## 3 Security of the Protocol  
In this section, we prove the unconditional security of our protocol. Since [PERSON] does not send any quantum and classical information to [PERSON] before the unveiling, the protocol is obviously secure against [PERSON]. It is because no matter what strategy [PERSON] takes (including the strategy that [PERSON] prepares any kind of quantum state), any local measurement performed in [PERSON]'s side does not change the reduced density matrix of the qubits in [PERSON]'s side, so [PERSON] cannot get any information about the committed qubit by only measuring the qubits in his side before the unveiling. From another perspective, the security against [PERSON] is guaranteed by the theorem of impossibility of superluminal signaling. Since if [PERSON] can learn something about the committed qubit before the unveiling, it means that an [PERSON] in a distant location can send information to [PERSON] simply by measuring the qubits on her side, which is essentially superluminal signaling. Hence, in our protocol, [PERSON] can learn nothing about [PERSON]'s choice until she chooses to unveil the qubit. The security of this protocol against [PERSON] should be discussed in different cases. The first case is that [PERSON] follows the honest strategy at point \(O\). In this case, she cannot change the collapsed states of qubits B again after \(O\), but she is free to pretend that one original input qubit (committed qubit) from \(\alpha|0\rangle_{\text{X}}\pm\beta|1\rangle_{\text{X}}\) and \(\beta|0\rangle_{\text{X}}\pm\alpha|1\rangle_{\text{X}}\) is any one of the three remaining ones of the four qubits just by properly interchanging her results of the collective measurement of qubits X and A. For example, she can pretend that the input qubit \(\alpha|0\rangle_{\text{X}}+\beta|1\rangle_{\text{X}}\) is \(\alpha|0\rangle_{\text{X}}-\beta|1\rangle_{\text{X}}\) by interchanging her measurement results \(|\phi_{1}\rangle_{\text{XA}}\) with \(|\phi_{2}\rangle_{\text{XA}}\), and \(|\phi_{3}\rangle_{\text{XA}}\) with \(|\phi_{4}\rangle_{\text{XA}}\). However, because of the prior agreement that any two of the four qubits \(\alpha|0\rangle_{\text{X}}\pm\beta|1\rangle_{\text{X}}\) and \(\beta|0\rangle_{\text{X}}\pm\alpha|1\rangle_{\text{X}}\) cannot be selected as the elements of the qubit set \(\{|\psi_{k}\rangle_{k=1}^{d}\}\) at the same time, [PERSON]'s cheating has no positive significance if qubit \(\alpha|0\rangle_{\text{X}}+\beta|1\rangle_{\text{X}}\) is one of the element of the qubit set. The reverse case is that [PERSON]'s input qubit \(\alpha|0\rangle_{\text{X}}+\beta|1\rangle_{\text{X}}\) is not the element of the qubit set while she unveils the right one, such as qubit \(\alpha|0\rangle_{\text{X}}-\beta|1\rangle_{\text{X}}\). From the perspective of the input, this is of course cheating. Although such cheating does not increase [PERSON]'s probability of successfully revealing any correct qubit, it cannot be prohibited. However, from the perspective of the output, no matter which of the four qubits [PERSON] inputs, the output is that qubits B on [PERSON]'s side collapse to \(\alpha|0\rangle_{\text{B}}\pm\beta|1\rangle_{\text{B}}\) and \(\beta|0\rangle_{\text{B}}\pm\alpha|1\rangle_{\text{B}}\) with equal probability, implying the four inputs are equivalent. In the second scenario, we ask if it is possible for [PERSON] to pretend that the input qubit \(\alpha|0\rangle_{\text{X}}+\beta|1\rangle_{\text{X}}\) is another qubit with parameters different from \(\alpha\) and \(\beta\), such as \(m|0\rangle_{\text{X}}+n|1\rangle_{\text{X}}\). If she tries to do this, she needs to pretend that the string of qubit \(|\varphi_{0}\rangle_{\text{B}}=\alpha|0\rangle_{\text{B}}+\beta|1\rangle_{ \text{B}}\) within the collapsed qubits B is the string of \(|\varphi_{1}\rangle_{\text{B}}=m|0\rangle_{\text{B}}+n|1\rangle_{\text{B}}\). The results of Ref.