$$\psi(x) = Ae^{ikx} \ + \ Be^{-ikx}$$

The above is not an eigenfunction and to obtain its momentum my professor applied the momentum operator separately to the first and second parts. The final result was that he obtained two different values of momentum, $$\hbar k \text{ and } -\hbar k\text{.}$$

My question: is this method correct? If yes, how to explain the separation of values?

  • 2
    $\begingroup$ What's to explain? Like you said, this is not an eigenfunction for momentum, hence it corresponds to multiple values of momentum. Likewise, it is not an eigenfunction for coordinate, hence it corresponds to multiple values of $x$. $\endgroup$ – Ivan Neretin Sep 6 '17 at 10:08
  • $\begingroup$ But then it will be like applying momentum operator to 2 different wave functions. $\endgroup$ – Anjan Sep 6 '17 at 10:15
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    $\begingroup$ Yes it will; so what? Any wave function which is not an eigenfunction can be thought of as a mixture of different eigenfunctions. $\endgroup$ – Ivan Neretin Sep 6 '17 at 10:20
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    $\begingroup$ If you don't want to separate the two states, then what is the single momentum of this function? And if you can't answer this question, then I think it highlights the need to understand why and why the linear combination of eigenstates should be treated separately. $\endgroup$ – Zhe Sep 6 '17 at 11:57

The idea of superposition of eigenstates is quantum 101, so you should probably go and read a proper book on QM. I personally recommend Griffiths.

In quantum mechanics, observables such as $p$ are represented by Hermitian operators such as $\hat{p} = -\mathrm{i}\hbar(\mathrm d/\mathrm dx)$.

The momentum of a particle is not well-defined unless it is in an eigenstate of the momentum operator $\hat{p}$. If it is, then a measurement of the momentum will always return the corresponding eigenvalue. For example, if the wavefunction of the particle is simply $\exp(ikx)$ , then a measurement of the momentum will always return the value $\hbar k$, and the particle is said to exist in a state of well-defined momentum.

Now, if the wavefunction of the particle is $A\exp(ikx) + B\exp(-ikx)$, then a measurement of the momentum will sometimes yield $\hbar k$ and sometimes yield $-\hbar k$. So, the actual momentum of the particle is not well-defined, as it could be either of the two values (before you observe it).

The expectation value of the momentum, though, is well-defined. It is simply

$$\langle p \rangle = P_+(\hbar k) + P_-(-\hbar k)$$

where $P_+$ and $P_-$ are the probabilities of measuring the momentum to be $\hbar k$ and $-\hbar k$ respectively. Assuming that your wavefunction is normalised, i.e. $|A|^2 + |B|^2 = 1$, basic quantum mechanics tells us

$$P_+ = |A|^2;\quad P_- = |B|^2$$

and so the expectation value in this case is found to be

$$\langle p \rangle = \hbar k (|A|^2 - |B|^2)$$

However, that simply means that if you prepare many particles in the state above, measure their momenta, and average their momenta, the average will tend toward this value. Each individual measurement is guaranteed to give either the value $\hbar k$ or $-\hbar k$.

And yes, each individual particle does not have one single well-defined momentum. That's quantum mechanics for you.

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