Why Must the State of a Quantum SHM Particle Be of This Form?

[MichalXC]

Homework Statement

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

\psi = \frac{1}{\sqrt2} \psi_0 + \frac{e^{i \phi}}{\sqrt2} \psi_1

where \psi_ and \psi_1 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 f_n is \mid c_n \mid ^2.

The Attempt at a Solution

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

\psi = c_1 \psi_0 + c_2 \psi_1

where

\mid c_1 \mid ^2 + \mid c_2 \mid ^2 =1 \rightarrow \mid c_n \mid ^2 = \frac{1}{2}

I have no idea where the factor of e^{i \phi} comes from in the final answer.

 

[ideasrule]

In the equation, c2=<br /> \frac{e^{i \phi}}{\sqrt2}<br />. What's the square of the absolute value of c2?

 

[MichalXC]

In the equation, c2=<br /> \frac{e^{i \phi}}{\sqrt2}<br />. What's the square of the absolute value of c2?

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

But we are supposed to derive the equation.

 

[fzero]

\mid c_1 \mid ^2 + \mid c_2 \mid ^2 =1 \rightarrow \mid c_n \mid ^2 = \frac{1}{2}

It doesn't follow from the normalization that

\mid c_n \mid ^2 = \frac{1}{2},

only that

|c_2|^2 = 1-|c_1|^2.

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

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

\alpha \psi~~\text{and} ~~\psi

are equivalent. This let's us reduce the number of phases to one.

 

[MichalXC]

Right.

The sum of the \mid c_n \mid ^2s equals 1 by normalization:

\mid c_1 \mid ^2 + \mid c_2 \mid ^2 =1

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

\mid c_1 \mid ^2 = \frac{1}{2}

and

\mid c_2 \mid ^2 = \frac{1}{2}

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 \psi finds the state \psi_1" is:

\psi = \frac{1}{\sqrt2} \psi_1 + \frac{1}{\sqrt2} \psi_2

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

 

[fzero]

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 \psi finds the state \psi_1" is:

\psi = \frac{1}{\sqrt2} \psi_1 + \frac{1}{\sqrt2} \psi_2

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

c_i = \frac{1}{\sqrt{2}} e^{i\theta_i},

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

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 |\psi\rangle) as

\langle A \rangle_\psi = \frac{ \langle \psi | A | \psi \rangle}{ \langle \psi | \psi \rangle }.

When |\psi\rangle 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 \alpha |\psi\rangle

\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 .

Since this formula is true for all operators A, we find that any observable measured in the state \alpha |\psi\rangle is the same as the measurement in the state |\psi\rangle. 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.

 

[MichalXC]

Nice! That all makes sense. Thank you.