 

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